Algebra Class 10th Mathematics Tamilnadu Board Solution

Class 10^th Mathematics Tamilnadu Board Solution

Exercise 3.1

1. x + 2y = 7, x - 2y = 1 Solve the following system of equation by elimination…
2. 3x + y = 8, 5x + y = 10 Solve the following system of equation by elimination…
3. x + y/2 = 4 , x/3 + 2y = 5 Solve the following system of equation by elimination…
4. 11x - 7y = xy, 9x - 4y = 6xy Solve the following system of equation by…
5. 3/x + 5/y = 20/xy , 2/x + 5/y = 15/xy , x not equal 0 , y not equal 0 Solve the…
6. 8x - 3y = 5xy, 6x - 5y = - 2xy Solve the following system of equation by…
7. 13x + 11y = 70, 11x + 13y = 74 Solve the following system of equation by…
8. 65x - 33y = 97, 33x - 65y = 1 Solve the following system of equation by…
9. 15/x + 2/y = 17 , 1/x + 1/y = 36/5 , x not equal 0 , y not equal 0 Solve the…
10. 2/x + 2/3y = 1/6 , 3/x + 2/y = 0 , x not equal 0 , y not equal 0 Solve the…

Exercise 3.10

1. Multiply the following and write your answer in lowest terms. (i) x^2 - 2x/x+2 x…
2. x/x+1 / x^2/x^2 - 1 Divide the following and write your answer in lowest terms.…
3. x^2 - 36/x^2 - 49 / x+6/x+7 Divide the following and write your answer in…
4. x^2 - 4x-5/x^2 - 25 / x^2 - 3x-10/x^2 + 7x+10 Divide the following and write…
5. x^2 + 11x+28/x^2 - 4x-77 / x^2 + 7x+12/x^2 - 2x-15 Divide the following and…
6. 2x^2 + 13x+15/x^2 + 3x-10 / 2x^2 - x-6/x^2 - 4x+4 Divide the following and…
7. 3x^2 - x-4/9x^2 - 16 / 4x^2 - 4/3x^2 - 2x-1 Divide the following and write your…
8. 2x^2 + 5x-3/2x^2 + 9x+9 / 2x^2 + x-1/2x^2 + x-3 Divide the following and write…

Exercise 3.11

1. Simplify the following as a quotient of two polynomials in the simplest form.…
2. Which rational expression should be added to x^3 - 1/x^2 + 2 to get 3x^3 + 2x^2…
3. Which rational expression should be subtracted from 4x^3 - 7x^2 + 5/2x-1 to get…
4. If p = x/x+y , q = y/x+y then find 1/p-q - 2q/p^2 - q^2

Exercise 3.12

1. 196a^6 b^8 c^10 Find the square root of the following
2. 289 (a-b)^4 (b-c)^6 Find the square root of the following
3. (x + 11)^2 -44x Find the square root of the following
4. (x-y)^2 + 4xy Find the square root of the following
5. 121x^8 y^6 */* 81x^4 y^8 Find the square root of the following
6. 64 (a+b)^4 (x-y)^8 (b-c)^6/25 (x+y)^4 (a-b)^6 (b+c)^10 Find the square root of…
7. 16x^2 -24x + 9 Find the square root of the following:
8. (x^2 - 25)(x^2 + 8x + 15)(x^2 -2x-15) Find the square root of the following:…
9. 4x^2 + 9y^2 + 25z^2 -12xy + 30yz-20zx Find the square root of the following:…
10. x^4 + 1/x^4 + 2 Find the square root of the following:
11. (6x^2 + 5x -6) (6x^2 -x-2)(4x^2 + 8x + 3) Find the square root of the…
12. (2x^2 -5x + 2) (3x^2 -5x-2) (6x^2 - x -1) Find the square root of the…

Exercise 3.13

1. x^4 -4x^3 + 10x^2 -12x + 9 Find the square root of the following polynomials by…
2. 4x^4 + 8x^3 + 8x^2 + 4x + 1 Find the square root of the following polynomials…
3. 9 x^4 -6 x^3 + 7 x^2 -2x + 1 Find the square root of the following polynomials…
4. 4 + 25x^2 12x-24x^3 + 16x^4 Find the square root of the following polynomials…
5. 4x^4 -12 x^3 + 37x^2 + ax + b Find the values of a and b if the following…
6. x^4 -4x^3 + 6x^2 -ax + b Find the values of a and b if the following…
7. ax^4 + bx^3 + 109x^2 -60x + 36 Find the values of a and b if the following…
8. a x^4 -bx^3 + 40 x^2 + 24x + 36 Find the values of a and b if the following…

Exercise 3.14

1. (2x + 3)^2 - 81 = 0 Solve the following quadratic equations by factorization…
2. 3x^2 -5x -12 = 0 Solve the following quadratic equations by factorization…
3. root 5x^2 + 2x-3 root 5 = 0 Solve the following quadratic equations by…
4. 3(x^2 - 6) =x (x + 7)-3 Solve the following quadratic equations by…
5. 3x - 8/x = 2 Solve the following quadratic equations by factorization method.…
6. x + 1/x = 26/5 Solve the following quadratic equations by factorization method.…
7. x/x+1 + x+1/x = 34/15 Solve the following quadratic equations by factorization…
8. a^2 b^2 x^2 - (a^2 + b^2) x + 1 =0 Solve the following quadratic equations by…
9. 2(x + 1)^2 -5 (x + 1) =12 Solve the following quadratic equations by…
10. 3(x-4)^2 - 5 (x - 4) = 12 Solve the following quadratic equations by…

Exercise 3.15

1. x^2 + 6x -7 = 0 Solve the following quadratic equations by completing the…
2. x^2 + 3x + 1 =0 Solve the following quadratic equations by completing the…
3. 2x^2 + 5x -3 = 0 Solve the following quadratic equations by completing the…
4. 4x^2 + 4bx - (a^2 - b^2) = 0 Solve the following quadratic equations by…
5. x^2 - (√3 + 1) x + √3 = 0 Solve the following quadratic equations by completing…
6. 5x+7/x-1 =3x + 2 Solve the following quadratic equations by completing the…
7. x^2 7x + 12= 0 Solve the following quadratic equations using quadratic formula.…
8. 15x^2 - 11x + 2 = 0 Solve the following quadratic equations using quadratic…
9. x + 1/x = 2 1/2 Solve the following quadratic equations using quadratic…
10. 3a^2 x^2 - ax -2b^2 = 0 Solve the following quadratic equations using quadratic…
11. a (x^2 + 1) = x (a^2 + 1) Solve the following quadratic equations using…
12. 36x^2 - 12ax + (a^2 - b^2) = 0 Solve the following quadratic equations using…
13. x-1/x+1 + x-3/x-4 = 10/3 Solve the following quadratic equations using…
14. a^2 x^2 + (a^2 - b^2) x - b^2 = 0 Solve the following quadratic equations using…

Exercise 3.16

1. The sum of a number and its reciprocal is 65/8 . Find the number.…
2. The difference of the squares of two positive numbers is 45. The square of the…
3. A farmer wishes to start a 100 sq. rectangular vegetable garden. Since he has…
4. A rectangular field is 20 m long and 14 m wide. There is a path of equal width…
5. A train covers a distance of 90 km at a uniform speed. Had the speed been 15…
6. The speed of a boat in still water is 15 km/hr. It goes 30 km upstream and…
7. One year ago, a man was 8 times as old as his son. Now his age is equal to the…
8. A chess board contains 64 equal squares and the area of each square is 6.25 cm2.…
9. A takes 6 day less than the time taken by B to finish a piece of work. If both A…
10. Two trains leave a railway station at the same time. The first train travels…

Exercise 3.17

1. x^2 - 8x + 12 = 0 Determine the nature of the roots of the equation.…
2. 2x^2 - 3x + 4 = 0 Determine the nature of the roots of the equation.…
3. 9x^2 + 12x + 4 = 0 Determine the nature of the roots of the equation.…
4. 3x^2 -2√6x + 2 = 0 Determine the nature of the roots of the equation.…
5. 3/5 x^2 - 12/3 x+1 = 0 Determine the nature of the roots of the equation.…
6. (x - 2a) (x - 2b) = 4ab Determine the nature of the roots of the equation.…
7. 2x^2 - 10x + k = 0 Find the values of k for which the roots are real and equal…
8. 12x^2 + 4kx + 3 = 0 Find the values of k for which the roots are real and equal…
9. x^2 + 2k (x - 2) + 5 = 0 Find the values of k for which the roots are real and…
10. (k + 1) x^2 - 2 (k - 1) x + 1 = 0 Find the values of k for which the roots are…
11. Show that the roots of the equation x^2 + 2(a + b) x + 2 (a^2 + b^2) = 0 are…
12. Show that the roots of the equation 3p^2 x^2 - 2pqx + q^2 = 0 are not real.…
13. If the roots of the equation (a^2 + b^2) x^2 - 2 (ac + bd) x + c^2 + d^2 = 0,…
14. Show that the roots of the equation (x - a) (x - b) + (x - b) (x - c) + (x - c)…
15. If the equation (1 + m^2) x^2 + 2mcx + c^2 - a^2 = 0 has equal roots, then prove…

Exercise 3.18

1. x^2 - 6x + 5 = 0 Find the sum and the product of the roots of the following…
2. kx^2 + ax + pk = 0 Find the sum and the product of the roots of the following…
3. 3x^2 - 5x = 0 Find the sum and the product of the roots of the following…
4. 8x^2 - 25 = 0 Find the sum and the product of the roots of the following…
5. Form a quadratic equation whose roots are (i) 3, 4 (ii) 3 + √7, 3 - √7 (iii) 4 +…
6. If α and β are the roots of the equation 3x^2 - 5x + 2= 0, then find the values…
7. If α and β are the roots of the equation 3x^2 - 6x + 4 = 0, find the value of…
8. If α, β are roots of 2x^2 - 3x - 5 = 0, from an equation whose roots are α^2 and…
9. If α, β are roots of x^2 - 3x + 2 = 0, form a quadratic equation whose roots are…
10. If α and β are roots of x^2 - 3x-1 = 0, then form a quadratic equation whose…
11. If α and β are roots of 3x^2 - 6x + 1 = 0, then form a quadratic equation whose…
12. Find a quadratic equation whose roots are the reciprocal of the roots of the…
13. If one root of the equation 3x^2 + kx - 81 = 0 is the square of the other, find…
14. If one root of the equation 2x^2 - ax + 64 = 0 is twice the other, then find…
15. If α and β are roots of 5x^2 - px + 1 = 0 and α - β = 1, then find P.…

Exercise 3.19

1. If the system 6x - 2y = 3, kx - y = 2 has a unique solution, thenA. k = 3 B. k ≠…
2. A system of two linear equations in two variables is consistent, if their…
3. The system of equations x -4y = 8 , 3x -12y = 24A. has infinitely many solutions…
4. If one zero of the polynomial p(x) = (k + 4)x^2 + 13x + 3k is reciprocal of the…
5. The sum of two zeros of the polynomial f(x) = 2x^2 + (p + 3)x + 5 is zero, then…
6. The remainder when x^2 - 2x + 7 is divided by x + 4 isA. 28 B. 29 C. 30 D. 31…
7. The quotient when x^3 - 5x^2 + 7x - 4 is divided by x-1 isA. x^2 + 4x + 3 B. x^2…
8. The GCD of (x^3 + 1) and x^4 - 1 isA. x^3 - 1 B. x^3 + 1 C. -(x + 1) D. x-1…
9. The GCD of x^2 - 2xy + y^2 and x^4 - y^4 isA. 1 B. x + y C. x - y D. x^2 - y^2…
10. The LCM of x^3 - a^3 and (x - a)^2 isA. (x^3 - a^3) (x + a) B. (x^3 - a^3) (x -…
11. The LCM of ak,ak + 3, ak + 5 where k inn isA. a k + 9 B. ak C. ak + 6 D. ak + 5…
12. The lowest form of the rational expression x^2 + 5x+6/x^2 - x-6 isA. x-3/x+3 B.…
13. If a+b/a-b and a^3 - b^3/a^3 + b^3 are the two rational expressions, then their…
14. On dividing x^2 - 25/x+3 by x+5/x^2 - 9 is equal toA. (x - 5) (x - 3) B. (x -…
15. If a^3/a-b is added with b^3/b-a , then the new expression isA. a^2 + ab + b^2…
16. The square root of 49 (x^2 - 2x + y^2)^2 isA. 7 |x - y| B. 7(x + y) (x - y) C.…
17. The square root of x^2 + y^2 + z^2 - 2xy + 2yz - 2zxA. |x + y - z| B. |x - y +…
18. The square root of 121 x^4 y^8 z^6 (l - m)^2 isA. 11x^2 y^4 z^4 |l - m| B.…
19. If ax^2 + bx + c = 0 has equal roots, then c is equalA. b^2/2a B. b^2/4a C. -…
20. If x^2 + 5kx + 16 = 0 has no real roots, thenA. k 8/5 B. k - 8/5 C. - 8/5 k 8/5…
21. A quadratic equation whose one root is 3 isA. x^2 - 6x - 5 = 0 B. x^2 + 6x - 5…
22. The common root of the equation x^2 - bx + c = 0 and x^2 + bx - a = 0 isA.…
23. If α, β are the roots of ax^2 + bx + c = 0 a ≠ 0, then the wrong statement isA.…
24. If α and β are the roots of ax^2 + bx + c = 0, then one of the quadratic…
25. Let b = a + c. Then the equation ax^2 + bx + c = 0 has equal roots, ifA. a = c…

Exercise 3.2

1. 3x + 4y = 24, 20x - 11y = 47 Solve the following systems of equation using…
2. 0.5x + 0.8y = 0.44, 0.8x + 0.6y = 0.5 Solve the following systems of equation…
3. 3x/2 - 5y/3 = - 2 , x/3 + y/2 = 13/6 Solve the following systems of equation…
4. 5/x - 4/y = - 2 , 2/x + 3/y = 13 Solve the following systems of equation using…
5. One number is greater than thrice the other number by 2. If 4 times the smaller…
6. The ratio of income of two persons is 9: 7 and the ratio of their expenditure…
7. A two digit number is seven times the sum of its digits. The number formed by…
8. Three chairs and two tables cost ₹ 700 and five chairs and three tables cost…
9. In a rectangle, if the length is increased and the breadth is reduced each by 2…
10. A train travelled a certain distance at a uniform speed. If the train had been…

Exercise 3.3

1. x^2 - 2x - 8 Find the zeros of the following quadratic polynomials and verify…
2. 4x^2 - 4x + 1 Find the zeros of the following quadratic polynomials and verify…
3. 6x^2 - 3 - 7x Find the zeros of the following quadratic polynomials and verify…
4. 4x^2 + 8x Find the zeros of the following quadratic polynomials and verify the…
5. x^2 - 15 Find the zeros of the following quadratic polynomials and verify the…
6. 3x^2 - 5x + 2 Find the zeros of the following quadratic polynomials and verify…
7. 2x^2 - 2√2 x + 1 Find the zeros of the following quadratic polynomials and…
8. x^2 + 2x - 143 Find the zeros of the following quadratic polynomials and verify…
9. 3, 1 Find a quadratic polynomial each with the given numbers as the sum and…
10. 2, 4 Find a quadratic polynomial each with the given numbers as the sum and…
11. 0, 4 Find a quadratic polynomial each with the given numbers as the sum and…
12. root 2 , 1/5 Find a quadratic polynomial each with the given numbers as the sum…
13. 1/3 , 1 Find a quadratic polynomial each with the given numbers as the sum and…
14. 1/2 ,-4 Find a quadratic polynomial each with the given numbers as the sum and…
15. 1/2 , - 1/3 Find a quadratic polynomial each with the given numbers as the sum…
16. root 3 , 2 Find a quadratic polynomial each with the given numbers as the sum…

Exercise 3.4

1. x^3 + x^2 - 3x + 5) ÷ (x - 1) Find the quotient and remainder using synthetic…
2. (3x^3 - 2x^2 + 7x - 5) ÷ (x + 3) Find the quotient and remainder using…
3. (3x^3 + 4x^2 - 10x + 6) ÷ (3x - 2) Find the quotient and remainder using…
4. (3x^3 - 4x^2 - 5) ÷ (3x + 1) Find the quotient and remainder using synthetic…
5. (8x^4 - 2x^2 + 6x + 5) ÷ (4x + 1) Find the quotient and remainder using…
6. (2x^4 - 7x^3 - 13x^2 + 63x - 48) ÷ (2x - 1) Find the quotient and remainder…
7. If the quotient on dividing x^4 + 10x^3 + 35x^2 + 50x + 29 by x + 4 is x^3 -…
8. If the quotient on dividing, 8x^4 - 2x^2 + 6x - 7 by 2x + 1 is 4x^3 + px^2 - qx…

Exercise 3.5

1. x3 - 2x2 - 5x + 6 Factorize each of the following polynomials.
2. 4x3 - 7x + 3 Factorize each of the following polynomials.
3. x3 - 23x2 + 142x - 120 Factorize each of the following polynomials.…
4. 4x3 - 5x2 + 7x - 6 Factorize each of the following polynomials.
5. x3 - 7x + 6 Factorize each of the following polynomials.
6. x^3 + 13x2 + 32x + 20 Factorize each of the following polynomials.…
7. 2x3 - 9x^2 + 7x + 6 Factorize each of the following polynomials.
8. x3 - 5x + 4 Factorize each of the following polynomials.
9. x3 - 10x2 - x + 10 Factorize each of the following polynomials.
10. 2x3 + 11x2 - 7x - 6 Factorize each of the following polynomials.
11. x3 + x^2 + x - 14 Factorize each of the following polynomials.
12. x3 - 5x2 - 2x + 24 Factorize each of the following polynomials.

Exercise 3.6

1. 7x2 yz4, 21x2 y^5 z^3 Find the greatest common divisor of
2. x2y, x3y, x2y^2 Find the greatest common divisor of
3. 25bc^4 d^3 , 35b2c^5 , 45c^3 d Find the greatest common divisor of…
4. 35x^5 y^3 z^4 , 49x^2 yz^3 , 14xy^2 z^2 Find the greatest common divisor of…
5. x^3 - x^2 + x - 1, x^4 - 1 Find the GCD of the following
6. c2 - d^2 , - c(c - d) Find the GCD of the following
7. x4 - 27a3 x,(x - 3a)^2 Find the GCD of the following
8. m2 - 3m - 18, m^2 + 5m + 6 Find the GCD of the following
9. x2 + 14x + 33, x3 + 10x2 - 11x Find the GCD of the following
10. x2 + 3xy + 2y^2 , x^2 + 5xy + 6y2 Find the GCD of the following
11. 2x2 - x - 1,4x2 + 8x + 3 Find the GCD of the following
12. x2 - x - 2,x2 + x - 6,3x2 - 13x + 14 Find the GCD of the following…
13. 24(6x4 - x^3 - 2x2),20(2x6 + 3x5 + x^4) Find the GCD of the following…
14. (a - 1)^5 (a + 3)^2 ,(a - 2)^2 (a - 1)^3 (a + 3)^4 Find the GCD of the…
15. x^3 - 9x^2 + 23x - 15, 4x^2 - 16x + 12 Find the GCD of the following pairs of…
16. 3x^3 + 18x^2 + 33x + 18, 3x^2 + 13x + 10 Find the GCD of the following pairs of…
17. 2x^3 + 2x^2 + 2x + 2, 6x^3 + 12x^2 + 6x + 12 Find the GCD of the following…
18. x^3 - 3x^2 + 4x - 12, x^4 + x^3 + 4x^2 + 4x Find the GCD of the following pairs…

Exercise 3.7

1. x^3 y^2 , xyz Find the LCM of the following
2. 3x^2 yz, 4x^3 y^3 Find the LCM of the following
3. a^2 bc, b^2 ca, c^2 ab Find the LCM of the following
4. 66a^4 b^2 c^3 , 44a^3 b^4 c^2 , 24a^2 b^3 c^4 Find the LCM of the following…
5. am + 1, am + 2, am + 3 Find the LCM of the following
6. x^2 y + xy^2 , x^2 + xy Find the LCM of the following
7. 3(a - 1), 2(a - 1)^2 , (a^2 - 1) Find the LCM of the following
8. 2x^2 - 18, 5x^2 y + 15xy^2 , x^3 + 27y^3 Find the LCM of the following…
9. (x + 4)^2 (x - 3)^3 , (x - 1)(x + 4)(x - 3)^2 Find the LCM of the following…
10. 10(9x^2 + 6xy + y^2), 12(3x^2 - 5xy - 2y^2), 14(6x^4 + 2x^3) Find the LCM of…

Exercise 3.8

1. x^2 - 5x + 6, x^2 + 4x - 12 whose GCD is x - 2. Find the LCM of each pair of…
2. x^4 + 3 x^3 + 6 x^2 + 5x + 3, x^4 + 2 x^2 + x + 2 whose GCD is x^2 + x + 1 Find…
3. 2x^3 + 15x^2 + 2x - 35, x^3 + 8x^2 + 4x - 21 whose GCD is x + 7. Find the LCM…
4. 2x^3 - 3x^2 - 9x + 5, 2x^4 - x^3 - 10x^2 - 11x + 8 whose GCD is 2x - 1 Find the…
5. (x + 1)^2 (x + 2)^2 , (x + 1) (x + 2), (x + 1)^2 (x + 2) Find the other…
6. (4x + 5)^3 (3x - 7)^3 , (4x + 5) (3x - 7)^2 , (4x + 5)^3 (3x - 7)^2 Find the…
7. (x^4 - y^4) (x^4 + x^2 y^2 + y^4), x^2 - y^2 , x^4 - y^4 . Find the other…
8. (x^3 - 4x) (5x + 1), (5 x^2 + x), (5 x^3 - 9 x^2 - 2x). Find the other…
9. (x - 1) (x - 2) (x^2 - 3x + 3), (x - 1), (x^3 - 4 x^2 + 6x - 3). Find the other…
10. 2(x + 1) (x^2 - 4), (x + 1), (x + 1) (x - 2). Find the other polynomial q(x) of…

Exercise 3.9

1. 6x^2 + 9x/3x^2 - 12x Simplify the following into their lowest forms.…
2. x^2 + 1/x^4 - 1 Simplify the following into their lowest forms.
3. x^3 - 1/x^2 + x+1 Simplify the following into their lowest forms.…
4. x^3 - 27/x^2 - 9 Simplify the following into their lowest forms.
5. x^4 + x^2 + 1/x^2 + x+1 (Hint : x^4 + x^2 + 1 =(x^2 + 1)^2 - x^2) Simplify the…
6. x^3 + 8/x^4 + 4x^2 + 16 Simplify the following into their lowest forms.…
7. 2x^2 + x-3/2x^2 + 5x+3 Simplify the following into their lowest forms.…
8. 2x^4 - 162/(x^2 + 9) (2x-6) Simplify the following into their lowest forms.…
9. (x-3) (x^2 - 5x+4)/(x-4) (x^2 - 2x-3) Simplify the following into their lowest…
10. (x-8) (x^2 - 5x-50)/(x+10) (x^2 - 13x+40) Simplify the following into their…
11. 4x^2 + 9x+5/8x^2 + 6x-5 Simplify the following into their lowest forms.…
12. (x-1) (x-2) (x^2 - 9x+14)/(x-7) (x^2 - 3x+2) Simplify the following into their…

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Exercise 3.1

Question 1.

Solve the following system of equation by elimination method.

x + 2y = 7, x – 2y = 1

Answer:

The given equations are

x + 2y = 7 … (1)

x – 2y = 1 … (2)

Adding (1) and (2),

⇒ x + 2y + x – 2y = 7 + 1

⇒ 2x = 8

⇒ x = 4

Substituting x = 4 in (1),

⇒ 4 + 2y = 7

⇒ 2y = 7 – 4 = 3

⇒ y = 

∴ (4,) is the solution to the given system.

Question 2.

Solve the following system of equation by elimination method.

3x + y = 8, 5x + y = 10

Answer:

The given equations are

3x + y = 8 … (1)

5x + y = 10 … (2)

Here, the coefficients of y in both equations are numerically equal.

Subtracting (2) from (1),

⇒ (3x + y) – (5x + y) = 8 – 10

⇒ 3x + y – 5x – y = – 2

⇒ – 2x = – 2

⇒ x = 1

Substituting x = 1 in (1),

⇒ 3 (1) + y = 8

⇒ y = 8 – 3

⇒ y = 5

∴ (1, 5) is the solution to the given system.

Question 3.

Solve the following system of equation by elimination method.

Answer:

The given equations are

x +  = 4 … (1)

 + 2y = 5 … (2)

(1) becomes 2x + y = 8

(2) becomes x + 6y = 15

Now, (2) × 2 – (1)

⇒ 2x + 12y – (2x + y) = 30 – 8

⇒ 2x + 12y – 2x – y = 22

⇒ 11y = 22

⇒ y = 2

Substituting y = 2 in (1),

⇒ 2x + 2 = 8

⇒ 2x = 6

⇒ x = 3

∴ (3, 2) is the solution to the given system.

Question 4.

Solve the following system of equation by elimination method.

11x – 7y = xy, 9x – 4y = 6xy

Answer:

The given equations are

11x – 7y = xy … (1)

9x – 4y = 6xy … (2)

Dividing both sides of the equation by xy,

⇒  –  = 1 i.e.  +  = 1 … (3)

⇒  –  = 6 i.e.  +  = 6 … (4)

Let a =  and b =.

Equations (3) and (4) become

⇒ – 7a + 11b = 1 … (5)

⇒ – 4a + 9y = 6 … (6)

Now, (6) × 7 – (5) × 4

⇒ – 28a + 63b – ( – 28a + 44b) = 42 – 4

⇒ – 28a + 63b + 28a – 44b = 38

⇒ 19b = 38

⇒ b = 2

Substituting b = 2 in (5),

⇒ – 7a + 11(2) = 1

⇒ – 7a = 1 – 22 = – 21

⇒ a = 3

When a = 3, we have  = 3. Thus, x = 

When b = 2, we have  = 2. Thus, y = 

∴ (,) is the solution for the given system.

Question 5.

Solve the following system of equation by elimination method.

Answer:

The given equations are

 +  =  … (1)

 +  =  … (2)

Multiplying both sides of the equation with xy,

3y + 5x = 20 i.e. 5x + 3y = 20 … (3)

2y + 5x = 15 i.e. 5x + 2y = 15 … (4)

Subtracting (4) from (3),

⇒ 5x + 3y – (5x + 2y) = 20 – 15

⇒ 5x + 3y – 5x – 2y = 5

⇒ y = 5

Substituting y = 5 in (3),

⇒ 5x + 3 (5) = 20

⇒ 5x = 20 – 15 = 5

⇒ x = 1

∴ (1, 5) is the solution for the given system.

Question 6.

Solve the following system of equation by elimination method.

8x – 3y = 5xy, 6x – 5y = – 2xy

Answer:

The given equations are

8x – 3y = 5xy … (1)

6x – 5y = – 2xy … (2)

Dividing both sides of the equation by xy,

⇒  –  = 5 i.e.  +  = 5 … (3)

⇒  –  = – 2 i.e.  +  = – 2 … (4)

Let a =  and b =.

Equations (3) and (4) become

⇒ – 3a + 8b = 5 … (5)

⇒ – 5a + 6y = – 2 … (6)

Now, (5) × 5 – (6) × 3

⇒ – 15a + 40b – ( – 15a + 18b) = 25 – ( – 6)

⇒ – 15a + 40b + 15a – 18b = 31

⇒ 22b = 31

⇒ b = 

Substituting b =  in (5),

⇒ – 3a + 8() = 5

⇒ – 3a = 5 –  = 

⇒ a = 

When a =, we have  =. Thus, x = 

When b =, we have  =. Thus, y = 

∴ (,) is the solution for the given system.

Question 7.

Solve the following system of equation by elimination method.

13x + 11y = 70, 11x + 13y = 74

Answer:

The given equations are

13x + 11y = 70 … (1)

11x + 13y = 74 … (2)

Adding (1) and (2),

⇒ 24x + 24y = 144

Dividing by 24,

⇒ x + y = 6 … (3)

Subtracting (2) from (1),

⇒ 2x + ( – 2y) = – 4

Dividing by 2,

⇒ x – y = – 4 … (4)

Solving (3) and (4),

⇒ x + y + (x – y) = 6 – 4

⇒ 2x = 2

⇒ x = 1

Substituting x = 1 in (3),

⇒ 1 + y = 6

⇒ y = 5

∴ (1, 5) is the solution to the given system.

Question 8.

Solve the following system of equation by elimination method.

65x – 33y = 97, 33x – 65y = 1

Answer:

The given equations are

65x – 33y = 97 … (1)

33x – 65y = 1 … (2)

Adding (1) and (2),

⇒ 98x – 98y = 98

Dividing by 98,

⇒ x – y = 1 … (3)

Subtracting (1) and (2),

⇒ 32x + 32y = 96

Dividing by 32,

⇒ x + y = 3 … (4)

Solving (3) and (4),

⇒ x – y + (x + y) = 1 + 3

⇒ 2x = 4

⇒ x = 2

Substituting x = 2 in (4),

⇒ 2 + y = 3

⇒ y = 1

∴ (2, 1) is the solution to the given system.

Question 9.

Solve the following system of equation by elimination method.

Answer:

The given equations are

 +  = 17 … (1)

 +  =  … (2)

Let a =  and b =.

⇒ 15a + 2b = 17 … (3)

⇒ a + b =  … (4)

Now, (3) – (4) × 2

⇒ 15a + 2b – (2a + 2b) = 17 – 

⇒ 15a + 2b – 2a – 2b = 

⇒ 13a = 

⇒ a = 

Substituting a =  in (4),

⇒  + b = 

⇒ b =  = 7

When a =,  =. Thus, x = 5.

When b = 7,  = 7. Thus, y = 

∴ (5,) is the solution to the given system.

Question 10.

Solve the following system of equation by elimination method.

Answer:

The given equations are

 +  =  … (1)

 +  = 0 … (2)

Let a =  and b =.

⇒ 2a + b =  … (3)

⇒ 3a + 2b = 0 … (4)

Now, (3) × 3 – (4) × 2

⇒ 6a + 2b – (6a + 4b) = 1/2 – 0

⇒ 6a + 2b – 6a – 4b = 1/2

⇒ – 2b = 1/2

⇒ b = 

Substituting b =  in (4),

⇒ 3a + 2() = 0

⇒ 3a = 1/2

⇒ a = 

When a =,  =. Thus, x = 6.

When b = ,  =. Thus, y = – 4

∴ (6, – 4) is the solution to the given system.

---

Exercise 3.10

Question 1.

Multiply the following and write your answer in lowest terms.
(i)  (ii) 

(iii)  (iv) 

(v)  (vi) 

Answer:

(i)

The like terms are cancelled.

=3x required solution

(ii)

We know a² – b² = (a-b) (a+b)

So,

Also,

x² + 6x + 8 = x² + 4x + 2x +8

= x(x+4) +2(x+4)

= (x+2)(x+4)

And

x² - 5x – 36 = x² - 9x + 4x – 36

= x(x-9)+4(x-9)

= (x+4)(x-9)

So,

The like terms are cancelled.

Required solution

(iii)

x² - 3x – 10 = x² - 5x + 2x – 10

= x(x-5)+2(x-5)

= (x+2)(x-5)

And,

x² - x - 20 = x² - 5x + 4x - 20

= x(x-5)+4(x-5)

= (x+4)(x-5)

We know the formula a³ + b³ = (a+b)(a² + b² - ab)

So x³ + 8 = x³ + 2³

= (x+2)(x² + 4 - 2x)

The like terms are cancelled.

Required solution

(iv)

We know a² – b² = (a-b) (a+b)

x² – 16 = (x-4) (x+4)

x² – 4 = (x-2) (x+2)

x² - 3x + 2 = x² – 2x – x + 2

= x(x-2)-1(x-2)

= (x-1)(x-2)

x² - 2x - 8 = x² – 4x + 2x - 8

= x(x-4)+2(x-4)

= (x-4)(x+2)

We know the formula a³ + b³ = (a+b)(a² + b² - ab)

So x³ + 64 = x³ + 4³

= (x+4)(x² + 16 - 4x)

The like terms are cancelled.

Required solution

(v) 

The like terms are cancelled.

Required solution

(vi)

We know a³ - b³ = (a-b) (a² + ab + b²)

The like terms are cancelled.

Required solution.

Question 2.

Divide the following and write your answer in lowest terms.

Answer:

The like terms are cancelled.

Required solution

Question 3.

Divide the following and write your answer in lowest terms.

Answer:

The like terms are cancelled.

Required solution.

Question 4.

Divide the following and write your answer in lowest terms.

Answer:

The like terms are cancelled.

Required solution.

Question 5.

Divide the following and write your answer in lowest terms.

Answer:

The like terms are cancelled.

Required solution

Question 6.

Divide the following and write your answer in lowest terms.

Answer:

The like terms are cancelled.

=1 required solution.

Question 7.

Divide the following and write your answer in lowest terms.

Answer:

The like terms are cancelled.

Required solution,

Question 8.

Divide the following and write your answer in lowest terms.

Answer:

The like terms are cancelled.

Required solution.

---

Exercise 3.11

Question 1.

Simplify the following as a quotient of two polynomials in the simplest form.

(i)  (ii) 

(iii)  (iv) 

_(v) (vi) 

_(vii) (viii) 

Answer:

(i) 

The like terms are cancelled.

=x² + 2x + 4 required solution.

(ii) 

The like terms are cancelled.

(iii) 

The like terms are cancelled.

Required solution

(iv) 

The like terms are cancelled.

Required solution.

(v)

The like terms are cancelled.

Required Solution 

(vi) 

Like terms are cancelled

Required solution

(vii) 

Like terms are cancelled

(viii) 

= 0

Question 2.

Which rational expression should be added to  to get 

Answer:

Let p(x)= , q(x)=

And the rational expression be r(x),

So according to question,

p(x) = q(x) + r(x)

or, r(x) = p(x) – q(x)

So required rational expression r(x) 

Question 3.

Which rational expression should be subtracted from

 to get 2x² – 5x + 1 ?

Answer:

Let p(x) = , q(x) =

And rational expression be r(x)

According to question,

p(x) – r(x)=q(x)

or, r(x) = p(x) – q(x)

r(x) =

So required rational expression 

Question 4.

If  then find 

Answer:

P – Q 

P² –Q² =(P + Q)(P – Q) 

=P – Q

So  (∵ P² –Q²=P – Q)

=1

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Exercise 3.12

Question 1.

Find the square root of the following

196a⁶b⁸c¹⁰

Answer:

In Square root the power of each term is divided by 2

√(196a⁶b⁸c¹⁰) = √(14² a⁶b⁸c¹⁰)

Square Root = |14a³b⁴c⁵|

Question 2.

Find the square root of the following

289 (a–b)⁴ (b–c)⁶

Answer:

289 = 17²

⇒Square Root = √[17²(a–b)⁴ (b–c)⁶]

Square Root = |17(a–b)²(b–c)³|

Question 3.

Find the square root of the following

(x + 11)²–44x

Answer:

(x + 11)²–44x

⇒ x² + 22x + 121–44x

⇒ x²–22x + 121 = (x–11)²

√[(x + 11)²–44x

Square root = |x–11|

Question 4.

Find the square root of the following

(x–y)² + 4xy

Answer:

(x–y)² + 4xy

⇒ x² + y²–2xy + 4xy

⇒ x² + y² + 2xy = (x + y)²

√[(x–y)² + 4xy]

Square Root = |x + y|

Question 5.

Find the square root of the following

121x⁸y⁶81x⁴y⁸

Answer:

Square Root 

⇒Square Root 

Square Root

Question 6.

Find the square root of the following

Answer:

Square Root 

Question 7.

Find the square root of the following:

16x² –24x + 9

Answer:

16x² –24x + 9

The above expression can be rewritten as

(4x)²–2×3×4x + 3²

It is in the form of (a–b)²

= (4x–3)²

Square root = √(4x–3)²

|4x–3|

Question 8.

Find the square root of the following:

(x² – 25)( x² + 8x + 15)( x² –2x–15)

Answer:

We factorize each of the above polynomials

x² – 25 = x² – 5²

Since it is in the form of a²–b² = (a–b)(a + b)

⇒ x² – 25 = (x–5)(x + 5) …(i)

x² + 8x + 15 = x² + 5x + 3x + 15

⇒ x² + 8x + 15 = x(x + 5) + 3(x + 5)

⇒ x² + 8x + 15 = (x + 3)(x + 5) …(ii)

x² –2x–15 = x² –5x + 3x–15

⇒ x² –2x–15 = x(x–5) + 3(x–5)

⇒ x² –2x–15 = (x + 3)(x–5) … (iii)

Combining (i), (ii) & (iii) we get

(x² – 25)( x² + 8x + 15)( x² –2x–15) = = (x–5)²(x + 5)²(x + 3)²

Square Root = √[(x–5)²(x + 5)²(x + 3)²]

|(x–5)(x + 5)(x + 3)|

Question 9.

Find the square root of the following:

4x² + 9y² + 25z²–12xy + 30yz–20zx

Answer:

The above expression can be rewritten as

(2x)² + (–3y)² + (–5z)² + 2((–3y)×(2x) + (–5z)×(–3y) + (2x)×(–5z))

The above expression is in the form of

(a–b–c)² = a² + b² + c² + 2(–ab + bc – ca)

So the expression becomes (2x–3y–5z)²

Square Root = √(2x–3y–5z)²

|2x–3y–5z|

Question 10.

Find the square root of the following:

Answer:

The equation can be written as

The above equation is in the form of (a + b)² = a² + b² + 2ab

So it becomes

Square root

Question 11.

Find the square root of the following:

(6x² + 5x –6) (6x²–x–2)(4x² + 8x + 3)

Answer:

We factorize each of the above polynomials

6x² + 5x –6 = 6x² + 9x –4x –6

⇒ 6x² + 5x –6 = 3x(2x + 3)–2(2x + 3)

⇒ 6x² + 5x –6 = (3x–2)(2x + 3) …(i)

6x²–x–2 = 6x²–4x + 3x–2

⇒ 6x²–x–2 = 2x(3x–2) + 1(3x–2)

⇒ 6x²–x–2 = (2x + 1)(3x–2) …(ii)

4x² + 8x + 3 = 4x² + 6x + 2x + 3

⇒ 4x² + 8x + 3 = 2x(2x + 3) + 1(2x + 3)

⇒ 4x² + 8x + 3 = (2x + 1)(2x + 3) …(iii)

Combining (i), (ii) & (iii) we get

(6x² + 5x –6) (6x²–x–2)(4x² + 8x + 3) = (3x–2)²(2x + 3)²(2x + 1)²

Square Root = √ (3x–2)²(2x + 3)²(2x + 1)²

| (3x–2)(2x + 3)(2x + 1)|

Question 12.

Find the square root of the following:

(2x² –5x + 2) (3x²–5x–2) (6x² – x –1)

Answer:

We factorize each of the above polynomials

2x² –5x + 2 = 2x²–4x–x + 2

⇒ 2x² –5x + 2 = 2x(x–2)–1(x–2)

⇒ 2x² –5x + 2 = (2x–1)(x–2) …(i)

3x²–5x–2 = 3x²–6x + x–2

⇒ 3x²–5x–2 = 3x(x–2) + 1(x–2)

⇒ 3x²–5x–2 = (3x + 1)(x–2) …(ii)

6x² – x –1 = 6x²– 3x + 2x–1

⇒ 6x² – x –1 = 3x(2x–1) + 1(2x–1)

⇒ 6x² – x –1 = (3x + 1)(2x–1) …(iii)

Combining (i), (ii) & (iii) we get

(2x² –5x + 2) (3x²–5x–2) (6x² – x –1) = (2x–1)²(x–2)²(3x + 1)²

Square Root = √(2x–1)²(x–2)²(3x + 1)²

|(2x–1)(x–2)(3x + 1)|

---

Exercise 3.13

Question 1.

Find the square root of the following polynomials by division method.

x⁴–4x³ + 10x²–12x + 9

Answer:

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

|x²–2x + 3|

Question 2.

Find the square root of the following polynomials by division method.

4x⁴ + 8x³ + 8x² + 4x + 1

Answer:

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

|2x² + 2x + 1|

Question 3.

Find the square root of the following polynomials by division method.

9 x⁴–6 x³ + 7 x²–2x + 1

Answer:

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient. To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

|3x²–x + 1|

Question 4.

Find the square root of the following polynomials by division method.

4 + 25x²12x–24x³ + 16x⁴

Answer:

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

|4x²–3x + 2|

Question 5.

Find the values of a and b if the following polynomials are perfect squares.

4x⁴–12 x³ + 37x² + ax + b

Answer:

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

Since it is a perfect square the remainder is 0

a = –42, b = 49

Question 6.

Find the values of a and b if the following polynomials are perfect squares.

x⁴–4x³ + 6x²–ax + b

Answer:

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

Since it is a perfect square the remainder is 0

a = 12, b = 9

Question 7.

Find the values of a and b if the following polynomials are perfect squares.

ax⁴ + bx³ + 109x²–60x + 36

Answer:

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

We rearrange the equation in the increasing order of power of x.

The polynomial becomes 36–60x + 109x² + bx³ + ax⁴

Since it is a perfect square the remainder is 0

a = 49, b = –70

Question 8.

Find the values of a and b if the following polynomials are perfect squares.

a x⁴–bx³ + 40 x² + 24x + 36

Answer:

Step 1: Find the algebraic expression whose square gives the first term.

Step 2: Add the divisor and quotient .To this sum add a suitable expression and add the same expression in the quotient such that the product gives the next term.

Step 3: Continue the process until all the terms are divided.

We rearrange the equation in the increasing order of power of x.

The polynomial becomes 36 + 24x + 40x²–bx³ + ax⁴

Since it is a perfect square the remainder is 0

a = 9, b = –12

---

Exercise 3.14

Question 1.

Solve the following quadratic equations by factorization method.

(2x + 3)² – 81 = 0

Answer:

(2x + 3)² – 81 = 0

= (2x)² + 2(2x)(3) + 3² – 81 = 0

= 4x² + 12x + 9 –81 = 0

= 4x² + 12x – 72 = 0

Divide by 4 both sides

⇒ 

= x² + 3x – 18 = 0

= x² + 6x – 3x – 18 = 0

= x (x + 6) – 3 (x + 6) = 0

= (x + 6) (x – 3) = 0

x + 6 = 0 or x – 3 = 0

x = –6 or x = 3

Question 2.

Solve the following quadratic equations by factorization method.

3x² –5x –12 = 0

Answer:

3x² – 5x –12 = 0

= 3x² – 9x + 4x – 12 = 0

= 3x (x – 3) + 4(x – 3) = 0

= (3x + 4) (x – 3) = 0

3x + 4 = 0 or x –3 = 0

3x = –4 or x = 3

 or x = 3

Question 3.

Solve the following quadratic equations by factorization method.

Answer:

√5x² + 2x – 3√5 = 0

= √5x² + 5x – 3x – 3√5 = 0

= √5x (x + √5) – 3(x + √5) = 0

= (√5x – 3) (x + √5) = 0

√5x – 3 = 0 or x + √5 = 0

√5x = 3 or x = –√5

or x = –√5

Question 4.

Solve the following quadratic equations by factorization method.

3(x² – 6) =x (x + 7)–3

Answer:

3(x² – 6) =x (x + 7)–3

= 3x² – 18 = x² + 7x – 3

= 3x² – 18 – x² – 7x + 3 = 0

= 2x² – 7x – 15 = 0

= 2x² – 10x + 3x – 15 = 0

= 2x(x – 5) + 3(x – 5) = 0

= (2x + 3)(x – 5) = 0

2x + 3 = 0 or x – 5 = 0

2x = –3 or x = 5

or x = 5

Question 5.

Solve the following quadratic equations by factorization method.

Answer:

= 3x² – 8 = 2x

= 3x² – 2x – 8 = 0

= 3x² – 6x + 4x – 8 =0

= 3x (x – 2) + 2(x – 2) = 0

= (3x + 2) (x – 2) = 0

3x + 2 = 0 or x – 2 = 0

3x = –2 or x = 2

or x = 2

Question 6.

Solve the following quadratic equations by factorization method.

Answer:

= 

= 5(x² + 1) = 26x

= 5x² + 5 = 26x

= 5x² –26x + 5 = 0

= 5x² – 25x – x + 5 = 0

= 5x (x – 5) – (x – 5) = 0

= (5x – 1) (x – 5) = 0

5x – 1 = 0 or x – 5 = 0

5x = 1 or x = 5

 or x = 5

Question 7.

Solve the following quadratic equations by factorization method.

Answer:

= 

= 

= 15(x² + x² + 2x + 1) = 34(x² + x)

= 15(2x² + 2x + 1) = 34(x² + x)

= 30x² + 30x + 15 = 34x² + 34x

= 34x² + 34x – 30x² –30x – 15 =0

= 4x² – 4x – 15 = 0

= 4x² – 10x + 6x – 15 = 0

= 2x(2x – 5) + 3(2x – 5) = 0

= (2x + 3) (2x – 5) = 0

2x + 3 =0 or 2x – 5 = 0

2x = –3 or 2x = 5

 or x =

Question 8.

Solve the following quadratic equations by factorization method.

a²b²x² – (a² + b²) x + 1 =0

Answer:

a²b²x² – (a² + b²) x + 1 = 0

= a²b²x² – a²x – b²x + 1 = 0

= a²x (b²x – 1) – (b²x – 1) = 0

= (a²x – 1) (b²x – 1) =0

a²x – 1 = 0 or b²x – 1 = 0

a²x = 1 or b²x = 1

Question 9.

Solve the following quadratic equations by factorization method.

2(x + 1)² –5 (x + 1) =12

Answer:

2(x + 1)² – 5(x + 1) = 12

= 2(x² + 2x + 1) – 5x – 5 = 12

= 2x² + 4x + 2 – 5x – 5 –12 = 0

= 2x² – x – 15 = 0

= 2x² – 6x + 5x – 15 = 0

= 2x (x – 3) + 5(x –3) = 0

= (2x + 5) (x – 3) = 0

2x + 5 = 0 or x – 3 =0

2x = –5 or x = 3

x =  or x = 3

Question 10.

Solve the following quadratic equations by factorization method.

3(x–4)² – 5 (x – 4) = 12

Answer:

3(x – 4)² – 5(x – 4) = 12

= 3(x² – 8x + 16) – 5x + 20 = 12

= 3x² – 24x + 48 –5x + 20 – 12 = 0

= 3x² – 29x + 56 = 0

= 3x² – 21x – 8x + 56 = 0

= 3x(x – 7) – 8(x – 7) =0

= (x – 7) (3x – 8) = 0

x – 7 = 0 or 3x – 8 = 0

x = 7 or 3x = 8

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Exercise 3.15

Question 1.

Solve the following quadratic equations by completing the square.

x² + 6x –7 = 0

Answer:

x² + 6x – 7 = 0

= x² + 6x = 7

Add 9 on both sides

= x² + 6x + 9 = 7 + 9

= x² + 2(3)(x) + 3² = 16

= (x + 3)² = 16

= x + 3 = √16

= x + 3 = ± 4

x + 3 = 4 or x + 3 = –4

x = 4 – 3 or x = – 4 – 3

x = 1 or x = – 7

Question 2.

Solve the following quadratic equations by completing the square.

x² + 3x + 1 =0

Answer:

x² + 3x + 1 =0

= x² + 3x = –1

Add  on both sides

Question 3.

Solve the following quadratic equations by completing the square.

2x² + 5x –3 = 0

Answer:

2x² + 5x – 3 = 0

= 2x² + 5x = 3

Add  on both sides

Question 4.

Solve the following quadratic equations by completing the square.

4x² + 4bx – (a² – b²) = 0

Answer:

4x² + 4bx – (a² – b²) = 0

Divide the whole equation by 4

Question 5.

Solve the following quadratic equations by completing the square.

x² – (√3 + 1) x + √3 = 0

Answer:

x² – (√3 + 1)x + √3 = 0

Question 6.

Solve the following quadratic equations by completing the square.

=3x + 2

Answer:

= 5x + 7 = (3x + 2)(x – 1)

= 5x + 7 = 3x(x – 1) + 2 (x – 1)

= 5x + 7 = 3x² – 3x + 2x –2

= 5x + 7 = 3x² – x – 2

= 3x² – x – 2 – 5x – 7 =0

= 3x² – 6x – 9 = 0

Divide whole equation by 3

= x² – 2x – 3 = 0

= x² – 3x + x – 3 = 0

= x (x – 3) + (x – 3) = 0

= (x – 3) (x + 1) = 0

x – 3 = 0 or x + 1 = 0

x = 3 or x = –1

Question 7.

Solve the following quadratic equations using quadratic formula.

x²7x + 12= 0

Answer:

x² – 7x + 12 = 0

(–7)²– 4 (1) (12)

=49 – 48

=1

Question 8.

Solve the following quadratic equations using quadratic formula.

15x² – 11x + 2 = 0

Answer:

15x² – 11x + 2 = 0

⇒ a = 15 , b = –11 and c = 2

(–11)²– 4 (15) (2)

=121 – 120

=1

Question 9.

Solve the following quadratic equations using quadratic formula.

Answer:

= 

⇒ 2(x² + 1) = 5x

⇒ 2x² + 2 = 5x

⇒ 2x² – 5x + 2 = 0

(–5)²– 4 (2) (2)

= 25 – 16

= 9

Question 10.

Solve the following quadratic equations using quadratic formula.

3a²x² – ax –2b² = 0

Answer:

3a²x² – abx – 2b² = 0

⇒ a = 3a² , b = –ab and c = –2b²

(–ab)²– 4 (3a²) (–2b²)

= a²b² + 24a²b²

=25a²b²

Question 11.

Solve the following quadratic equations using quadratic formula.

a (x² + 1) = x (a² + 1)

Answer:

a(x² + 1) = x(a² + 1)

⇒ ax² + a = a²x + x

⇒ ax² + a – a²x – x = 0

⇒ ax² – x(a² + 1) + a = 0

⇒ a = a , b = – a² – 1 and c = a

( –a² – 1 )²– 4 (a)(a)

= a⁴ + 2a² + 1 – 4a²

= (a²)² – 2a² + 1

= (a² – 1 )²

Question 12.

Solve the following quadratic equations using quadratic formula.

36x² – 12ax + (a² – b²) = 0

Answer:

36x² – 12ax + (a² –b²) = 0

⇒ a = 36, b = –12a and c = a² – b²

(–12a)²– 4 (36)(a² – b²)

= 144a² – 144(a² – b²)

= 144a² – 144a² + 144b²

= 144 b²

Question 13.

Solve the following quadratic equations using quadratic formula.

Answer:

⇒ 

⇒ 

⇒ 3(2x² – 7x + 1) = 10(x² – 3x – 4)

⇒ 6x² – 21x + 3 = 10x² – 30x – 40

⇒ 10x² – 6x² – 30x + 21x – 40 –3 = 0

⇒ 4x² – 9x – 43 = 0

(–9)²– 4 (4) (–43)

= 91 + 688

= 769

Question 14.

Solve the following quadratic equations using quadratic formula.

a²x² + (a² – b²) x – b² = 0

Answer:

a²x² + (a² – b²)x – b² =0

⇒ a = a² , b = a² – b² and c = –b²

(a² – b²)²– 4 (a²)(–b²)

= (a²)² – 2a²b² + (b²)² + 4a²b²

= (a²)² + 2a²b² + (b²)²

= (a² + b²)²

---

Exercise 3.16

Question 1.

The sum of a number and its reciprocal is . Find the number.

Answer:

Let x be the required number. Then, the reciprocal is .

⇒ sum of a number and its reciprocal is 

⇒ 

⇒ 

⇒ 8(x² + 1) = 65x

⇒ 8x² + 8 = 65x

⇒ 8x² – 65x + 8 = 0

⇒ 8x² – x – 64x + 8 = 0

⇒ x(8x – 1) – 8(8x – 1) = 0

⇒ (x – 8) (8x – 1) = 0

x – 8 = 0 or 8x – 1 = 0

x = 8 or 8x = 1

Therefore, the two required numbers are 8 and .

Question 2.

The difference of the squares of two positive numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.

Answer:

Let ‘x’ be the larger number and ‘y’ be the smaller number.

x² – y² = 45 …(1)

y² = 4x …(2)

Now, put the value of y² in equation (1).

x² – 4x = 45

⇒ x² – 4x – 45 = 0

⇒ x²– 9x + 5x – 45 = 0

⇒ x(x – 9) + 5(x – 9) = 0

⇒ (x + 5)(x – 9) =0

x + 5 = 0 or x – 9 = 0

x = –5 or x = 9

Here positive 9 only admissible. From this we need to find the value of y for that we are going to aplly this value in the second equation.

y² = 4 x

⇒ y² = 4 × 9

⇒ y² = 36

⇒ y = √36

⇒ y = ± 6

Here, positive 6 only admissible.

Therefore, the required numbers are 6 and 9.

Question 3.

A farmer wishes to start a 100 sq. rectangular vegetable garden. Since he has only 30 m barbed wire, he fences the sides of the rectangular garden letting his house compound wall act as the fourth side fence. Find the dimension of the garden.

Answer:

Let ‘x’ and ‘y’ are the dimension of the vegetable garden.

Area of rectangle = Length × Width

x × y = 100

we are going to cover the barbed wire for fencing only. So, it must be the perimeter of vegetable garden. Usually perimeter always covers all the four side. Bute here we are going to cover only three sides, because one side of the vegetable garden will act as the compound wall.

x + x + y = 30

⇒ 2x + y = 30

⇒ 

⇒ 

⇒ 200 + y² = 30y

⇒ y² – 30y + 200 = 0

⇒ y² – 10y –20y + 200 = 0

⇒ y(y – 10) – 20(y – 10) = 0

⇒ (y – 10)(y – 20) =0

y – 10 = 0 or y – 20 = 0

y = 10 or y = 20

Now we are going to apply these values in to get the values of x.

If y = 10 if y = 20

x = 10 x = 5

Therefore, the required dimensions are 10m and 10m or 20m and 5m.

Question 4.

A rectangular field is 20 m long and 14 m wide. There is a path of equal width all around it having an area of 111 sq. meters. Find the width of the path on the outside.

Answer:

Length of the rectangular field = 20m

Breadth of the rectangular field = 14m

Let x be the uniform width all around the path.

Length of the rectangular field including the path

= 20 + x + x

= 20 + 2x

Width of the rectangular field including path

= 14 + x + x

= 14 + 2x

Area of path = area of rectangular field including path – area of rectangular field

⇒ 111 = (20 + 2x)(14 + 2x) – (20 × 14)

⇒ 111 = 280 + 40x + 28x + 4x² – 280

⇒ 111 = 68x + 4x²

⇒ 4x² + 68x – 111 = 0

⇒ 4x² + 74x – 6x – 111 = 0

⇒ 2x(2x + 37) – 3(2x + 37) = 0

⇒ (2x + 37)(2x – 3) = 0

2x + 37 = 0 or 2x – 3 = 0

2x = –37 or 2x = 3

x = –18.5 or x = 1.5

Therefore, width of the path = 1.5m

Question 5.

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

Answer:

Let x be the usual speed of the train.

Let T1 be the time taken to cover the distance 90 km in the speed x km/hr.

Let T2 be the time taken to cover the distance 90 km in the speed x + 15 km/hr.

By using the given condition

Taking 90commonly from two fractions

⇒ 

⇒ 

⇒ 

⇒ 15 × 180 = x² + 15x

⇒ x² + 15x = 2700

⇒ x² + 15x – 2700 = 0

⇒ x² + 60x –45x – 2700 = 0

⇒ x(x + 60) –45(x + 60) = 0

⇒ (x – 45)(x + 60) = 0

x – 45 = 0 or x + 60 = 0

x = 45 or x= –60

therefore speed of the train is 45 km/hr.

Question 6.

The speed of a boat in still water is 15 km/hr. It goes 30 km upstream and return downstream to the original point in 4 hrs 30 minutes. Find the speed of the stream.

Answer:

Let x km/hr be the speed of water

Speed of boat is 15km/hr.

So, speed in upstream = (15 + x) km/hr.

speed in downstream = (15 – x) km/hr.

Let T1 be the time taken to cover the distance 30 km in upstream.

Let T2 be the time taken to cover the distance 30 km in downstream.

T1 + T2 = 4hours 30minutes

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 900 × 2 = 9(225 – x²)

Now, let us divide the entire equation by 9.

So, that we will get,

200 = 225 – x²

200 + x² = 225

x² = 225 – 200

x² = 25

x = √25

x = ± 5

Speed must be positive so x =5 is the requied speed.

Speed of water = 5km/hr.

Question 7.

One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.

Answer:

Let x be the present age of son

Let ‘y’ be the present age of father

So, x – 1 be the age of son one year age

y – 1 be the age of father one year ago.

By using the given information

y = x²

y – 1 = 8(x – 1)

⇒ y = 8x – 8 + 1

⇒ y = 8x – 7

⇒ x² = 8x – 7

⇒ x² – 8x + 7 = 0

⇒ x² –x – 7x + 7 = 0

⇒ x(x – 1) –7(x – 1) = 0

⇒ (x – 1)(x – 7) = 0

X – 1 = 0 or x – 7 = 0

x = 1 or x = 7

Therefore, age of father is 49.

Question 8.

A chess board contains 64 equal squares and the area of each square is 6.25 cm². An order around the board is 2 cm wide. Find the length of the side of the chess board

Answer:

Let x be the side length of the square board

Area of one square in the chess board = 6.25cm²

Area of 64square = 64×6.25

(x – 4)² = 400

⇒ x – 4 = √400

⇒ x – 4 = ± 20

x – 4 = 20 or x – 4 = –20

x = 20 + 4 or x = –20 + 4

x = 24 or x = –16

Therefore, side length of square shaped chess board is 24cm.

Question 9.

A takes 6 day less than the time taken by B to finish a piece of work. If both A and B together can finish it in 4 days, find the time that B would take to finish this work by himself.

Answer:

Let x be the tune taken by A to finish the work.

So, x – 6 be the time taken by B to finish the work.

Work done by A in one day = 

Work done by B in one day = 

Number of days taken by both to finish the work = 

⇒ 

⇒ 

⇒ 4(2x – 6) = x² – 6x

⇒ 8x – 24 = x² – 6x

⇒ x² – 6x – 8x + 24 = 0

⇒ x² – 14x + 24 = 0

⇒ x² – 12x – 2x + 24 = 0

⇒ x(x – 12) –2(x – 12) = 0

⇒ (x – 12)(x – 2) = 0

x – 12 = 0 or x – 2 = 0

x = 12 or x = 2

Here, 2 is not admissible.

So, B is taking 12 days to finish the work.

Question 10.

Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after two hours, they are 50 km apart, find the average speed of each train.

Answer:

Let x km/hr. be the speed of second train.

So, speed of first train will be (x + 15) km/hr.

Distance covered by first train in 2 hours = 2(x + 5)

Distance covered by the second train in 2 hours = 2x

By using Pythagoras theorem

[2(x + 5)]² + (2x)² = 50²

⇒ (2x + 10)² + (2x)² = 50²

⇒ (4x² + 100 + 40x) + 4x² = 2500

⇒ 8x² + 40x + 100 = 2500

⇒ 8x² + 40x + 100 –2500 = 0

⇒ 8x² + 40x – 2400 = 0

Divide by 8 both sides

⇒ x² + 5x – 300 = 0

⇒ x² + 20x – 15x – 300 = 0

⇒ x(x + 20) – 15 (x + 20) = 0

⇒ (x – 15)(x + 20) = 0

x – 15 = 0 or x + 20 = 0

x = 15 or x = –20

Therefore, speed of the second train is 15 km/hr.

---

Exercise 3.17

Question 1.

Determine the nature of the roots of the equation.

x² – 8x + 12 = 0

Answer:

x² – 8x + 12 = 0

⇒ a = 1 , b = –8 and c = 12

 = (–8)²– 4(1)(12)

 64 – 48

= 16

∴ b² – 4ac > 0. hence, roots are real.

Question 2.

Determine the nature of the roots of the equation.

2x² – 3x + 4 = 0

Answer:

2x² – 3x + 4 = 0

⇒ a = 2, b = –3 and c = 4

 = (–3)²– 4(2)(4)

 9 – 32

= –23

∴ b² – 4ac < 0. hence, roots are not real.

Question 3.

Determine the nature of the roots of the equation.

9x² + 12x + 4 = 0

Answer:

9x² – 12x + 4 = 0

⇒ a = 9, b = –12 and c = 4

 = (–12)²– 4(9)(4)

 144 – 144

= 0

∴ b² – 4ac > 0. hence, roots are real and equal.

Question 4.

Determine the nature of the roots of the equation.

3x² –2√6x + 2 = 0

Answer:

3x² – 2√6 x + 2 =

⇒ a = 3, b = –2√6 and c = 2

 = (–2√6 )²– 4(3)(2)

 24 – 24

= 0

∴ b² – 4ac = 0. hence, roots are real and equal.

Question 5.

Determine the nature of the roots of the equation.

Answer:

⇒ 

⇒ 9x² – 60x + 15 = 0

⇒ a = 9, b = –60 and c = 15

 = (–60)²– 4(9)(15)

 3600 – 540

= 3060

∴ b² – 4ac > 0. hence, roots are real.

Question 6.

Determine the nature of the roots of the equation.

(x – 2a) (x – 2b) = 4ab

Answer:

(x – 2a)(x – 2b) = 4ab

⇒ x(x –2b) – 2a(x – 2b) = 4ab

⇒ x² – 2bx – 2ax + 4ab – 4ab = 0

⇒ x² – 2x(b + a) = 0

⇒ a = 1, b = –b – a and c = 0

 = (–b – a )²– 4(1)(0)

 b² + a² + 2ab

∴ b² – 4ac > 0. hence, roots are real.

Question 7.

Find the values of k for which the roots are real and equal in each of the following equations

2x² – 10x + k = 0

Answer:

2x² – 10x + k = 0

 ∴ 

⇒ 100 – 8k = 0

⇒ 100 = 8k

 = 

∴ 

Question 8.

Find the values of k for which the roots are real and equal in each of the following equations

12x² + 4kx + 3 = 0

Answer:

12x² + 4kx + 3 = 0

 ∴ 

⇒ 16k² – 144 = 0

⇒ 16k² = 144

⇒ 

⇒ 

∴ k = 3 and k = –3

Question 9.

Find the values of k for which the roots are real and equal in each of the following equations

x² + 2k (x – 2) + 5 = 0

Answer:

x² + 2k (x – 2) + 5 = 0

⇒ x² + 2kx – 4k + 5 = 0

 ∴ 

⇒ 4k² + 16k – 20 =0

Divide by 4

⇒ k² + 4k – 5 = 0

⇒ k² + 5k – k – 5 = 0

⇒ k(k + 5) – (k + 5) =0

⇒ (k + 5)(k – 1) = 0

k + 5 = 0 or k – 1 = 0

k = –5 or k = 1

∴ k = –5 and k = 1

Question 10.

Find the values of k for which the roots are real and equal in each of the following equations

(k + 1) x² – 2 (k – 1) x + 1 = 0

Answer:

(k + 1) x² – 2 (k – 1) x + 1 = 0

 ∴ 

⇒ 4k² – 8k + 4 – 4k – 4 = 0

⇒ 4k² – 12k = 0

Divide by 4

⇒ k² – 3k = 0

⇒ k(k – 3) = 0

⇒ k = 0 or k– 3 = 0

k = 0 or k = 3

∴ k = 0 and k = 3

Question 11.

Show that the roots of the equation x² + 2(a + b) x + 2 (a² + b²) = 0 are unreal.

Answer:

x² + 2(a + b) x + 2 (a² + b²) = 0

Compare this equation with ax² + bx + c = 0

∴ a = 1, b = 2(a + b) and c = 2(a² + b²)

b² – 4ac = (2a + 2b)² – 4(1)[2(a² + b²)]

= 4a² + 8ab + 4b² – 8a² – 8b²

= 8ab – 4a² – 4b²

= – 4a² + 8ab – 4b²

= –4(a² – 2ab + b²)

= –4 (a – b)²

Since squared quantity is always positive.

Hence, (a – b)² ≥ 0

Now, it is given a ≠ b, so (a – b)² > 0

So, D = –4(a – b)² will be negative.

Hence the equation has no real roots.

Question 12.

Show that the roots of the equation 3p²x² – 2pqx + q² = 0 are not real.

Answer:

3p²x² – 2pqx + q² = 0

Compare this equation with ax² + bx + c = 0

∴ a = 3p² , b = 2pq and c = q²

b² – 4ac = (2pq)² – 4(3p²)[q²]

= 4p²q² – 12p²q²

= – 8p²q²

Since squared quantity is always positive.

Hence, p²q² ≥ 0

Now, it is given p ≠ q, so p²q² > 0

So, D = –8p²q² will be negative.

Hence the equation has no real roots.

Question 13.

If the roots of the equation (a² + b²) x² – 2 (ac + bd) x + c² + d² = 0, where a, b, c and d ≠ 0, are equal, prove that .

Answer:

Given: (a² + b²) x² – 2 (ac + bd) x + c² + d² = 0

To prove: 

Proof:

We know that,

D = b² – 4ac

If roots are equal, then b² = 4ac

⇒ {–2(ac + bd)}² = 4{(a² + b²)( c² + d²)}

⇒ 4(a²c² + b²d² + 2acbd) = 4 (a²c² + a²d² + b²c² + b²d²)

⇒ 2acbd = a²d² + b²c²

⇒ a²d² + b²c² – 2acbd = 0

⇒ (ad – bc)² = 0

⇒ ad – bc = 0

⇒ ad = bc

⇒ 

Hence proved.

Question 14.

Show that the roots of the equation

(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are always real and they cannot be unless a = b = c.

Answer:

(x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0

⇒ x(x – b) – a(x – b) + x(x – c) – b(x – c) + x(x – a) – c(x – a) = 0

⇒ x² – bx – ax + ab + x² – cx – bx + bc + x² – ax – cx + ac = 0

⇒ 3x² – 2x(a + b + c) + ab + bc + ac = 0

D = b² – 4ac

D = (a + b + c)² – 4(3)(ab + bc + ac) = 0

D = 4(a² + b² + c² + 2ab + 2bc + 2ca – 3ab – 3bc – 3ca)

D = 4(a² + b² + c² – ab – bc – ca)

D = 2[(a –b)² + (b – c)² + (c – a)²]

Which is always greater than zero so the roots are real.

Roots are equal if D = 0

i.e. (a – b)² + (b + c)² + (c – a)² = 0

since sum of three perfect square is equal to zero so each of them separately equal to zero.

So, a – b = 0, b – c = 0, c – a = 0

a = b , b = c, c = a

so, a = b = c.

Question 15.

If the equation (1 + m²) x² + 2mcx + c² – a² = 0 has equal roots, then prove that c² = a² (1 + m²)

Answer:

Given: (1 + m²) x² + 2mcx + c² – a² = 0

To prove: c² = a (1 + m²)

Proof: it is being that equation has equal roots, therefore

D = b² – 4ac = 0 …(1)

From the equation, we have

a = (1 + m²), b = 2mc, c = c² – a²

putting values of a, b and c in (1), we get

D = (2mc)² – 4(1 + m²)(c² – a²)

⇒ 4m²c² – 4(c² + c²m² – a² – a²m²) = 0

⇒ 4m²c² – 4c² – 4c²m² + 4a² + 4a²m² = 0

⇒ –4c² + 4a² + 4a²m² = 0

⇒ 4c² = 4a² + 4a²m²

⇒ c² = a² + a²m²

⇒ c² = a² (1 + m²)

Hence proved.

---

Exercise 3.18

Question 1.

Find the sum and the product of the roots of the following equation.

x² – 6x + 5 = 0

Answer:

x² – 6x + 5 = 0

Compare this equation with ax² + bx + c = 0

a = 1, b = –6 and c = 5

Question 2.

Find the sum and the product of the roots of the following equation.

kx² + ax + pk = 0

Answer:

kx² + ax + pk = 0

Compare this equation with ax² + bx + c = 0

a = k, b = –a and c = pk

Question 3.

Find the sum and the product of the roots of the following equation.

3x² – 5x = 0

Answer:

3x² – 5x = 0

Compare this equation with ax² + bx + c = 0

a = 3, b = –5 and c = 0

Question 4.

Find the sum and the product of the roots of the following equation.

8x² – 25 = 0

Answer:

8x² – 25 = 0

Compare this equation with ax² + bx + c = 0

a = 8, b = 0 and c = –25

Question 5.

Form a quadratic equation whose roots are

(i) 3, 4 (ii) 3 + √7, 3 – √7 (iii) 

Answer:

i: 3 and 4

Let 

∴ 

∴

∴ 

∴ 

ii: 3 + √7 and 3 – √7

Let α=3 + √7 and β=3–√7

∴ α + β=3 + √7 + 3–√7=6 and αβ=(3 + √7)(3–√7)

=9–7=2

∴

∴ 

∴ 

iii: 

⇒ 

⇒ 

Question 6.

If α and β are the roots of the equation 3x² – 5x + 2= 0, then find the values of

(i)  (ii) α – β (iii) 

Answer:

3x² – 5x + 2 = 0 compare this with ax² – bx + c = 0

∴ a = 3 , b = –5 and c = 2

i). 

⇒ 

⇒ 

⇒ = 

⇒  = 

⇒  = 

ii). α – β

α – β = √(α + β)² – 4α β

iii). 

Question 7.

If α and β are the roots of the equation 3x² – 6x + 4 = 0, find the value of α² – β².

Answer:

3x² – 6x + 4 = 0 compare this with ax² – bx + c = 0

∴ a = 3 , b = –6 and c = 4

Question 8.

If α, β are roots of 2x² – 3x – 5 = 0, from an equation whose roots are α² and β².

Answer:

2x² – 3x – 5 = 0 compare this with ax² – bx + c = 0

∴ a = 2 , b = –3 and c = –5

Here α = α² and β = β²

General form of quadratic equation whose roots are α² and β²

⇒ x² – (α² + β²) x + α²β² = 0

⇒ x² – (α² + β²) x + (αβ)² = 0

α² + β² = (α + β)² – 2(αβ)

x² – (α² + β²) x + (αβ)² = 0

4x² – 29x + 25 = 0

Therefore the required equation is 4x² – 29x + 25 = 0

Question 9.

If α, β are roots of x² – 3x + 2 = 0, form a quadratic equation whose roots are –α and –β

Answer:

x² – 3x + 2 = 0 compare this with ax² – bx + c = 0

∴ a = 1 , b = –3 and c = 2

Here α = –αand β = – β

General form of quadratic equation whose roots are α² and β²

⇒ x² – (–α – β) x + (–α) (–β) = 0

⇒ x² + (α + β) x + (αβ) = 0

⇒ x² + (3)x + (2) = 0

Therefore, the required quadratic equation is x² – 3x + 2 = 0

Question 10.

If α and β are roots of x² – 3x–1 = 0, then form a quadratic equation whose roots are

Answer:

x² – 3x – 1 = 0 compare this with ax² – bx + c = 0

∴ a = 1 , b = –3 and c = –1

Here 

General form of quadratic equation whose roots are α² and β²

⇒ 

⇒ 

⇒ 

= 3²– 2 × (–1)

=9 + 2

= 11

⇒ 

⇒ 

⇒ x² – 11x + 1 = 0

Therefore, the required equation is x² + 11x + 1 = 0

Question 11.

If α and β are roots of 3x² – 6x + 1 = 0, then form a quadratic equation whose roots are

(i)  (ii) α²β, β²α (iii) 2α + β, 2β + α

Answer:

3x² – 6x + 1 = 0 compare this with ax² – bx + c = 0

∴ a = 3 , b = –6 and c = 1

i). Here 

General form of quadratic equation whose roots are α² and β²

⇒ 

⇒ 

⇒ 

⇒ 

⇒ x² – 6x + 3 = 0

Therefore, required equation is x² – 6x + 3 = 0

ii). Here, α = α²β and β = β² α

General form of quadratic equation whose roots are α²β and β² α

x² – (α²β + β² α) x + (α²β)(β² α) = 0

⇒ x² – αβ (α + β) x + (α³β³) = 0

⇒ x² – αβ (α + β) x + (αβ)³ = 0

⇒ 

⇒ 

⇒ 

⇒ 27x² – 18x + 1 = 0

Therefore, required equation is 27x² – 18x + 1 = 0

iii). Here, α = 2α + β and β = 2β + α

General form of equation whose roots are 2α + β and 2β + α

x² – (2α + β + 2β + α)x + (2α + β)(2β + α) = 0

x² – (3α + 3β) x + (4αβ + 2α² + 2β² + αβ) = 0

x² – (3α + 3β) x + (2(α² + β² ) + 5αβ) = 0

α² + β² = (α + β)² – 2αβ

x² – (3α + 3β) x + (2(α² + β² ) + 5αβ) = 0

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 3x² –18x + 70 = 0

Therefore, the required equation is 3x² –18x + 70 = 0

Question 12.

Find a quadratic equation whose roots are the reciprocal of the roots of the equation

4x² – 3x – 1 = 0

Answer:

4x² – 3x – 1 = 0 compare this with ax² – bx + c = 0

∴ a = 4 , b = –3 and c = –1

Here 

General form of quadratic equation whose roots are α² and β²

⇒ 

⇒ 

⇒ 

⇒ 

⇒ x² – 3x – 4 = 0

Therefore, required equation is x² – 3x – 4 = 0

Question 13.

If one root of the equation 3x² + kx – 81 = 0 is the square of the other, find k.

Answer:

Two roots of any quadratic equation are α and β.

Here, one root is square of the other i.e α = β²

3x² – kx – 81 = 0 compare this with ax² – bx + c = 0

∴ a = 3, b = –k and c = –81

αβ = –27

β²(β) = –27

β³ = –27

β³ = (–3)³

β = –3

now, we are going to apply in first equation

⇒ 

⇒ 

⇒ 

⇒ 6 × 3 = k

⇒ k = 18

Question 14.

If one root of the equation 2x² – ax + 64 = 0 is twice the other, then find the value of a

Answer:

Roots of any quadratic equation are α and β

Here one root is twice the other.

α = 2β

by comparing the given equation with general form of quadratic equation we get,

a = 2, b = –a and c = 64

_{α + β =}

 …(1)

αβ = 32

2β(β) = 32

2β² = 32

⇒ 

⇒ β² = 16

⇒ β = √16

⇒ β =± 4

⇒ 

⇒ 

⇒ 12 × 2 = a

⇒ a = 24

Question 15.

If α and β are roots of 5x² – px + 1 = 0 and α – β = 1, then find P.

Answer:

Roots of any quadratic equation are α and β

By comparing the given equation with ax² + bx + c = 0

a = 5, b = –p and c = 1

α – β = 1

α – β = (α + β)² – 4αβ

⇒ 

⇒ 

⇒ p² – 20 = 25

⇒ p² = 25 + 20

⇒ p² = 45

⇒ p = √45

⇒ p = 3√5

---

Exercise 3.19

Question 1.

If the system 6x – 2y = 3, kx – y = 2 has a unique solution, then
A. k = 3

B. k ≠ 3

C. k = 4

D. k ≠ 4

Answer:

Given: Two equations: 6x – 2y = 3 and kx – y = 2

Required: To find the value of k such that system of equations have unique solutions

To have unique solution to the System of equations, the required condition is 

∴ 

⇒ 

⇒ 

⇒ 

∴ For every value of k except 3 the system equations have unique solutions.

∴ Correct option is – Option (B)

Question 2.

A system of two linear equations in two variables is consistent, if their graphs
A. coincide

B. intersect only at a point

C. do not intersect at any point

D. cut the x-axis

Answer:

Given: A system of two linear equations in two variables is consistent

We know that if a system of two linear equations in two variables is consistent then their graph do not intersect at any point

∴ Correct option is – Option (C)

Question 3.

The system of equations x –4y = 8 , 3x –12y = 24
A. has infinitely many solutions

B. has no solution

C. has a unique solution

D. may or may not have a solution

Answer:

Given: system of equations x –4y = 8 , 3x –12y = 24

Here,

a₁ = 1, b₁ = -4, c₁ = 8 and a₂ = 3, b₂ = -12, c₂ = 24

Now,

 , ,

Here, we can clearly see that  which is the condition for infinitely many solutions.

∴ The system of equations have infinitely many solutions.

∴ Correct option is – Option (A)

Question 4.

If one zero of the polynomial p(x) = (k + 4)x² + 13x + 3k is reciprocal of the other, then k is equal to
A. 2

B. 3

C. 4

D. 5

Answer:

Given: A Quadratic equation p(x) = (k + 4)x² + 13x + 3k

Required: To find the value of k

Let the roots of the given Quadratic equation be:  and 

∴ Product of roots of the given Quadratic equation is 

We know that, Product of roots of a given Quadratic equation is 

∴ 

⇒ 

⇒ 3k = k + 4

⇒ 2k = 4

⇒ k = 2

∴ The value of k is 2

∴ Correct option is – Option (A)

Question 5.

The sum of two zeros of the polynomial f(x) = 2x² + (p + 3)x + 5 is zero, then the value of p is
A. 3

B. 4

C. –3

D. –4

Answer:

Given: A Quadratic equation f(x) = 2x² + (p + 3)x + 5 and sum of roots is zero.

Required: to find the value of p

We know that, sum of roots = 

∴  = 0

⇒ -(p + 3) = 0

⇒ P + 3 = 0

∴ p = –3

∴ The value of p is –3

∴ Correct option is – Option (C)

Question 6.

The remainder when x² – 2x + 7 is divided by x + 4 is
A. 28

B. 29

C. 30

D. 31

Answer:

Given: x² – 2x + 7

Required: Reminder of x² – 2x + 7 at x + 4

By synthetic division we can find reminder of

x² – 2x + 7 at x = -4

∴ Reminder of x² – 2x + 7 at x + 4 = 31

∴ Correct option is –Option (D)

Question 7.

The quotient when x³ – 5x² + 7x – 4 is divided by x–1 is
A. x² + 4x + 3

B. x² – 4x + 3

C. x² – 4x -3

D. x² + 4x – 3

Answer:

Given: x² – 2x + 7

Required: Reminder of x³ – 5x² + 7x – 4 at x – 1

By synthetic division we can find Quotient of x³ – 5x² + 7x – 4 at x = 1

∴ Quotient of x² – 2x + 7 when divided by x–1 = x²–4x + 3

∴ Correct option is –Option(B)

Question 8.

The GCD of (x³ + 1) and x⁴ – 1 is
A. x³ – 1

B. x³ + 1

C. -(x + 1)

D. x–1

Answer:

Given two polynomials: (x³ + 1) and x⁴ – 1

Required: To find GCD of the given two polynomials

Let f(x) = x³ + 1 and g(x) = X⁴ – 1

Here, degree of g(x) > f(x) ∴ Devisor = x³ + 1

Here reminder = –(x + 1) (Reminder ≠ zero)

∴ The GCD of given two polynomials is -(x + 1)

∴ Correct Option is - Option(C)

Question 9.

The GCD of x² – 2xy + y² and x⁴ – y⁴ is
A. 1

B. x + y

C. x – y

D. x² – y²

Answer:

Given two polynomials: x² – 2xy + y² and x⁴ – y⁴

Required: To find GCD of the given two polynomials

Let f(x) = x² – 2xy + y² and g(x) = x⁴ – y⁴

f(x) = x² – 2xy + y²

⇒ f(x) = (x–y)²

g(x) = x⁴ – y⁴

⇒ g(x) = (x²)² – (y²)²

⇒ g(x) = (x² – y²)(x² + y²) (∵ a²–b² = (a–b)(a + b))

⇒ g(x) = (x–y)(x + y)(x² + y²) (∵ a²–b² = (a–b)(a + b))

∴ The GCD of given two polynomials is (x–y)

∴ Correct Option is - Option (C)

Question 10.

The LCM of x³ – a³ and (x - a)² is
A. (x³ – a³) (x + a)

B. (x³ – a³) (x - a)²

C. (x - a)² (x² + ax + a²)

D. (x + a)²(x² + ax + a²)

Answer:

Given two polynomials: x³ – a³ and (x - a)²

Required: To find LCM of the given two polynomials

Let f(x) = x³ – a³ and g(x) = (x - a)²

here,

f(x) = x³ – a³

⇒f(x) = (x-a)(x² + a² + xa)

g(x) = (x - a)²

⇒g(x) = (x-a)(x-a)

∴ LCM = (x-a) (x² + a² + xa) (x-a) = (x-a)²(x² + a² + xa)

∴ Correct Option is - Option (C)

Question 11.

The LCM of a^k,a^{k + 3}, a^{k + 5} where is
A. a ^{k + 9}

B. a^k

C. a^{k + 6}

D. a^{k + 5}

Answer:

Given three polynomials: a^k , a^{k + 3} and a^{k + 5}

Required: To find LCM of the given two polynomials

Let f(x) = a^k , g(x) = a^{k + 3} and h(x) = a^{k + 5}

here,

f(x) = a^k

⇒f(x) = a^k

g(x) = a^{k + 3}

⇒g(x) = a^k ×a³

h(x) = a^{k + 5}

⇒h(x) = a^k ×a^{3 + 2} = a^k×a³×a²

∴ LCM = a^k×a³×a² = a^{k + 5}

∴ Correct Option is - Option (D)

Question 12.

The lowest form of the rational expression  is
A. 

B. 

C. 

D. 

Answer:

Given: Rational Expression: 

Required: The lowest form of the given Rational Expression.

Let f(x) = 

⇒ f(x) =  (∵ factorization)

⇒ f(x) = 

⇒ f(x) = 

⇒ f(x) = 

∴ The lowest form of the given rational expression is:

∴ Correct Option is - Option (B)

Question 13.

If and are the two rational expressions, then their product is
A. 

B. 

C. 

D. 

Answer:

Given: Two rational expressions: and 

Required: Product of the given two rational numbers

Let f(x) =  and g(x) = 

Now, f(x)×g(x) = 

⇒ f(x)×g(x) = 

⇒ f(x)×g(x) = 

∴ Product of the given rational expressions is:

∴ Correct Option is - Option (A)

Question 14.

On dividing by is equal to
A. (x – 5) (x – 3)

B. (x – 5) (x + 3)

C. (x + 5) (x – 3)

D. (x + 5) (x + 3)

Answer:

Given: Two polynomials and 

Required: Divide by 

Let f(x) =  and g(x) = 

Now, 

⇒ 

⇒ 

⇒  (∵ a²–b² = (a–b)(a + b))

⇒ 

∴ Quotient when f(x) is divided g(x) we get is: (x-5)(x-3)

∴ Correct Option is - Option (A)

Question 15.

If is added with , then the new expression is
A. a² + ab + b²

B. a² – ab + b²

C. a³ + b³

D. a³ – b³

Answer:

Given: Two polynomials: and 

Required: Two add the given two polynomials

Let f(x) = and g(x) = 

Now, f(x) + g(x) =  + 

⇒ f(x) + g(x) = 

⇒ f(x) + g(x) = 

⇒ f(x) + g(x) = 

⇒ f(x) + g(x) = 

⇒ f(x) + g(x) =  (∵ a³–b³ = (a–b)(a² + ab + b²))

∴ f(x) + g(x) = a² + ab + b²

That is, the sum of given two polynomials is a² + ab + b²

∴ Correct Option is - Option (A)

Question 16.

The square root of 49 (x² – 2x + y²)² is
A. 7 |x – y|

B. 7(x + y) (x – y)

C. 7(x + y)²

D. 7(x – y)²

Answer:

Given: A polynomial: 49 

Required: to find the square root of the given polynomial

Let f(x) = 49 

Now, 

⇒  (∵ (x–y)² = x²-2xy + y²)

⇒ 

⇒ 

∴ Square root of the given polynomial is:

∴ Correct Option is - Option (D)

Question 17.

The square root of x² + y² + z² – 2xy + 2yz – 2zx
A. |x + y – z|

B. |x – y + z|

C. |x + y + z|

D. |x – y – z|

Answer:

Given: A polynomial: 

Required: to find the square root of the given polynomial

Let f(x) = 

Now, 

⇒  (∵ (x–y–z)² = x² + y² + z²-2xy + 2yz-2zx)

⇒ 

∴ Square root of the given polynomial is:

∴ Correct Option is - Option (D)

Question 18.

The square root of 121 x⁴y⁸z⁶ (l – m)² is
A. 11x²y⁴z⁴|l – m|

B. 11x⁴y⁴|z³(l – m)

C. 11x²y⁴z⁶|(l – m)|

D. 11x²y⁴|z³(l – m)|

Answer:

Given: A polynomial: 121 

Required: to find the square root of the given polynomial

Let f(x) = 121 

Now, 

⇒ 

⇒ 

∴ Square root of the given polynomial is:

∴ Correct Option is - Option (D)

Question 19.

If ax² + bx + c = 0 has equal roots, then c is equal
A. 

B. 

C. 

D. 

Answer:

Given: The quadratic equation has equal roots

Here,

b²–4ac = 0 (∵ The quadratic equation has equal roots)

⇒ b²–4ac = 0

⇒ b² = 4ac

⇒ 4ac = b²

⇒ c = 

∴ The value of c = 

∴ Correct Option is - Option (B)

Question 20.

If x² + 5kx + 16 = 0 has no real roots, then
A. 

B. 

C. 

D. 

Answer:

Given: The quadratic equation  and has no real roots.

Required: To find the value of k

Here,

b²–4ac<0 (∵ Quadratic equation has no real roots)

⇒ (5k)²–4(1)(16)<0

⇒ 25k²<64

⇒ k²<

Applying Square root on both sides

⇒ √k² < 

⇒ 

∴ The value of k is 

∴ Correct Option is - Option (C)

Question 21.

A quadratic equation whose one root is 3 is
A. x² – 6x – 5 = 0

B. x² + 6x – 5 = 0

C. x² – 5x – 6 = 0

D. x² – 5x + 6 = 0

Answer:

Given: One of the root of the Quadratic Equation that is 3

Required: To find the Quadratic equation, which satisfies the given root.

Now,

We substitute the zero in every equation given in the options

∴ Case (i): x²–6x-5 = 0

⇒ (3)²–6(3)–5 = 0

⇒ 9–18–5 = –14≠0

Case (ii): x² + 6x–5 = 0

⇒ (3)² + 6(3)–5 = 0

⇒ 9 + 18–5 = 22≠0

Case(iii): x²–5x–6 = 0

⇒ (3)²–5(3)–6 = 0

⇒ 9–15–6 = –12≠0

Case(iv): x²–5x + 6 = 0

⇒ (3)²–5(3) + 6 = 0

⇒ 9–15 + 6 = 0

∴ The Quadratic equation with one root as 3 is x²–5x–6 = 0

∴ Correct Option is - Option (D)

Question 22.

The common root of the equation x² – bx + c = 0 and x² + bx – a = 0 is
A. 

B. 

C. 

D. 

Answer:

Given: Two Quadratic Equations and 

Required: to find the common root of the given Quadratic Equations

Let the common root be α

Α is the root of x²–bx + c = 0

∴ x²–bx + c = 0

⇒ α²–bα + c = 0 -eq(1)

Also, α is the root of x² + bx–a = 0

∴ α² + bα–a = 0 -eq(2)

Here, eq(1) = eq(2)

∴ α² + bα–a = α²–bα + c

⇒ 2bα = c + a

⇒ α = 

∴ The common root is α = 

∴ Correct Option is - Option(A)

Question 23.

If α, β are the roots of ax² + bx + c = 0 a ≠ 0, then the wrong statement is
A. 

B. 

C. 

D. 

Answer:

Given: are the roots of 

Required: To find the wrong statement in the given options

We know that sum of roots is given by :

∴ 

We, also know that product of roots is given by: 

∴ 

Now, 

⇒ 

⇒ 

⇒ 

⇒ 

Now, 

∴ We can clearly see in the option that  is wrong.

∴ Correct Option is - Option (C)

Question 24.

If α and β are the roots of ax² + bx + c = 0, then one of the quadratic equations whose roots are is
A. ax² + bx + c = 0

B. bx² + ax + c = 0

C. cx² + bx + a = 0

D. cx² + ax + b = 0

Answer:

Given: and  are roots of 

Required:- Quadratic equation with roots and 

Sum of roots of given quadratic equation = 

∴  =  -eq(1)

Product of roots of given quadratic equation = 

∴  =  -eq(2)

Sum of roots of required quadratic equation = 

Product of roots of required quadratic equation = 

Here,

Dividing eq(1) by eq(2) we get,

∴ Sum of roots of the required quadratic equation = 

Again by making the reciprocal of eq(2), we get

∴ Product of roots of the required quadratic equation = 

We know that, when roots of the quadratic equation are known, we can calculate the quadratic equation as:

x²-(sum of roots)x + (product of roots) = 0

∴ Required quadratic equation: x² –() + () = 0

⇒  = 0

⇒ cx² + bx + a = 0

∴ Required quadratic equation is: cx² + bx + a = 0

∴ Correct option is -Option(C)

Question 25.

Let b = a + c. Then the equation ax² + bx + c = 0 has equal roots, if
A. a = c

B. a = - c

C. a = 2 c

D. a = -2c

Answer:

Given: b = a + c and the quadratic equation has equal roots

Here,

b²–4ac = 0 (∵ The quadratic equation has equal roots)

⇒ b²–4ac = 0

⇒ (a + c)²–4ac = 0 (∵ b = a + c)

⇒ a² + 2ac + c²–4ac = 0

⇒ a²–2ac + c² = 0

⇒ (a–c)² = 0

Applying sq.rt on both sides

⇒ √(a–c)² = √0

⇒ a–c = 0

⇒ a = c

∴ a = c

∴ Correct option is -Option (A)

---

Exercise 3.2

Question 1.

Solve the following systems of equation using cross multiplication method.

3x + 4y = 24, 20x – 11y = 47

Answer:

The given system of equations is

3x + 4y – 24 = 0 and

20x – 11y – 47 = 0

For cross multiplication method, we write the coefficients as

Hence, we get  =  = 

⇒  =  = 

⇒  =  = 

⇒ x =  = 4

⇒ y =  = 3

∴ (4, 3) is the solution to the given system.

Question 2.

Solve the following systems of equation using cross multiplication method.

0.5x + 0.8y = 0.44, 0.8x + 0.6y = 0.5

Answer:

The given equations are

0.5x + 0.8y – 0.44 = 0

0.8x +0.6y – 0.5 = 0

For cross multiplication method, we write the coefficients as

Hence, we get  =  = 

⇒  =  = 

⇒  =  = 

⇒ x =  = 0.4

⇒ y =  = 0.3

∴ (0.4, 0.3) is the solution to the given system.

Question 3.

Solve the following systems of equation using cross multiplication method.

Answer:

The given equations are

 –  = – 2 … (1)

 +  =  … (2)

By taking LCM,

(1) becomes 9x – 10y + 12 = 0

(2) becomes 2x + 3y – 13 = 0

For cross multiplication method, we write the coefficients as

Hence, we get  =  = 

⇒  =  = 

⇒  =  = 

⇒ x =  = 2

⇒ y =  = 3

∴ (2, 3) is the solution to the given system.

Question 4.

Solve the following systems of equation using cross multiplication method.

Answer:

The given equations are

 –  = – 2 … (1)

 +  = 13 … (2)

Let a =  and b =.

⇒ 5a – 4b + 2 = 0 … (3)

⇒ 2a + 3b – 13 = 0 … (4)

For cross multiplication method, we write the coefficients as

Hence, we get  =  = 

⇒  =  = 

⇒  =  = 

⇒ a =  = 2

⇒ b =  = 3

When a = 2, = 2. Thus, x = 1/2.

When b = 3,  = 3. Thus, y =.

∴ (,) is the solution to the given system.

Question 5.

Formulate the following problems as a pair of equations, and hence find their solutions:

One number is greater than thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, find the numbers.

Answer:

Let x be the greater number and y be the smaller number.

First condition is x = 3y + 2

Equation is x – 3y – 2 = 0 … (1)

Second condition is x = 4y – 5

Equation is x – 4y + 5 = 0 … (2)

For cross multiplication method, we write the coefficients as

Hence, we get  =  = 

⇒  =  = 

⇒  =  = 

⇒ x =  = 23

⇒ y =  = 7

∴ The numbers are 23 and 7.

Question 6.

Formulate the following problems as a pair of equations, and hence find their solutions:

The ratio of income of two persons is 9: 7 and the ratio of their expenditure is 4: 3. If each of them manages to save ₹ 2000 per month, find their monthly income.

Answer:

Let the incomes be x and expenditure be y.

We know that savings = income – expenditure

First condition is

9x – 4y = 2000

Second condition is

7x – 3y = 2000

For cross multiplication method, we write the coefficients as

Hence, we get  =  = 

⇒  =  = 

⇒  =  = 

⇒ x = 2000

⇒ y = 4000

∴ Income of first person = 9x = 9 × 2000 = Rs. 18, 000

Income of second person = 7x = 7 × 2000 = Rs. 14, 000

Question 7.

Formulate the following problems as a pair of equations, and hence find their solutions:

A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.

Answer:

Let x denote the digit in the tenth place and y denote the digit in the unit place. So the number may be written as 10x + y.

When digits are reversed, the number becomes 10y + x.

First condition is

10x + y = 7 (x + y)

⇒ 10x + y – 7x – 7y = 0

⇒ 3x – 6y = 0

⇒ x – 2y = 0

Second condition is

10y + x = (10x + y) – 18

⇒ 10y + x – 10x – y + 18 = 0

⇒ – 9x + 9y + 18 = 0

⇒ – x + y + 2 = 0

For cross multiplication method, we write the coefficients as

Hence, we get  =  = 

⇒  =  = 

⇒  =  = 

⇒ x =  = 4

⇒ y =  = 2

∴ The number is 10x + y = 10 (4) + 2 = 40 + 2 = 42.

Question 8.

Formulate the following problems as a pair of equations, and hence find their solutions:

Three chairs and two tables cost ₹ 700 and five chairs and three tables cost ₹1100. What is the total cost of 2 chairs and 3 tables?

Answer:

Let chairs be x and tables be y.

Then the equations are

3x + 2y = 700 i.e. 3x + 2y – 700 = 0

5x + 3y = 1100 i.e. 5x + 3y – 1100 = 0

For cross multiplication method, we write the coefficients as

Hence, we get  =  = 

⇒  =  = 

⇒  =  = 

⇒ x =  = 100 (Cost of one chair)

⇒ y =  = 200 (Cost of one table)

Now, total cost of two chairs and three tables,

2x + 3y = 2 (100) + 3 (200) = 200 + 600 = Rs. 800

Question 9.

Formulate the following problems as a pair of equations, and hence find their solutions:

In a rectangle, if the length is increased and the breadth is reduced each by 2 cm then the area is reduced by 28 cm². If the length is reduced by 1 cm and the breadth increased by 2 cm , then the area increases by 33 cm². Find the area of the rectangle.

Answer:

Let length be l and breadth be b.

Then the first condition is

(l + 2) (b – 2) = lb – 28

⇒ lb – 2l + 2b – 4 – lb + 28 = 0

⇒ – 2l + 2b + 24 = 0

⇒ – l + b + 12 = 0

The second condition is

(l – 1) (b + 2) = lb + 33

⇒ lb + 2l – b – 2 – lb – 33 = 0

⇒ 2l – b – 35 = 0

For cross multiplication method, we write the coefficients as

Hence, we get  =  = 

⇒  =  = 

⇒  =  = 

⇒ l =  = 23

⇒ b =  = 11

∴ Area of rectangle= l × b = 23 × 11 = 253 cm²

Question 10.

Formulate the following problems as a pair of equations, and hence find their solutions:

A train travelled a certain distance at a uniform speed. If the train had been 6 km/hr faster, it would have taken 4 hours less than the scheduled time. If the train were slower by 6 km/hr, then it would have taken 6 hours more than the scheduled time. Find the distance covered by the train.

Answer:

Let the speed be x km/hr and distance travelled be y km.

We know that distance = speed × time

Scheduled time to cover distance = y/x hr

Then the first condition is

⇒  =  – 4

⇒  = 

⇒ xy = (x + 6) (y – 4x)

⇒ xy = xy – 4x² + 6y – 24x

⇒ 4x² – 6y + 24x = 0

⇒ 2x² – 3y + 12x = 0 i.e. 12x – 3y + 2x² = 0

Second condition is

⇒  =  + 6

⇒  = 

⇒ xy = (x – 6) (y + 6x)

⇒ xy = xy + 6x² – 6y – 36x

⇒ 6x² – 6y – 36x = 0

⇒ x² – y – 6x = 0 i.e. – 6x – y + x² = 0

For cross multiplication method, we write the coefficients as

Hence, we get  =  = 

⇒  =  = 

⇒  =  = 

⇒  = 

⇒ – 30x = – x²

⇒ 30km/hr = x

Now,  = 

⇒  = 

⇒ y = 24 × 30 = 720 km

∴ The distance covered by train = 720 km

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Exercise 3.3

Question 1.

Find the zeros of the following quadratic polynomials and verify the basic relationships between the zeros and the coefficients.

x² – 2x – 8

Answer:

Let f(x) = x² – 2x – 8

To find out zeros of the given polynomial.

We put f(x) = 0

⇒ x² – 2x – 8 = 0

To find out roots of this polynomial we use splitting of middle term method.

According to this method we need to find two numbers whose sum is – 2 and product is 8.

∴ x² – (4 – 2)x – 8 = 0

⇒ x² – 4x + 2x – 8 = 0

⇒ x(x – 4) + 2(x – 4) = 0

⇒ (x + 2)(x – 4) = 0

∴ x = – 2 and x = 4.

⇒ Our zeros are α = – 2 and β = 4.

⇒ sum of zeros = α + β = – 2 + 4 = 2.

⇒ Product of zeros = αβ = ( – 2) × 4 = – 8.

⇒ Comparing f(x) = x² – 2x – 8 with standard equation ax² + bx + c = 0.

We get, a = 1, b = – 2 and c = – 8

We can verify,

⇒ Sum of zeros = 

i.e. α + β = 

∴ α + β = 2

⇒ Product of zeros = 

αβ = 

αβ = – 8.

Hence, relationship between zeros and coefficient is verified.

Question 2.

Find the zeros of the following quadratic polynomials and verify the basic relationships between the zeros and the coefficients.

4x² – 4x + 1

Answer:

Let f(x) = 4x² – 4x + 1

To find out zeros of the given polynomial.

We put f(x) = 0

⇒ 4x² – 4x + 1 = 0

To find out roots of this polynomial we use splitting of middle term method.

According to this method we need to find two numbers whose sum is – 4 and product is 4.

∴ 4x² – (2 + 2)x + 1 = 0

⇒ 4x² – 2x – 2x + 1 = 0

⇒ 2x(2x – 1) – 1(2x – 1) = 0

⇒ (2x – 1)(2x – 1) = 0

∴ 2x – 1 = 0

∴ x =  .

Again, 2x – 1 = 0

∴ x = 

⇒ Our zeros are α =  and β = .

⇒ sum of zeros = α + β =  +  = 1.

⇒ Product of zeros = αβ = .

Now, Comparing f(x) = 4x² – 4x + 1 with standard equation ax² + bx + c = 0.

We get, a = 4, b = – 4 and c = 1

We can verify,

⇒ Sum of zeros = 

i.e. α + β = 

∴ α + β = 1

⇒ Product of zeros = 

αβ = 

Hence, relationship between zeros and coefficient is verified.

Question 3.

Find the zeros of the following quadratic polynomials and verify the basic relationships between the zeros and the coefficients.

6x² – 3 – 7x

Answer:

Let f(x) = 6x² – 3 – 7x

Arranging equation in proper form.

Now, f(x) = 6x² – 7x – 3

To find out zeros of the given polynomial.

We put f(x) = 0

⇒ 6x² – 7x – 3 = 0

To find out roots of this polynomial we use splitting of middle term method.

According to this method we need to find two numbers whose sum is – 7 and product is – 18.

∴ 6x² – (9 – 2)x – 3 = 0

⇒6x² – 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (3x + 1)(2x – 3) = 0

∴ 3x + 1 = 0

∴ x =  .

Again, 2x – 3 = 0

∴ x = 

⇒ Our zeros are α =  and β = .

⇒ sum of zeros = α + β =  + 

⇒ sum of zeros = α + β = 

⇒ Product of zeros = αβ = .

Now, Comparing f(x) = 6x² – 7x – 3 with standard equation ax² + bx + c.

We get, a = 6, b = – 7 and c = – 3.

We can verify,

⇒ Sum of zeros = 

i.e. α + β = 

∴ α + β = 

⇒ Product of zeros = 

αβ = 

Hence, relationship between zeros and coefficient is verified.

Question 4.

Find the zeros of the following quadratic polynomials and verify the basic relationships between the zeros and the coefficients.

4x² + 8x

Answer:

Let f(x) = 4x² + 8x

Arranging equation in proper form.

Now, f(x) = 4x² + 8x + 0

To find out zeros of the given polynomial.

We put f(x) = 0

⇒ f(x) = 4x² + 8x + 0 = 0

∴4x² + 8x = 0

⇒4x(x + 2) = 0

Now, 4x = 0

∴ x = 0

When, (x + 2) = 0

Then, x = – 2

⇒ Our zeros are α = 0 and β = – 2.

⇒ sum of zeros = α + β = 0 + ( – 2)

⇒ sum of zeros = α + β = – 2.

⇒ Product of zeros = αβ = 0 × ( – 2) = 0.

Now, Comparing f(x) = 4x² + 8x + 0 with standard equation ax² + bx + c.

We get, a = 4, b = 8 and c = 0.

We can verify,

⇒ Sum of zeros = 

i.e. α + β = 

∴ α + β = – 2

⇒ Product of zeros = 

αβ = 

Hence, relationship between zeros and coefficient is verified.

Question 5.

Find the zeros of the following quadratic polynomials and verify the basic relationships between the zeros and the coefficients.

x² – 15

Answer:

Let f(x) = x² – 15

Arranging equation in proper form.

Now, f(x) = x² + 0x – 15

To find out zeros of the given polynomial.

We put f(x) = 0

⇒ x² – 15 = 0

∴ x² –  = 0

So, (x + )(x – ) = 0

When, (x + ) = 0

Then, x = –  .

When. (x – ) = 0

Then, x = 

⇒ Our zeros are α = –  and β = .

⇒ sum of zeros = α + β = –  + 

⇒ sum of zeros = α + β = 0

⇒ Product of zeros = αβ = –  = – 15

Now, Comparing f(x) = x² + 0x – 15 with standard equation ax² + bx + c.

We get, a = 1, b = 0 and c = – 15.

We can verify,

⇒ Sum of zeros = 

i.e. α + β = 

∴ α + β = 0

⇒ Product of zeros = 

αβ = 

Hence, relationship between zeros and coefficient is verified.

Question 6.

Find the zeros of the following quadratic polynomials and verify the basic relationships between the zeros and the coefficients.

3x² – 5x + 2

Answer:

Let f(x) = 3x² – 5x + 2.

To find out zeros of the given polynomial.

We put f(x) = 0

⇒ 3x² – 5x + 2 = 0

To find out roots of this polynomial we use splitting of middle term method.

According to this method we need to find two numbers whose sum is – 5 and product is 6.

∴ 3x² – (3 + 2)x + 2 = 0

⇒3x² – 3x – 2x + 2 = 0

⇒ 3x(x – 1) – 2(x – 1) = 0

⇒ (3x – 2)(x – 1) = 0

When, 3x – 2 = 0

Then, x =  .

Again when, x – 1 = 0

∴ then, x = 1

⇒ Our zeros are α =  and β = 1.

⇒ sum of zeros = α + β =  + 1

⇒ sum of zeros = α + β = 

⇒ Product of zeros = αβ = .

Now, Comparing f(x) = 3x² – 5x + 2 with standard equation ax² + bx + c.

We get, a = 3, b = – 5 and c = 2.

We can verify,

⇒ Sum of zeros = 

i.e. α + β = 

∴ α + β = 

⇒ Product of zeros = 

αβ = 

Hence, relationship between zeros and coefficient is verified.

Question 7.

Find the zeros of the following quadratic polynomials and verify the basic relationships between the zeros and the coefficients.

2x² – 2√2 x + 1

Answer:

Let f(x) = 2x² – 2x + 1

To find out zeros of the given polynomial.

We put f(x) = 0

⇒ 2x² – 2x + 1 = 0

To find out roots of this polynomial we use splitting of middle term method.

According to this method we need to find two numbers whose sum is – 2 and product is 2.

∴ 2x² – ()x + 1 = 0

⇒2x² – x – x + 1 = 0

⇒ x(x – 1) – 1(x – 1) = 0

⇒ (x – 1)(x – 1) = 0

⇒ (x – 1)² = 0

∴ x = ,

⇒ Our zeros are α =  and β = .

⇒ sum of zeros = α + β =  + 

⇒ sum of zeros = α + β = 

⇒ Product of zeros = αβ = .

Now, Comparing f(x) = 2x² – 2x + 1 with standard equation ax² + bx + c.

We get, a = 2, b = – 2 and c = 1.

We can verify,

⇒ Sum of zeros = 

i.e. α + β = 

∴ α + β = 

⇒ Product of zeros = 

αβ = 

Hence, relationship between zeros and coefficient is verified.

Question 8.

Find the zeros of the following quadratic polynomials and verify the basic relationships between the zeros and the coefficients.

x² + 2x – 143

Answer:

Let f(x) = x² + 2x – 143

To find out zeros of the given polynomial.

We put f(x) = 0

⇒ x² + 2x – 143 = 0

To find out roots of this polynomial we use splitting of middle term method.

According to this method we need to find two numbers whose sum is 2 and product is – 143.

∴ x² + (13 – 11)x – 143 = 0

⇒x² + 13x – 11x – 143 = 0

⇒ x (x + 13) – 11(x + 13) = 0

⇒ (x – 11) (x + 13) = 0

∴ When, (x – 11) = 0

∴ Then, x = 11.

Again, When, (x + 13) = 0

∴ Then, x = – 13

⇒ Our zeros are α = 11 and β = – 13.

⇒ sum of zeros = α + β = 11 + ( – 13)

⇒ sum of zeros = α + β = – 2

⇒ Product of zeros = αβ = 11 × ( – 13) = – 143

Now, Comparing f(x) = x² + 2x – 143 with standard equation ax² + bx + c.

We get, a = 1, b = 2 and c = – 143.

We can verify,

⇒ Sum of zeros = 

i.e. α + β = 

∴ α + β = – 2

⇒ Product of zeros = 

αβ = – 143

Hence, relationship between zeros and coefficient is verified.

Question 9.

Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.

3, 1

Answer:

Formula for quadratic equation is,

x² – (sum of roots) x + Product of roots = 0

Given, sum of roots = 3 and product of roots = 1

∴ Quadratic equation is,

⇒ x² – 3x + 1 = 0

Hence, Quadratic equation is x² – 3x + 1 = 0.

Question 10.

Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.

2, 4

Answer:

Formula for quadratic equation is,

x² – (sum of roots) x + Product of roots = 0

Given, sum of roots = 2 and product of roots = 4

∴ Quadratic equation is,

⇒ x² – 2x + 4 = 0

Hence, Quadratic equation is x² – 2x + 4 = 0.

Question 11.

Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.

0, 4

Answer:

Formula for quadratic equation is,

x² – (sum of roots) x + Product of roots = 0

Given, sum of roots = 0 and product of roots = 4

∴ Quadratic equation is,

⇒ x² – 0x + 4 = 0

⇒ x² + 4 = 0

Hence, Quadratic equation is x² + 4 = 0.

Question 12.

Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.

Answer:

Formula for quadratic equation is,

x² – (sum of roots) x + Product of roots = 0

Given, sum of roots =  and product of roots = 

∴ Quadratic equation is,

⇒ x² –  x +  = 0

Hence, Quadratic equation is x² –  x +  = 0.

Question 13.

Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.

Answer:

Formula for quadratic equation is,

x² – (sum of roots) x + Product of roots = 0

Given, sum of roots =  and product of roots = 1

∴ Quadratic equation is,

⇒ x² – x + 1 = 0

⇒ 

Hence, Quadratic equation is 

Question 14.

Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.

Answer:

formula for quadratic equation is,

x² – (sum of roots) x + Product of roots = 0

Given, sum of roots =  and product of roots = – 4

∴ Quadratic equation is,

⇒ x² – x – 4 = 0

⇒ 

Hence, Quadratic equation is .

Question 15.

Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.

Answer:

Formula for quadratic equation is,

x² – (sum of roots) x + Product of roots = 0

Given, sum of roots =  and product of roots = 

∴ Quadratic equation is,

⇒ x² – x –  = 0

⇒  = 0

Hence, Quadratic equation is  = 0.

Question 16.

Find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.

Answer:

Formula for quadratic equation is,

x² – (sum of roots) x + Product of roots = 0

Given, sum of roots =  and product of roots = 2

∴ Quadratic equation is,

⇒ x² – x + 2 = 0

Hence, Quadratic equation is x² – x + 2 = 0.

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Exercise 3.4

Question 1.

Find the quotient and remainder using synthetic division.

x³ + x² – 3x + 5) ÷ ( x – 1)

Answer:

Let p(x) = x³ + x² – 3x + 5 be the dividend. Arranging p(x) according to the descending powers of x.

p(x) = x³ + x² – 3x + 5

Divisor, q(x) = x – 1

⇒ To find out Zero of the divisor –

q(x) = 0

x – 1 = 0

x = 1

So, zero of divisor is 1.

⇒ p(x) = x³ + x² – 3x + 5

Put zero for the first entry in the second row.

∴ Quotient = x² + 2x – 1

Hence, when p(x) is divided by (x – 1) the quotient is x² + 2x – 1 and remainder is 4.

Question 2.

Find the quotient and remainder using synthetic division.

(3x³ – 2x² + 7x – 5) ÷ ( x + 3)

Answer:

Let p(x) = 3x³ – 2x² + 7x – 5 be the dividend and arranging p(x) according to the descending powers of x.

Divisor, q(x) = x + 3

⇒ To find out Zero of the divisor –

q(x) = 0

x + 3 = 0

x = – 3

So, zero of divisor is – 3.

And, p(x) = 3x³ – 2x² + 7x – 5

Put zero for the first entry in the second row.

∴ Quotient = 3x² – 11x + 40

Hence, when p(x) is divided by (x – 1) the quotient is 3x² – 11x + 40 and remainder is – 125.

Question 3.

Find the quotient and remainder using synthetic division.

(3x³ + 4x² – 10x + 6) ÷ ( 3x – 2)

Answer:

Let p(x) = 3x³ + 4x² – 10x + 6 be the dividend and arranging p(x) according to the descending powers of x.

Divisor, q(x) = 3x – 2

⇒ To find out Zero of the divisor –

q(x) = 0

3x – 2= 0

x = 

So, zero of divisor is .

And, p(x) = 3x³ + 4x² – 10x + 6

Put zero for the first entry in the second row.

∵ p(x) = (Quotient)×q(x) + remainder.

So, 3x³ + 4x² – 10x + 6 = (x – )(3x² + 6x – 6) + 2

= (3x – 2)(3x² + 6x – 6) + 2

Thus, the Quotient = (3x² + 6x – 6)= x² + 2x – 2 and remainder is 2.

Hence, when p(x) is divided by (3x – 2) the quotient is x² + 2x – 2 and remainder is 2.

Question 4.

Find the quotient and remainder using synthetic division.

(3x³ – 4x² – 5) ÷ (3x + 1)

Answer:

Let p(x) = 3x³ – 4x² – 5 be the dividend. Arranging p(x) according to the descending powers of x and insert zero for missing term.

p(x) = 3x³ – 4x² + 0x – 5

Divisor, q(x) = 3x + 1

⇒ To find out Zero of the divisor –

q(x) = 0

3x + 1 = 0

x = 

zero of divisor is .

And, p(x) = 3x³ – 4x² + 0x – 5

Put zero for the first entry in the 2^nd row.

∵ p(x) = (Quotient)×q(x) + remainder.

So, 3x³ – 4x² – 5 = (x + )(3x² – 5x + ) + ()

= (3x + 1)(3x² – 5x + ) 

Thus, the Quotient = (3x² – 5x + )= (x² – x + ) and remainder is .

Hence, when p(x) is divided by (3x + 1) the quotient is (x² – x + ) and remainder is .

Question 5.

Find the quotient and remainder using synthetic division.

(8x⁴ – 2x² + 6x + 5) ÷ (4x + 1)

Answer:

Let p(x) = 8x⁴ – 2x² + 6x – 5 be the dividend. Arranging p(x) according to the descending powers of x and insert zero for missing term.

p(x) = 8x⁴ + 0x³ – 2x² + 6x – 5

Divisor, q(x) = 4x + 1

⇒ To find out Zero of the divisor –

q(x) = 0

4x + 1 = 0

x = 

zero of divisor is .

And, p(x) = 8x⁴ + 0x³ – 2x² + 6x – 5

Put zero for the first entry in the 2^nd row.

∵ p(x) = (Quotient)×q(x) + remainder.

So, 8x⁴ – 2x² + 6x – 5 = (x + )( 8x³ – 2x² – x + ) + ()

= (4x + 1)(8x³ – 2x² – x + ) 

Thus, the Quotient = (8x³ – 2x² – x + )= (2x³ – x² – x + ) and remainder is .

Hence, when p(x) is divided by (4x + 1) the quotient is (2x³ – x² – x + ) and remainder is .

Question 6.

Find the quotient and remainder using synthetic division.

(2x⁴ – 7x³ – 13x² + 63x – 48) ÷ (2x – 1)

Answer:

Let p(x) = 2x⁴ – 7x³ – 13x² + 63x – 48 be the dividend. Arranging p(x) according to the according descending powers of x.

p(x) = 2x⁴ – 7x³ – 13x² + 63x – 48

Divisor, q(x) = 2x – 1

⇒ To find out Zero of the divisor –

q(x) = 0

2x – 1 = 0

x = 

zero of divisor is .

And, p(x) = 2x⁴ – 7x³ – 13x² + 63x – 48

Put zero for the first entry in the 2^nd row.

∴ Quotient = 2x³ – 6x² – 16x + 55

∵ p(x) = (Quotient)×q(x) + remainder.

So, 2x⁴ – 7x³ – 13x² + 63x – 48

= (x – )( 8x³ – 2x² – x + ) + ()

= (2x – 1)(2x³ – 6x² – 16x + 55) 

Thus, the Quotient = (2x³ – 6x² – 16x + 55) = (x³ – 3x² – 8x + ) and remainder is .

Hence, when p(x) is divided by (2x – 1) the quotient is (x³ – 3x² – 8x + ) and remainder is .

Question 7.

If the quotient on dividing x⁴ + 10x³ + 35x² + 50x + 29 by x + 4 is x³ – ax² + bx + 6, then find a, b and also the remainder.

Answer:

Let p(x) = x⁴ + 10x³ + 35x² + 50x + 29 be the dividend. Arranging p(x) according to the descending powers of x.

p(x) = x⁴ + 10x³ + 35x² + 50x + 29

Divisor, q(x) = x + 4

⇒ To find out Zero of the divisor –

q(x) = 0

x + 4 = 0

x = – 4

zero of divisor is – 4.

And, p(x) = x⁴ + 10x³ + 35x² + 50x + 29

Put zero for the first entry in the 2^nd row.

∴ Quotient = x³ + 6x² + 11x + 6

Hence, when p(x) is divided by (x + 4) the quotient is x³ + 6x² + 11x + 6 and remainder is 5.

Comparing x³ + 6x² + 11x + 6 with x³ – ax² + bx + 6 we get,

a = – 6 and b = 11.

Question 8.

If the quotient on dividing, 8x⁴ – 2x² + 6x – 7 by 2x + 1 is 4x³ + px² – qx + 3,then find p, q and also the remainder.

Answer:

Let p(x) = 8x⁴ – 2x² + 6x – 7 be the dividend. Arranging p(x) according to the descending powers of x and write zero in place of missing term.

p(x) = 8x⁴ + 0x³ – 2x² + 6x – 7

Divisor, q(x) = 2x + 1

⇒ To find out Zero of the divisor –

q(x) = 0

2x + 1 = 0

x = .

zero of divisor is .

And, p(x) = 8x⁴ + 0x³ – 2x² + 6x – 7

Put zero for the first entry in the 2^nd row.

∴ Quotient = 8x³ – 4x² + 0x + 6

Hence, when p(x) is divided by (2x + 1) the quotient is 8x³ – 4x² + 0x + 6

6 and remainder is – 10.

Comparing 8x³ – 4x² + 0x + 6 with 4x³ + px² – qx + 3 we get,

p = – 4 and q = 0.

---

Exercise 3.5

Question 1.

Factorize each of the following polynomials.

x³ – 2x² – 5x + 6

Answer:

Given,

x³ – 2x² – 5x + 6, put x = 1

then, 1 – 2 – 5 + 6 = 0, since,

this equation is divisible by (x – 1)

 according to the question,

x³ – 2x² – 5x + 6 = (x – 1)(x² – x – 6)

= (x – 1)[x² – 3x + 2x – 6]

= (x – 1)[x(x – 3) + 2(x – 3)]

= (x – 1)(x + 2)(x – 3)

Question 2.

Factorize each of the following polynomials.

4x³ – 7x + 3

Answer:

Given,

4x³ – 7x + 3,put x = 1

Then, 4 × 1 – 7 + 3 = 0.

since, this equation is divisible by (x – 1).

 according to the question,

4x³ – 7x + 3 = (x – 1)(4x² + 4x – 3)

= (x – 1)[4x² + 6x – 2x – 3]

= (x – 1)[2x(2x + 3) – 1(2x + 3)]

= (x – 1)(2x + 3)(2x – 1)

Question 3.

Factorize each of the following polynomials.

x³ – 23x² + 142x – 120

Answer:

Given,

x³ – 23x² + 142x – 120, put x = 1

then, 1 – 23 × 1 + 142 – 120 = – 143 + 143 = 0

since, this equation is divisible by (x – 1).

 according to the question,

x³ – 23x² + 142x – 120 = (x – 1)[x² – 22x + 120]

= (x – 1)[x² – 12x – 10x + 120]

= (x – 1)[x(x – 12) – 10(x – 12)]

= (x – 1)(x – 12)(x – 10).

Question 4.

Factorize each of the following polynomials.

4x³ – 5x² + 7x – 6

Answer:

Given,

4x³ – 5x² + 7x – 6,put x = 1

Then, 4 × 1 – 5 + 7 – 6 = 0

since, this equation is divisible by (x – 1).

 according to the question,

4x³ – 5x² + 7x – 6 = (x – 1)[4x² – x + 6]

Question 5.

Factorize each of the following polynomials.

x³ – 7x + 6

Answer:

Given,

x³ – 7x + 6,

put x = 1

then, 1 – 7 + 6 = 0

since, this equation is divisible by (x – 1).

 according to the question,

x³ – 7x + 6 = (x – 1)(x² + x – 6)

= (x – 1)[x² + 3x – 2x – 6]

= (x – 1)[x(x + 3) – 2(x + 3)]

= (x – 1)(x – 2)(x + 3)

Question 6.

Factorize each of the following polynomials.

x³ + 13x² + 32x + 20

Answer:

Given,

x³ + 13x² + 32x + 20,

put x = – 1

then, – 1 + 13 – 32 + 20 = 0

since, this equation is divisible by (x + 1).

 according to the question,

x³ + 13x² + 32x + 20 = (x + 1)(x² + 12x + 20)

= (x + 1)[x² + 10x + 2x + 20]

= (x + 1)[x(x + 10) + 2(x + 10)]

= (x + 1)(x + 2)(x + 10)

Question 7.

Factorize each of the following polynomials.

2x³ – 9x² + 7x + 6

Answer:

Given,

2x³ – 9x² + 7x + 6,

put x = 2

Then, 16 – 36 + 14 + 6 = 0

since, this equation is divisible by (x – 2).

 according to the question,

2x³ – 9x² + 7x + 6 = (x – 2)(2x² – 5x – 3)

= (x – 2)[2x² – 6x + x – 3]

= (x – 2)[2x(x – 3) + 1(x – 3)]

= (x – 2)(x – 3)(2x + 1)

Question 8.

Factorize each of the following polynomials.

x³ – 5x + 4

Answer:

Given,

x³ – 5x + 4,put x = 1

then,1 – 5 + 4 = 0

since, this equation is divisible by (x – 1).

 according to the question,

x³ – 5x + 4 = (x – 1)(x² + x – 4).

Question 9.

Factorize each of the following polynomials.

x³ – 10x² – x + 10

Answer:

Given,

x³ – 10x² – x + 10,put x = 1

then, 1 – 10 – 1 + 10 = 0

since, this equation is divisible by (x – 1).

 according to the question,

x³ – 10x² – x + 10 = (x – 1)(x² – 9x – 10)

= (x – 1)[x² – 10x + x – 10]

= (x – 1)[x(x – 10) + 1(x – 10)]

= (x – 1)(x + 1)(x – 10)

Question 10.

Factorize each of the following polynomials.

2x³ + 11x² – 7x – 6

Answer:

Given,

2x³ + 11x² – 7x – 6,

put x = 1

Then, 2 + 11 – 7 – 6 = 0

since, this equation is divisible by (x – 1).

 according to the question,

2x³ + 11x² – 7x – 6 = (x – 1)(2x² + 13x + 6)

= (x – 1)[2x² + 12x + x + 6]

= (x – 1)[2x(x + 6) + 1(x + 6)]

= (x – 1)(2x + 1)(x + 6)

Question 11.

Factorize each of the following polynomials.

x³ + x² + x – 14

Answer:

Given,

x³ + x² + x – 14,

put x = 2

then, 8 + 4 + 2 – 14 = 0

since, this equation is divisible by (x – 2).

 according to the question,

x³ + x² + x – 14 = (x – 2)(x² + 3x + 7).

Question 12.

Factorize each of the following polynomials.

x³ – 5x² – 2x + 24

Answer:

Given,

x³ – 5x² – 2x + 24,put x = – 2

then, – 8 – 20 + 4 + 24 = 0

since, this equation is divisible by (x + 2).

 according to the question,

x³ – 5x² – 2x + 24 = (x + 2)(x² – 7x + 12)

= (x + 2)[x² – 4x – 3x + 12]

= (x + 2)[x(x – 4) – 3(x – 4)]

= (x + 2)(x – 4)(x – 3)

---

Exercise 3.6

Question 1.

Find the greatest common divisor of

7x² yz⁴, 21x² y⁵ z³

Answer:

Given,

7x² yz⁴ = 7x² yz³ × z

21x² y⁵ z³ = 3 × 7x² y × y⁴ z³

Greatest common divisor = 7x²yz³

Question 2.

Find the greatest common divisor of

x²y, x³y, x²y²

Answer:

Given,

x² y = x × x × y

x³ y = x × x × x × y

x² y² = x × x × y × y

Greatest common divisor = x²y

Question 3.

Find the greatest common divisor of

25bc⁴ d³, 35b2c⁵, 45c³ d

Answer:

Given,

25bc⁴ d³ = 5 × 5 × b × c³ × c × d³

35b2c⁵ = 5 × 7 × b × 2 × c³ × c²

45c³ d = 5 × 3 × 3 × c³ × d

Greatest common divisor = 5c³

Question 4.

Find the greatest common divisor of

35x⁵ y³ z⁴, 49x² yz³, 14xy² z²

Answer:

Given,

35x⁵ y³ z⁴ = 5 × 7 × x × x⁴ × y² × y × z² × z²

49x² yz³ = 7 × 7 × x² × y × z × z²

14xy² z² = 2 × 7 × x × y × y × z²

Greatest common divisor = 7xyz²

Question 5.

Find the GCD of the following

x³ – x² + x – 1, x⁴ – 1

Answer:

x³ – x² + x – 1, put x = 1

Then,1 – 1 + 1 – 1 = 0

Since, this equation is divisible by x – 1

(x – 1)(x² + 1) = (x + 1)(x² + 1)

In second equation,

x⁴ – 1 = (x²)² – 1 = (x² – 1)(x² + 1) = (x + 1)(x – 1)(x² + 1)

[using a² – b² = (a – b)(a + b)]

Greatest common divisor = (x + 1)(x² + 1)

Question 6.

Find the GCD of the following

c² – d², – c(c – d)

Answer:

Given,

c² – d² = (c + d) × (c – d)

– c(c – d) = – c × (c – d)

Greatest common divisor = (c – d)

Question 7.

Find the GCD of the following

x⁴ – 27a³ x,(x – 3a)²

Answer:

Given,

x⁴ – 27a³ x = x[x³ – (3a)³]

= x[(x – 3a)(x² + 9a² + 3ax)]

= x(x – 3a)(x² + 9a² + 3ax)

(x – 3a)² = (x – 3a)(x – 3a)

Greatest common divisor = (x – 3a)

Question 8.

Find the GCD of the following

m² – 3m – 18, m² + 5m + 6

Answer:

Given,

m² – 3m – 18 = m² – 6m + 3m – 18 = m(m – 6) + 3(m – 6)

= (m + 3)(m – 6)

m² + 5m + 6 = m² + 3m + 2m + 6 = m(m + 3) + 2(m + 3)

= (m + 3)(m + 2)

Greatest common divisor = (m + 3)

Question 9.

Find the GCD of the following

x² + 14x + 33, x³ + 10x² – 11x

Answer:

Given,

X² + 14x + 33 = x² + 3x + 11x + 33 = x(x + 3) + 11(x + 3)

= (x + 11)(x + 3)

X³ + 10x² – 11x = x(x² + 10x – 11) = x(x² + 11x – x – 11)

= x[x(x + 11) – 1(x + 11)]

= x(x + 11)(x – 1)

Greatest common divisor = (x + 11)

Question 10.

Find the GCD of the following

x² + 3xy + 2y², x² + 5xy + 6y²

Answer:

Given,

X² + 3xy + 2y² = x² + xy + 2xy + 2y²

= x(x + y) + 2y(x + y)

= (x + 2y)(x + y)

x² + 5xy + 6y² = x² + 3xy + 2xy + 6y²

= x(x + 3y) + 2y(x + 3y)

= (x + 2y)(x + 3y)

Greatest common divisor = (x + 2y)

Question 11.

Find the GCD of the following

2x² – x – 1,4x² + 8x + 3

Answer:

Given,

2x² – x – 1 = 2x² – 2x + x – 1 = 2x(x – 1) + 1(x – 1)

= (2x + 1)(x – 1)

4x² + 8x + 3 = 4x² + 2x + 6x + 3 = 2x(x + 1) + 3(2x + 1)

= (2x + 1)(2x + 3)

Greatest common divisor = (2x + 1)

Question 12.

Find the GCD of the following

x² – x – 2,x² + x – 6,3x² – 13x + 14

Answer:

Given,

x² – x – 2 = x² – 2x + x – 2 = x(x – 2) + 1(x – 2) = (x + 1)(x – 2)

x² + x – 6 = x² + 3x – 2x – 6

= x(x + 3) – 2(x + 3) = (x – 2)(x + 3)

3x² – 13x + 14 = 3x² – 6x – 7x + 14 = 3x(x – 2) – 7(x – 2)

= (x – 2)(3x – 7)

Greatest common divisor = (x – 2)

Question 13.

Find the GCD of the following

24(6x⁴ – x³ – 2x²),20(2x⁶ + 3x⁵ + x⁴)

Answer:

Given,

24(6x⁴ – x³ – 2x²) = 2 × 2 × 2 × 3[x²(6x² – x – 2)]

= 2 × 2 × 2 × 3[x²(6x² – 4x + 3x – 2)]

= 2 × 2 × 2 × 3 × x² [2x(3x – 2) + 1(3x – 2)]

= 2 × 2 × 2 × 3 × x² (2x + 1)(3x – 2)

20(2x⁶ + 3x⁵ + x⁴) = 2 × 2 × 5[x⁴(2x² + 3x + 1)]

= 2 × 2 × 5 × x² × x²[2x(x + 1) + 1(x + 1)]

= 2 × 2 × 5 × x² × x²(2x + 1)(x + 1)

Greatest common divisor = 2 × 2 × x² × (2x + 1) = 4x²(2x + 1)

Question 14.

Find the GCD of the following

(a – 1)⁵ (a + 3)²,(a – 2)² (a – 1)³(a + 3)⁴

Answer:

Given,

(a – 1)⁵ (a + 3)² = (a – 1)³(a – 1)²(a + 3)²

(a – 2)²(a – 1)³(a + 3)⁴ = (a – 2)²(a – 1)³(a + 3)²(a + 3)²

Greatest common divisor = (a – 1)³(a + 3)²

Question 15.

Find the GCD of the following pairs of polynomials using division algorithm.

x³ – 9x² + 23x – 15, 4x² – 16x + 12

Answer:

x³ – 9x² + 23x – 15

put x = 1 in polynomials equations is 1 – 9(1) + 23 – 15 = 0

 This equation is divisible by (x – 1),

then using division algorithm method.

Since,

(x – 1)(x² – 8x + 15) = (x – 1)[x² – 5x – 3x + 15]

= (x – 1)[x(x – 5) – 3(x – 5)] = (x – 1)(x – 3)(x – 5)

4x² – 16x + 12 = 4x² – 12x – 4x + 12 = 4x(x – 3) – 4(x – 3)

= 4(x – 1)(x – 3)

Greatest common divisor = (x – 1)(x – 3) = x² – 4x + 3

Question 16.

Find the GCD of the following pairs of polynomials using division algorithm.

3x³ + 18x² + 33x + 18, 3x² + 13x + 10

Answer:

According to the question,

3x³ + 18x² + 33x + 18 then put x = – 1 in that equation

 put x = – 1

then, – 3 + 18 – 33 + 18 = 0

Then the equation is divisible by x + 1, using division algorithm method.

(x + 1)(3x² + 15x + 18) = (x + 1)[3x² + 9x + 6x + 18]

= (x + 1)[3x(x + 3) + 6(x + 3)]

= 3(x + 1)(x + 2)(x + 3)

3x² + 13x + 10 = 3x² + 3x + 10x + 10 = 3x(x + 1) + 10(x + 1)

= (x + 1)(3x + 10)

Greatest common divisor = (x + 1)

Question 17.

Find the GCD of the following pairs of polynomials using division algorithm.

2x³ + 2x² + 2x + 2, 6x³ + 12x² + 6x + 12

Answer:

2x³ + 2x² + 2x + 2 then put x = – 1

Since, – 2 + 2 – 2 + 2 = 0

Then the equation is divisible by x + 1, using division algorithm method.

(x + 1)(2x² + 2) = 2(x + 1)(x² + 1)

In equation second,

6x³ + 12x² + 6x + 12,put x = – 2

Then, – 48 + 48 – 12 + 12 = 0

Then the equation is divisible by x + 2, using division algorithm method.

(x + 2)(6x² + 6) = 2 × 3(x + 2)(x² + 1)

Greatest common divisor = 2(x² + 1)

Question 18.

Find the GCD of the following pairs of polynomials using division algorithm.

x³ – 3x² + 4x – 12, x⁴ + x³ + 4x² + 4x

Answer:

According to the question,

x³ – 3x² + 4x – 12,then put x = 3

then,27 – 27 + 12 – 12 = 0

Then the equation is divisible by x – 3, using division algorithm method.

(x – 3)(x² + 4) = (x – 3)(x² + 4)

In equation second,

x⁴ + x³ + 4x² + 4x,put x = – 1

then, 1 – 1 + 4 – 4 = 0,

Then the equation is divisible by x + 1, using division algorithm method.

(x + 1)(x³ + 4x) = x(x + 1)(x² + 4)

Greatest common divisor = (x² + 4)

---

Exercise 3.7

Question 1.

Find the LCM of the following

x³ y² , xyz

Answer:

Given terms: –

x³, y² , xyz

Formula used: –

LCM = Least Common Multiple

Means it is the lowest term by which every element must be divided completely;

x³ = x × x × x

y² = y × y

xyz = x × y × z

⇒ first find the common factors in all terms

Common factor = x × y

⇒ then multiply the remaining factors of terms in common

factor to get the LCM

= (x × y) × [(x²)(y)(z)]

= x³y²z

Conclusion: –

The LCM of given terms [x³ , y² , xyz] is x³y²z

Question 2.

Find the LCM of the following

3x²yz, 4x³y³

Answer:

Given terms: –

3x²yz, 4x³y³

Formula used: –

LCM = Least Common Multiple

Means it is the lowest term by which every element must be

divided completely;

3x²yz, = 3 × x × x × y × z

4x³y³ = 4 × x × x × x × y × y × y

⇒ first find the common factors in all terms

Common factor = x × x × y

⇒ then multiply the remaining factors of terms in common

factor to get the LCM

= (x × y × x) × [(3yz)(4xy²)]

= 12x³y³z

Conclusion: –

The LCM of given terms [3x²yz, 4x³y³] is 12x³y³z

Question 3.

Find the LCM of the following

a²bc, b²ca, c²ab

Answer:

Given terms: –

a²bc, b²ca, c²ab

Formula used: –

LCM = Least Common Multiple

Means it is the lowest term by which every element must be

divided completely;

a²bc = a × a × b × c

b²ca = a × b × b × c

c²ab = a × b × c × c

⇒ first find the common factors in all terms

Common factor = a × b × c

⇒ then multiply the remaining factors of terms in common

factor to get the LCM

= ( a × b × c) × [(a)(b)(c)]

= a²b²c²

Conclusion: –

The LCM of given terms [a²bc, b²ca, c²ab] is a²b²c²

Question 4.

Find the LCM of the following

66a⁴b²c³, 44a³b⁴c², 24a²b³c⁴

Answer:

Given terms: –

66a⁴b²c³, 44a³b⁴c², 24a²b³c⁴

Formula used: –

LCM = Least Common Multiple

Means it is the lowest term by which every element must be

divided completely;

66a⁴b²c³ = 3 × 2 × 11 × a × a × a × a × b × b × c × c × c

44a³b⁴c² = 2 × 2 × 11 × a × a × a × b × b × b × b × c × c

24a²b³c⁴ = 2 × 2 × 2 × 3 × a × a × b × b × b × c × c × c × c

⇒ first find the common factors in all terms

Common factor in all terms = 2 × a × a × b × b × c × c

Common factors from any 2 terms

2a²b²c² × [(3 × 11 × a × a × c)( 2 × 11 × a × b × b)( 2 × 2 × 3 × b × c × c)]

2a²b²c² × (3 × 11 × 2 × a × b × c)[(a)(b)(2c)]

⇒ then multiply the remaining factors of terms in common

factor to get the LCM

= 2a²b²c² × (66abc) × (2abc)

(66 × 2 × 2)( a²b²c² × abc × abc )

264 a⁴b⁴c⁴

Conclusion: –

The LCM of given terms [66a⁴b²c³, 44a³b⁴c², 24a²b³c⁴] is 264 a⁴b⁴c⁴

Question 5.

Find the LCM of the following

a^{m + 1}, a^{m + 2}, a^{m + 3}

Answer:

Given terms: –

a^{m + 1}, a^{m + 2}, a^{m + 3}

Formula used: –

LCM = Least Common Multiple

Means it is the lowest term by which every element must be

divided completely;

a^{m + 1} = a^m × a

a^{m + 2} = a^m × a × a

a^{m + 3} = a^m × a × a × a

⇒ first find the common factors in all terms

Common factor = a^m × a

Common factor from any 2 terms

= ( a^m × a) × [(1)(a)(a²)]

= ( a^m × a) × a[(1)(1)(a)]

⇒ then multiply the remaining factors of terms in common

factor to get the LCM

= ( a^m × a²)(a)

= a^ma³

Conclusion: –

The LCM of given terms [a^{m + 1}, a^{m + 2}, a^{m + 3}] is a^ma³

Question 6.

Find the LCM of the following

x²y + xy², x² + xy

Answer:

Given terms: –

x²y + xy², x² + xy

Formula used: –

LCM = Least Common Multiple

Means it is the lowest term by which every element must be

divided completely;

x²y + xy² = x × y × x + x × y × y

= x × y × (x + y)

x² + xy = x × x + x × y

= x × (x + y)

⇒ first find the common factors in all terms

Common factor = (x + y) × x

⇒ then multiply the remaining factors of terms in common

factor to get the LCM

= (x + y) × x × (x)

= x³ + yx²

Conclusion: –

The LCM of given terms [x²y + xy², x² + xy] is x³ + yx²

Question 7.

Find the LCM of the following

3(a – 1), 2(a – 1)², (a² – 1)

Answer:

Given terms: –

3(a – 1), 2(a – 1)², (a² – 1)

Formula used: –

LCM = Least Common Multiple

Means it is the lowest term by which every element must be

divided completely;

3(a – 1) = 3 × (a – 1)

2(a – 1)² = 2 × (a – 1) × (a – 1)

(a² – 1) = (a² – 1²) = (a – 1) × (a + 1)

⇒ first find the common factors in all terms

Common factor = (a – 1)

⇒ then multiply the remaining factors of terms in common

factor to get the LCM

= (a – 1) × [(3) × (2(a – 1)) × (a + 1)]

= 6(a + 1)(a – 1)²

Conclusion: –

The LCM of given terms [3(a – 1), 2(a – 1)², (a² – 1)] is

6(a + 1)(a – 1)²

Question 8.

Find the LCM of the following

2x² – 18, 5x²y + 15xy², x³ + 27y³

Answer:

Given terms: –

2x² – 18, 5x²y + 15xy², x³ + 27y³

Formula used: –

LCM = Least Common Multiple

Means it is the lowest term by which every element must be

divided completely;

2x² – 18y² = 2 × (x² – 9y²) = 2(x² – (3y)²) = 2 × (x – 3y) × (x + 3y)

5x²y + 15xy² = 5 × x × y × (x + 3y)

x³ + 27y³ = (x³ + (3y)³) = (x + 3y)(x² – 3xy + 9y²)

⇒ first find the common factors in all terms

Common factor = (x + 3y)

⇒ then multiply the remaining factors of terms in common

factor to get the LCM

= (x + 3y) × [2 × (x – 3y) × 5xy × ( x² – 3xy + 9y²)]

= 10xy(x + 3y)(x – 3y)( x² – 3xy + 9y²)]

Conclusion: –

The LCM of given terms [2x² – 18, 5x²y + 15xy², x³ + 27y³] is

10xy(x + 3y)(x – 3y)( x² – 3xy + 9y²)]

Question 9.

Find the LCM of the following

(x + 4)² (x – 3)³, (x – 1)(x + 4)(x – 3)²

Answer:

Given terms: –

(x + 4)² (x – 3)³, (x – 1)(x + 4)(x – 3)²

Formula used: –

LCM = Least Common Multiple

Means it is the lowest term by which every element must be divided completely;

(x + 4)² (x – 3)³ = (x + 4) × (x + 4) × (x – 3) × (x – 3) × (x – 3)

(x – 1)(x + 4)(x – 3)² = (x – 1) × (x + 4) × (x – 3) × (x – 3)

⇒ first find the common factors in all terms

Common factor = (x + 4) × (x – 3) × (x – 3)

⇒ then multiply the remaining factors of terms in common

factor to get the LCM

= (x + 4)(x – 3)(x – 3) × [(x + 4) × (x – 3) × (x – 1)]

= (x + 4)² (x – 3)³(x – 1)

Conclusion: –

The LCM of given terms [(x + 4)² (x – 3)³, (x – 1)(x + 4)(x – 3)²] is

(x + 4)² (x – 3)³(x – 1)

Question 10.

Find the LCM of the following

10(9x² + 6xy + y²), 12(3x² – 5xy – 2y²), 14(6x⁴ + 2x³)

Answer:

Given terms: –

10(9x² + 6xy + y²), 12(3x² – 5xy – 2y²), 14(6x⁴ + 2x³)

Formula used: –

LCM = Least Common Multiple

Means it is the lowest term by which every element must be

divided completely;

10(9x² + 6xy + y²) = 2 × 5 × ((3x)² + 2 × 3x × y + y²)

= 2 × 5 × (3x + y)²

= 2 × 5 × (3x + y) × (3x + y)

12(3x² – 5xy – 2y²) = 2 × 2 × 3 × (3x² – 6xy + xy – 2y²)

= 2 × 2 × 3 × (3x(x – 2y) + y(x – 2y))

= 2 × 2 × 3 × (x – 2y) × (3x + y)

14(6x⁴ + 2x³) = 2 × 7 × 2 × x × x × x × (3x + 1)

⇒ first find the common factors in all terms

Common factor = 2

Common factors in any 2 terms

2 × [(5(3x + 4)²)(2 × 3 × (x – 2y)(3x + y))(7 × 2 × x³ × (3x + 1))]

2 × 2 × (3x + y)[(5(3x + 4))(3 × (x – 2y))(7 × x³ × (3x + 1))]

⇒ then multiply the remaining factors of terms in common

factor to get the LCM

= 2 × 2 × 5 × 3 × 7 × x³ × (3x + y)(3x + y)(x – 2y)(3x + 1)

= 420x³(3x + y)²(x – 2y)(3x + 1)

Conclusion: –

The LCM of given terms [10(9x² + 6xy + y²), 12(3x² – 5xy – 2y²), 14(6x⁴ + 2x³)] is 420x³(3x + y)²(x – 2y)(3x + 1)

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Exercise 3.8

Question 1.

Find the LCM of each pair of the following polynomials.

x² – 5x + 6, x² + 4x – 12 whose GCD is x – 2.

Answer:

Given: –

Polynomials x² – 5x + 6, x² + 4x – 12

And GCD[Greatest Common Divisor] = (x – 2)

Formula used: –

The product of 2 polynomial is equal to product of their LCM

and GCD.

Product of 2 polynomial = LCM × GCD

Product of 2 polynomial = (x² – 5x + 6) × (x² + 4x – 12)

= ( x² – 2x – 3x + 6)(x² + 6x – 2x – 12)

= (x(x – 2) – 3(x – 2))(x(x + 6) – 2(x + 6))

= (x – 3)(x – 2)(x – 2)(x + 6)

Product of 2 polynomial = LCM × GCD

LCM = 

LCM = 

LCM = (x – 3)(x – 2)(x + 6)

Conclusion: –

The LCM of polynomial [x² – 5x + 6, x² + 4x – 12] is

(x – 3)(x – 2)(x + 6)

Question 2.

Find the LCM of each pair of the following polynomials.

x⁴ + 3 x³ + 6 x² + 5x + 3, x⁴ + 2 x² + x + 2 whose GCD is x² + x + 1

Answer:

Given: –

Polynomials x⁴ + 3 x³ + 6 x² + 5x + 3 , x⁴ + 2 x² + x + 2

And GCD[Greatest Common Divisor] = (x² + x + 1)

Formula used: –

The product of 2 polynomial is equal to product of their LCM

and GCD.

Product of 2 polynomial = LCM × GCD

Product of 2 polynomial = (x⁴ + 3x³ + 6x² + 5x + 3) × (x⁴ + 2x² + x + 2)

Product of 2 polynomial = LCM × GCD

LCM = 

LCM = 

LCM = 

LCM = (x² + 2x + 3)(x⁴ + 2x² + x + 2)

Conclusion: –

The LCM of polynomial [x⁴ + 3 x³ + 6 x² + 5x + 3, x⁴ + 2 x² + x + 2] is

(x² + 2x + 3)(x⁴ + 2x² + x + 2)

Question 3.

Find the LCM of each pair of the following polynomials.

2x³ + 15x² + 2x – 35, x³ + 8x² + 4x – 21 whose GCD is x + 7.

Answer:

Given: –

Polynomials 2x³ + 15x² + 2x – 35 , x³ + 8x² + 4x – 21

And GCD[Greatest Common Divisor] = (x + 7)

Formula used: –

The product of 2 polynomial is equal to product of their LCM

and GCD.

Product of 2 polynomial = LCM × GCD

Product of 2 polynomial = (2x³ + 15x² + 2x – 35) × (x³ + 8x² + 4x – 21)

Product of 2 polynomial = LCM × GCD

LCM = 

LCM = 

LCM = 

LCM = (2x² + x – 5)(x³ + 8x² + 4x – 21)

Conclusion: –

The LCM of given polynomials [2x³ + 15x² + 2x – 35 , x³ + 8x² + 4x – 21] is (2x² + x – 5)(x³ + 8x² + 4x – 21)

Question 4.

Find the LCM of each pair of the following polynomials.

2x³ – 3x² – 9x + 5, 2x⁴ – x³ – 10x² – 11x + 8 whose GCD is 2x – 1

Answer:

Given: –

Polynomials 2x³ – 3x² – 9x + 5, 2x⁴ – x³ – 10x² – 11x + 8

And GCD[Greatest Common Divisor] = (x + 7)

Formula used: –

The product of 2 polynomial is equal to product of their LCM

and GCD.

Product of 2 polynomial = LCM × GCD

Product of 2 polynomial = (2x³ – 3x² – 9x + 5) × (2x⁴ – x³ – 10x² – 11x + 8)

Product of 2 polynomial = LCM × GCD

LCM = 

LCM = 

LCM = 

LCM = (x³ – 5x – 8)( 2x³ – 3x² – 9x + 5)

Conclusion: –

The LCM of given polynomials [2x³ – 3x² – 9x + 5, 2x⁴ – x³ – 10x² – 11x + 8] is (x³ – 5x – 8)( 2x³ – 3x² – 9x + 5)

Question 5.

Find the other polynomial q(x) of each of the following, given that LCM and GCD and one polynomial p(x) respectively.

(x + 1)² (x + 2)², (x + 1) (x + 2), (x + 1)² (x + 2)

Answer:

Given: –

Polynomials p(x) = (x + 1)² (x + 2)

And GCD[Greatest Common Divisor] = (x + 1) (x + 2)

And LCM[Lowest Common Multiple] = (x + 1)² (x + 2)²

Formula used: –

The product of 2 polynomial is equal to product of their LCM

and GCD.

Product of 2 polynomial = LCM × GCD

p(x) × q(x) = LCM × GCD

LCM × GCD = (x + 1)² × (x + 2)² × (x + 1) × (x + 2)

p(x) × q(x) = LCM × GCD

q(x) = 

q(x) = 

q(x) = 

q(x) = (x + 1)(x + 2)²

Conclusion: –

The other polynomial term q(x) is (x + 1)(x + 2)²

Question 6.

Find the other polynomial q(x) of each of the following, given that LCM and GCD and one polynomial p(x) respectively.

(4x + 5)³ (3x – 7)³, (4x + 5) (3x – 7)², (4x + 5)³ (3x – 7)²

Answer:

Given: –

Polynomials p(x) = (4x + 5)³ (3x – 7)²

And GCD[Greatest Common Divisor] = (4x + 5) (3x – 7)²

And LCM[Lowest Common Multiple] = (4x + 5)³ (3x – 7)³

Formula used: –

The product of 2 polynomial is equal to product of their LCM

and GCD.

Product of 2 polynomial = LCM × GCD

p(x) × q(x) = LCM × GCD

LCM × GCD = (4x + 5)³ × (3x – 7)³ × (4x + 5) × (3x – 7)²

= (4x + 5)⁴(3x – 7)⁵

p(x) × q(x) = LCM × GCD

q(x) = 

q(x) = 

q(x) = 

q(x) = (4x + 5)(3x – 7)³

Conclusion: –

The other polynomial term q(x) is (4x + 5)(3x – 7)³

Question 7.

Find the other polynomial q(x) of each of the following, given that LCM and GCD and one polynomial p(x) respectively.

(x⁴ – y⁴) (x⁴ + x²y² + y⁴), x² – y², x⁴ – y⁴.

Answer:

Given: –

Polynomials p(x) = x⁴ – y⁴

And GCD[Greatest Common Divisor] = x² – y²

And LCM[Lowest Common Multiple] = (x⁴ – y⁴)(x⁴ + x²y² + y⁴)

Formula used: –

The product of 2 polynomial is equal to product of their LCM

and GCD.

Product of 2 polynomial = LCM × GCD

p(x) × q(x) = LCM × GCD

LCM × GCD = (x⁴ – y⁴) (x⁴ + x²y² + y⁴) × (x² – y²)

p(x) × q(x) = LCM × GCD

q(x) = 

q(x) = 

q(x) = 

q(x) = (x⁴ + x²y² + y⁴)(x² – y²)

Conclusion: –

The other polynomial term q(x) is (x⁴ + x²y² + y⁴)(x² – y²)

Question 8.

Find the other polynomial q(x) of each of the following, given that LCM and GCD and one polynomial p(x) respectively.

(x³ – 4x) (5x + 1), (5 x² + x), (5 x³ – 9 x² – 2x).

Answer:

Given: –

Polynomials p(x) = (5x³ – 9x² – 2x)

And GCD[Greatest Common Divisor] = (5x² + x)

And LCM[Lowest Common Multiple] = (x³ – 4x)(5x + 1)

Formula used: –

The product of 2 polynomial is equal to product of their LCM

and GCD.

Product of 2 polynomial = LCM × GCD

p(x) × q(x) = LCM × GCD

LCM × GCD = (x³ – 4x) (5x + 1) × (5x² + x)

= x(x² – 4)(5x + 1) × x(5x + 1)

= x²(x + 2)(x – 2)(5x + 1)(5x + 1)

p(x) = (5 x³ – 9 x² – 2x)

= x(5x² – 9x – 2)

= x(5x² – 10x + x – 2)

= x[5x(x – 2) + 1(x – 2)]

= x(5x + 1)(x – 2)

p(x) × q(x) = LCM × GCD

q(x) = 

q(x) = 

q(x) = 

q(x) = x(x + 2)(5x + 1)

Conclusion: –

The other polynomial term q(x) is x(x + 2)(5x + 1)

Question 9.

Find the other polynomial q(x) of each of the following, given that LCM and GCD and one polynomial p(x) respectively.

(x – 1) (x – 2) (x² – 3x + 3), (x – 1), (x³ – 4 x² + 6x – 3).

Answer:

Given: –

Polynomials p(x) = (x³ – 4 x² + 6x – 3)

And GCD[Greatest Common Divisor] = (x – 1)

And LCM[Lowest Common Multiple] = (x – 1)(x – 2)(x² – 3x + 3)

Formula used: –

The product of 2 polynomial is equal to product of their LCM

and GCD.

Product of 2 polynomial = LCM × GCD

p(x) × q(x) = LCM × GCD

LCM × GCD = (x – 1) (x – 2) (x² – 3x + 3) × (x – 1)

= (x – 1)² (x – 2) (x² – 3x + 3)

p(x) × q(x) = LCM × GCD

q(x) = 

q(x) =

q(x) =

q(x) =

q(x) = (x – 1)(x – 2)

Conclusion: –

The other polynomial term q(x) is (x – 1)(x – 2)

Question 10.

Find the other polynomial q(x) of each of the following, given that LCM and GCD and one polynomial p(x) respectively.

2(x + 1) (x² – 4), (x + 1), (x + 1) (x – 2).

Answer:

Given: –

Polynomials p(x) = (x + 1)(x – 2)

And GCD[Greatest Common Divisor] = (x + 1)

And LCM[Lowest Common Multiple] = 2(x + 1) (x² – 4)

Formula used: –

The product of 2 polynomial is equal to product of their LCM

and GCD.

Product of 2 polynomial = LCM × GCD

p(x) × q(x) = LCM × GCD

LCM × GCD = 2(x + 1) (x² – 4) × (x + 1)

= 2(x + 1)² (x² – 2²)

= 2(x + 1)² (x – 2)(x + 2)

p(x) × q(x) = LCM × GCD

q(x) = 

q(x) = 

q(x) = 

q(x) = 2(x + 2)(x + 1)

Conclusion: –

The other polynomial term q(x) is 2(x + 2)(x + 1)

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Exercise 3.9

Question 1.

Simplify the following into their lowest forms.

Answer:

[dividing numerator and denominator by 3x]

Question 2.

Simplify the following into their lowest forms.

Answer:

The like terms are cancelled.

 which is the required answer.

Question 3.

Simplify the following into their lowest forms.

Answer:

The like terms are cancelled.

=x – 1 which is the required answer.

Question 4.

Simplify the following into their lowest forms.

Answer:

The like terms are cancelled.

 which is the required answer.

Question 5.

Simplify the following into their lowest forms.

 (Hint : x⁴ + x² + 1 =(x² + 1)² – x²)

Answer:

The like terms are cancelled.

=x² + x + 1 which is the required answer.

Question 6.

Simplify the following into their lowest forms.

Answer:

The like terms are cancelled.

Required solution.

Question 7.

Simplify the following into their lowest forms.

Answer:

The like terms are cancelled.

Required solution.

Question 8.

Simplify the following into their lowest forms.

Answer:

The like terms are cancelled.

=x + 3

Required solution.

Question 9.

Simplify the following into their lowest forms.

Answer:

The like terms are cancelled.

Required answer.

Question 10.

Simplify the following into their lowest forms.

Answer:

The like terms are cancelled.

=1 required solution.

Question 11.

Simplify the following into their lowest forms.

Answer:

The like terms are cancelled.

Required solution

Question 12.

Simplify the following into their lowest forms.

Answer:

The like terms are cancelled.

=x – 2 required solution.