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