# RS Aggarwal Solutions Chapter 10 Quadratic Equation MCQ Class 10 Maths

 Chapter Name RS Aggarwal Chapter 10 Quadratic Equation Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 10AExercise 10BExercise 10CExercise 10DExercise 10E Related Study NCERT Solutions for Class 10 Maths

1. Which of the following is a quadratic equation?

(a) x3 - 3√x + 2 = 0

(b) x + 1/x = x2

(c) x2 + 1/x2 = 5

(d) 2x2 – 5x = (x – 1)2

Solution

(d) 2x2 – 5x = (x – 1)2

A quadratic equation is the equation with degree 2.

∵ 2x2 – 5x = (x – 1)2

⇒ 2x2 – 5x = x2 – 2x + 1

⇒ 2x2 – 5x – x2 + 2x – 1 = 0

⇒ x2 – 3x – 1 = 0, which is a quadratic equation.

2. Which of the following is a quadratic equation?

(a) (x2 + 1) = (2 – x)2 + 3

(b) (x3 – x2) = (x – 1)3

(c) 2x2 + 3 = (5 + x)(2x – 3)

(d) None of these

Solution

(b) x3 – x2 = (x – 1)3

∵ x3 – x2 = (x – 1)3

⇒ x3 – x2 = x3 – 3x2 + 3x – 1

⇒ 2x2 – 3x + 1 = 0, which is a quadratic equation

3. Which of the following is not a quadratic equation?

(a) 3x – x2 = x2 + 5

(b) (x + 2) = 2(x2 – 5)

(c) (√2x + 3)2 = 2x2 + 6

(d) (x – 1)2 = 3x2 + x – 2

Solution

(c) (√2x + 3)2 = 2x2 + 6

∵ (√2x + 3)2 = 2x2 + 6

⇒ 2x2 + 9 + 6√2x = 2x2 + 6

⇒ 6√2x + 3 = 0, which is not a quadratic equation

4. If x = 3 is a solution of the equation 3x2 + (k – 1)x + 9 = 0, then k = ?

(a) 11

(b) -11

(c) 13

(d) -13

Solution

(b) – 11

It is given that x = 3 is a solution of 3x2 + (k – 1)x + 9 = 0; therefore, we have:

3(3)2 + (k – 1) × 3 + 9 = 0

⇒ 27 + 3(k – 1) + 9 = 0

⇒ 3(k – 1) = - 36

⇒ (k – 1) = - 12

⇒ k = - 11

5. If one root of the equation 2x2 + ax + 6 = 0 is 2 then a = ?

(a) 7

(b) -7

(c) 7/2

(d) -7/2

Solution

(b) – 7

It is given that one root of the equation 2x2 + ax + 6 = 0 is 2.

∴ 2 × 22 + a × 2 + 6 = 0

⇒ 2a + 14 = 0

⇒ a = - 7

6. The sum of the roots of the equation x2 – 6x + 2 = 0 is

(a) 2

(b) -2

(c) 6

(d) – 6

Solution

(b) – 2

Sum of roots of the equation x2 – 6x + 2 = 0 is

α + β = -b/a = - (-6)/1 = 6, where α and β are the roots of the equation.

7. If the product of the roots of the equation x2 – 3x + k = 10 is -2 then the value of k is

(a) -2

(b) – 8

(c) 8

(d) 12

Solution

It is given that the product of the roots of the equation x2 – 3x + k = 10 is -2.

The equation can be rewritten as:

x2 – 3x + (k – 10) = 0

Product of the roots of a quadratic equation = c/a

⇒ c/a = -2

⇒ (k – 10)/1 = -2

⇒ k = 8

8. The ratio of the sum and product of the roots of the equation 7x2 – 12x + 18 = 0 is

(a) 7:12

(b) 7:18

(c) 3:2

(d) 2:3

Solution

(d) 2:3

Given:

7x2 – 12 + 18 = 0

∴ α + β = 12/7 and β = 18/7, where α and β are the roots of the equation

∴ Ratio of the sum and product of the roots = 12/7 : 18/7

= 12 : 18

= 2 : 3

9. If one root of the equation 3x2 – 10x + 3 = 0 is 1/3 then the other root is

(a) -1/3

(b) 1/3

(c) – 3

(d) 3

Solution

(d) 3

Given:

3x2 – 10x + 3 = 0

One root of the equation is 1/3.

Let the other root be α

Product of the roots = c/a

⇒ 1/3 × α = 3/3

⇒ α = 3

10. If one root of 5x2 + 13x + k = 0 be the reciprocal of the other root then the value of k is

(a) 0

(b) 1

(c) 2

(d) 5

Solution

(d) 5

Let the roots of the equation -2/3 be α and 1/α.

∴ Product of the roots = c/a

⇒ α × 1/α = k/5

⇒ α = k/5

⇒ k = 5

11. If the sum of the roots of the equation kx2 + 2x + 3k = 0 is equal to their product then the value of k

(a) 1/3

(b) -1/3

(c) 2/3

(d) -2/3

Solution

(d) -2/3

Given:

kx2 + 2x + 3k = 0

Sum of the roots = Product of the roots

⇒ -2/k = 3k/k

⇒ 3k = - 2

⇒ k = -2/3

12. The roots of a quadratic equation are 5 and -2. Then, the equation is

(a) x2 – 3x + 10 = 0

(b) x2 – 3x – 10 = 0

(c) x2 + 3x – 10 = 0

(d) x2 + 3x + 10 = 0

Solution

(b) x2 – 3x – 10 = 0

It is given that the roots of the quadratic equation are 5 and -2.

Then, the equation is:

x2 – (5 – 2)x + 5 × (-2) = 0

⇒ x2 – 3x – 10 = 0

13. If the sum of the roots of a quadratic equation is 6 and their product is 6, the equation is

(a) x2 – 6x + 6 = 0

(b) x2 + 6x + 6 = 0

(c) x2 – 6x – 6 =

(d) x2 + 6x + 6 = 0

Solution

(a) x2 – 6x + 6 = 0

Given:

Sum of roots = 6

Product of roots = 6

Thus, the equation is:

x2 – 6x + 6 = 0

14. If α and β are the roots of the equation 3x2 + 8x + 2 = 0 then (1/α + 1/β) = ?

(a) -3/8

(b) 2/3

(c) -4

(d) 4

Solution

(c) -4

It is given that α and β are the roots of the equation 3x2 + 8x + 2 = 0

∴ α + β = -8/3 and αβ = 2/3

1/α + 1/β = (α + β)/αβ = -(8/3)/-(2/3)

= - 4

15. The roots of the equation ax2 + bx + c = 0 will be reciprocal each other if

(a) a = b

(b) b = c

(c) c = a

(d) none of these

Solution

(c) c = a

Let the roots of the equation (ax2 + bx + c = 0) be α and 1/α.

∴ Product of the roots = α × 1/α = 1

⇒ c/a = 1

⇒ c = a

16. If the roots of the equation ax2 + bx + c = 0 are equal then c = ?

(a) -b/2a

(b) b/2a

(c) -b2/4a

(d) b2/4a

Solution

(d) b2/4a

It is given that the roots of the equation (ax2 + bx + c = 0) are equal.

∴ (b2 – 4ac) = 0

⇒ b2 = 4ac

⇒ c = b2/4a

17. If the equation 9x2 + 6kx + 4 = 0 has equal roots then k = ?

(a) 0 or 0

(b) -2 or 0

(c) 2 or – 2

(d) 0 only

Solution

(c) 2 or – 2

It is given that the roots of the equation (9x2 + 6kx + 4 = 0) are equal.

∴ (b2 – 4ac) = 0

⇒ (6k)2 – 4 × 9 × 4 = 0

⇒ 36k2 = 144

⇒ k2 = 4

⇒ k = ±2

18. If the equation x2 + 2(k + 2)x + 9k = 0 has equal roots then k = ?

(a) 1 or 4

(b) -1 or 4

(c) 1 or – 4

(d) -1 or -4

Solution

It is given that the roots of the equation (x2 + 2(k + 2)x + 9k = 0) are equal.

∴ (b2 – 4ac) = 0

⇒ {2(k + 2)}2 – 4 × 1 × 9k = 0

⇒ 4(k2 + 4k + 4) – 36k = 0

⇒ 4k2 + 16k + 16 – 36k = 0

⇒ 4k2 – 20 + 16 = 0

⇒ k2 – 5k + 4 = 0

⇒ k2 – 4k – k + 4 = 0

⇒ k(k – 4) – (k – 4) = 0

⇒ (k – 4)(k – 1) = 0

⇒ k = 4 or k = 1

19. If the equation 4x2 – 3kx + 1 = 0 has equal roots then value of k = ?

(a) ± 2/3

(b) ± 1/3

(c) ± 3/4

(d) ± 4/3

Solution

(d) ± 4/3

It is given that the roots of the equation (4x2 – 3kx + 1 = 0) are equal.

∴ (b2 – 4ac) = 0

⇒ (3k)2 – 4 × 4 × 1 = 0

⇒ 9k2 = 16

⇒ k2 = 16/9

⇒ k = ± 4/3

20. The roots of ax2 + bx + c = 0, a ≠ 0 are real and unequal, if (b2 – 4ac) is

(a) > 0

(b) = 0

(c) < 0

(d) none of these

Solution

(a) > 0

The roots of the equation are real and unequal when (b2 – 4ac) > 0.

21. In the equation ax2 + bx + c = 0, it is given that D = (b2 – 4ac) > 0. Then, the roots of the equation are

(a) real and equal

(b) real and unequal

(c) imaginary

(d) none of these

Solution

(b) real and unequal

We know that when discriminant, D > 0, the roots of the given quadratic equation are real and unequal.

22. The roots of the equation 2x2 – 6x + 7 = 0 are

(a) real, unequal and rational

(b) real, unequal and irrational

(c) real and equal

(d) imaginary

Solution

(d) imaginary

∵ D = (b2 – 4ac)

= (-6)2 - 4×2×7

= 36 – 56

= - 20 < 0

Thus, the roots of the equation are imaginary

23. The roots of the equation 2x2 – 6x + 3 = 0 are

(a) real, unequal and rational

(b) real, unequal and irrational

(c) real and equal

(d) imaginary

Solution

(b) real, unequal and irrational

∵ D = (b2 – 4ac)

= (-6)2 – 4 × 2 × 3

= 36 – 24

= 12

12 is greater than 0 and it is not a perfect square; therefore, the roots of the equation are real, unequal and irrational.

24. If the roots of 5x2 – k + 1 = 0 are real and distinct then

(a) -2√5 < k < 2√5

(b) k > 2√5 only

(c) k < -2√ 5

(d) either k > 2√5 or k < -2√5

Solution

(d) either k > 2√5 or k < -2√5

It is given that the roots of the equation (5x2 – k + 1 = 0) are real and distinct.

∴ (b2 – 4ac) > 0

⇒ (-k)2 – 4 × 5 × 1 > 0

⇒ k2 – 20 > 0

⇒ k2 > 20

25. If the equation x2 + 5kx + 16 = 0 has no real roots then

(a) k > 8/5

(b) k < -8/5

(c) -8/5

(d) None of these

Solution

(c) -8/5 < k < 8/5

It is given that the equation (x2 + 5kx + 16 = 0) has no real roots.

∴ (b2 – 4ac) < 0

⇒ (5k)2 – 4 × 1 × 16 < 0

⇒ 25k2 – 64 < 0

⇒ k2 < 64/25

⇒ -8/5 < k < 8/5

26. If the equation x2 – kx +1 = 0 has no real roots then

(a) k < -2

(b) k > 2

(c) – 2 < k < 2

(d) None of these

Solution

(c) -2 < k < 2

It is given that the equation x2 – kx + 1 = 0 has no real roots.

∴ (b2 – 4ac) < 0

⇒ (-k)2 – 4 × 1 × 1 < 0

⇒ k2 < 4

⇒ -2 < k < 2

27. For what value of k, the equation kx2 – 6x – 2 = 0 has real roots?

(a) k ≤ -9/2

(b) k ≥ -9/2

(c) k ≤ -2

(d) None of these

Solution

(b) k ≥ -9/2

It is given that the roots of the equation (kx2 – 6x – 2 = 0) are real.

∴ D ≥ 0

⇒ (b2 – 4ac) ≥ 0

⇒ (-6)2 – 4 × k × (-2) ≥ 0

⇒ 36 + 8k ≥ 0

⇒ k ≥ -36/8

⇒ k ≥ -9/2

28. The sum of a number and its reciprocal is 2.1/20. The number is

(a) 5/4 or 4/5

(b) 4/3 or 3/4

(c) 5/6 or 6/5

(d) 1/6 or 6

Solution

(a) 5/4 or 4/5

Let the required number be x.

According to the question:

x + 1/x = 41/20

⇒ (x2 + 1)/x = 41/20

⇒ 20x2 – 41x + 20 = 0

⇒ 20x2 – 25x – 16x + 20 = 0

⇒ 5x(4x – 5) – 4(4x – 5) = 0

⇒ (4x – 5)(5x – 4) = 0

⇒ x = 5/4 or x = 4/5

29. The perimeter of a rectangle is 82 m and its area is 400 m2. The breadth of the rectangle is

(a) 25 m

(b) 20 m

(c) 16 m

(d) 9 m

Solution

(c) 16 m

Let the length and breadth of the rectangle be l and b.

Perimeter of the rectangle = 82 m

⇒ 2×(l + b) = 82

⇒ l + b = 41

⇒ l = (41 – b) ...(i)

Area of rectangle = 400 m2

⇒ l × b = 400 m2

⇒ (41 – b)b = 400 (using (i))

⇒ 41b – b2 = 400

⇒ b2 – 41b + 400 = 0

⇒ b2 – 25b – 16b + 400 = 0

⇒ b(b – 25) – 16(b – 25) = 0

⇒ (b – 25)(b – 16) = 0

⇒ b = 25 or b = 16

If b = 25, we have:

l = 41 – 25 = 16

Since, l cannot be less than b,

∴ b = 16 m

30. The length of a rectangular field exceeds its breadth by 8 m and the area of the field is 240 m2. The breadth of the field is

(a) 20 m

(b) 30 m

(c) 12 m

(d) 16 m

Solution

Let the breadth of the rectangular field be x m.

∴ Length of the rectangular field = (x + 8) m

Area of the rectangular field = 240 m2 (Given)

∴ (x + 8) × x = 20 (Area = Length × Breadth)

⇒ x2 + 8x – 240 = 0

⇒ x2 + 20x – 12x – 240 = 0

⇒ x(x + 20) - 12(x + 20) = 0

⇒ (x + 20)(x – 12) = 0

⇒ x = - 20 or x = 12

∴ x = 12 (Breadth cannot be negative)

Thus, the breadth of the field is 12 m

Hence, the correct answer is option C.

31. The roots of the quadratic equation 2x2 – x – 6 = 0. are

(a) -2, 3/2

(b) 2, -3/2

(c) -2, -3/2

(d) 2, 3/2

Solution

(b) 2, -3/2

The given quadratic equation is 2x2 – x – 6 = 0

2x2 – x – 6 = 0

⇒ 2x2 – 4x + 3x – 6 = 0

⇒ 2x(x – 2) + 3(x – 2) = 0

⇒ (x – 2)(2x + 3) = 0

⇒ x – 2 = 0 or 2x + 3 = 0

⇒ x = 2 or x = -3/2

Thus, the roots of the given equation are 2 and -3/2.

Hence, the correct answer is option B.

32. The sum of two natural numbers is 8 and their product is 15. Find the numbers.

Solution

Let the required natural numbers be x and (8 – x).

It is given that the product of the two numbers is 15.

∴ x(8 – x) = 15

⇒ 8x – x2 = 15

⇒ x2 – 8x + 15 = 0

⇒ x2 – 5x – 3x + 15 = 0

⇒ x(x – 5) – 3(x – 5) = 0

⇒ (x – 5)(x – 3) = 0

⇒ x – 5 = 0 or x – 3 = 0

⇒ x – 5 or x = 3

Hence, the required numbers are 3 and 5.

33. Show that x = - 3 is a solution of x2 + 6x + 9 = 0

Solution

The given equation is x2 + 6x + 9 = 0

Putting x = -3 in the given equation, we get

LHS = (-3)2 + 6 × (-3) + 9

= 9 – 18 + 9 = 0

= RHS

∴ x = - 3 is a solution of the given equation.

34. Show that x = - 2 is a solution of 3x2 + 13x + 14 = 0

Solution

The given equation is 3x2 + 13x + 14 = 0.

Putting x = - 2 in the given equation, we get

LHS = - 3 × (-2)2 + 13 × (-2) + 14

= 12 – 26 + 14

= 0

= RHS

∴ x = -2 is a solution of the given equation.

35. If x = -1/2 is a solution of the quadratic equation 3x2 + 2kx – 3 = 0. Find the value of k.

Solution

It is given that x = -1/2 is a solution of the quadratic equation 3x2 + 2kx – 3 = 0.

∴ 3 × (-1/2)2 + 2k × (-1/2) – 3 = 0

⇒ 3/4 – k – 3 = 0

⇒ k = (3 – 12)/4 = - (9/4)

Hence, the value of k is – 9/4.

36. Find the roots of the equation 2x2 – x – 6 = 0.

Solution

The given quadratic equation is 2x2 – x – 6 = 0.

2x2 – x – 6 = 0

⇒ 2x2 – 4x + 3x – 6 = 0

⇒ 2x(x – 2) + 3(x – 2) = 0

⇒ (x – 2)(2x + 3) = 0

⇒ x – 2 = 0 or 2x + 3 = 0

⇒ x = 2 or x = -(3/2)

Hence, the roots of the given equation are 2 and – (3/2).

37. Find the solution of the quadratic equation 3√3x2 + 10x + √3 = 0

Solution

The given quadratic equation is √3x2 + 10x + √3 = 0

3√3x2 + 10x + √3 = 0

⇒ 3√3x2 + 9x + x + √3 = 0

⇒ 3√3x (x + √3) + 1(x + √3) = 0

⇒ (x + √3)(3√3x + 1) = 0

⇒ x + √3 = 0 or 3√3x + 1 = 0

⇒ x = - √3 or x = - (1/3√3) = -(√3/9)

Hence, -√3 and -√3/9 are the solutions of the given equation.

38. If the roots of the quadratic equation 2x2 + 8x + k = 0 are equal then find the value of k.

Solution

It is given that the roots of the quadratic equation 2x2 + 8x + k = 0 are equal.

∴ D = 0

⇒ 82 – 4 × 2 × k = 0

⇒ 64 – 8k = 0

⇒ k = 8

Hence, the value of k is 8.

39. If the quadratic px2 - 2√5px + 15 = 0 has two equal roots then find the value of p.

Solution

It is given that the quadratic equation px2 - 2√5px + 15 = 0 has two equal roots.

∴ D = 0

⇒ (-2√p)2 – 4 × p × 15 = 0

⇒ 20p2 – 60p = 0

⇒ p = 0 or p – 3 = 0

⇒ p = 0 or p = 3

For p = 0, we get 15 = 0, which is not true.

∴ p ≠ 0

Hence, the value of p is 3.

40. If 1 is a root of the equation ay2 + ay + 3 = 0 and y2 + y + b = 0 then find the value of ab.

Solution

It is given that y = 1 is a root of the equation ay2 + ay + 3 = 0

∴ a × (1)2 + a × 1 + 3 = 0

⇒ a + a + 3 = 0

⇒ 2a + 3 = 0

⇒ a = -(3/2)

Also, y = 1 is a root of the equation y2 + y + b = 0.

∴ (1)2 + l + b = 0

⇒ l + l + b = 0

⇒ b + 2 = 0

⇒ b = - 2

∴ ab = (-3/2) × (-2) = 3

Hence, the value of ab is 3.

41. If one zero of the polynomial x2 – 4x + 1 is (2 + √3), write the other zero.

Solution

Let the other zero of the given polynomial be α.

Now,

Sum of zeros of the given polynomial = -(-4)/1 = 4

∴ α + (2 + √3) = 4

⇒ α = 4 – 2 - √3

= 2 - √3

Hence, the other zero of the given polynomial is (2 - √3).

42. If one root of the quadratic equation 3x2 – 10x + k = 0 is reciprocal of the other, find the value of k.

Solution

Let α and β be the roots of the equation 3x2 + 10x + k = 0.

∴ α = 1/β (Given)

⇒ αβ = 1

⇒ k/3 = 1 (Product of the roots = c/a)

⇒ k = 3

Hence, the value of k is 3.

43. If the roots of the quadratic equation px(x – 2) + 0 are equal, find the value of p.

Solution

It is given that the roots of the quadratic equation px2 – 2px + 6 = 0 are equal.

∴ D = 0

⇒ (-2p)2 – 4 × p × 6 = 0

⇒ 4p2 – 24p = 0

⇒ 4p(p – 6) = 0

⇒ p = 0 or p – 6 = 0

⇒ p = 0 or p = 6

For p = 0, we get 6 = 0, which is not true.

∴ p ≠ 0

Hence, the value of p is 6.

44. Find the value of k so that the quadratic equation x2 – 4kx + k = 0 has equal roots.

Solution

It is given that the quadratic equation x2 – 4kx + k = 0 has equal roots.

∴ D = 0

⇒ (-4k)2 – 4 × 1 × k = 0

⇒ 16k2 – 4k = 0

⇒ 4k(k – 1) = 0

⇒ k = 0 or 4k – 1 = 0

⇒ k = 0 or k = 1/4

Hence, 0 and 1/4 are the required values of k.

45. Find the value of k for which the quadratic equation 9x2 – 3kx + k = 0 has equal roots.

Solution

It is given that the quadratic equation 9x2 – 3kx + k = 0 has equal roots.

∴ D = 0

⇒ (-3k)2 – 4 × 9 × k = 0

⇒ 9k2 – 36k = 0

⇒ 9k(k – 4) = 0

⇒ k = 0 or k – 4 = 0

⇒ k = 0 or k = 4

Hence, 0 and 4 are the required values of k.

46. Solve x2 – (√3 + 1)x + √3 = 0

Solution

x2 – (√3 + 1)x + √3 = 0

⇒ x2 - √3x – x + √3 = 0

⇒ x(x - √3) – 1(x - √3) = 0

⇒ (x - √3)(x – 1) = 0

⇒ x - √3 = 0 or x – 1 = 0

⇒ x = √3 or x = 1

Hence, 1 and √3 are roots of the given equation.

47. Solve 2x2 + ax – a2 = 0

Solution

2x2 + ax – a2 = 0

⇒ 2x2 + 2ax – ax – a2 = 0

⇒ 2x(x + a) – a(x + a) = 0

⇒ (x + a)(2x – a) = 0

⇒ x + a = 0 or 2x – a = 0

⇒ x = - a or x = a/2

Hence, -a and a/2 are the roots of the given equation.

48. Solve: 3x2 + 5√5x – 10 = 0

Solution

3x2 + 5√5x – 10 = 0

⇒ 3x2 + 6√5x - √5x – 10 = 0

⇒ 3x(x + 2√5) - √5(x + 2√5) = 0

⇒ (x + 2√5)(3x - √5) = 0

⇒ x + 2√5 = 0 or 3x - √5 = 0

⇒ x = -2√5 or x = √5/3

Hence, -2√5 and √5/3 are the toots of the given equation.

49. Solve √3x2 + 10x - 8√3 = 0

Solution

√3x2 + 10x - 8√3 = 0

⇒ 3x2 + 12x – 2x - 8√3 = 0

⇒ √3x(x + 4√3) – 2(x + 4√3) = 0

⇒ (x + 4√3)(√3x – 2) = 0

⇒ x + 4√3 = 0 or √3x – 2 = 0

⇒ x = -4√3 or x = 2/√3 = 2√3/3

Hence, -4√3 and 2√3/3 are the roots of the given equation.

50. Solve: √3x2 - 2√2x - 2√3 = 0

Solution

√3x2 - 2√2x - 2√3 = 0

⇒ √3x2 - 3√2x + √2x - 2√3 = 0

⇒ √3x(x - √6) + √2(x - √6) = 0

⇒ (x - √6)( √3 + √2) = 0

⇒ x = √6 or x = -(√2/√3) = -(√6/3)

Hence, √6 and - (√6/3) are the roots of the given equation.

51. Solve 4√3x2 + 5x - 2√3 = 0

Solution

4√3x2 + 5x - 2√3 = 0

⇒ 4√3x2 + 8x – 3x - 2√3 = 0

⇒ 4x(√3x + 2) - √3(√3x + 2) = 0

⇒ (√3x + 2)(4x - √3) = 0

⇒ √3x + 2 = 0 or 4x - √3 = 0

⇒ x = -(2/√3) = -(2√3/3) or x = √3/4

Hence, -(2√3)/3 and √3/4 are the roots of the given equation.

52. Solve: 4x2+ 4bx – (a2 – b2) = 0

Solution

4x2 + 4bx – (a2 - b2) = 0

⇒ 4x2 + 4bx – (a – b)(a + b) = 0

⇒ 4x2 + 2[(a + b) – (a – b)]x – (a – b)(a + b) = 0

⇒ 4x2 + 2(a + b)x – 2(a – b)x – (a – b)(a + b) = 0

⇒ 2x[2x + (a + b)] – (a – b)[2x + (a + b)] = 0

⇒ [2x + (a + b)][2x – (a – b)] = 0

⇒ 2x + (a + b) = 0 or 2x – (a – b) = 0

⇒ x = -(a + b)/2 or x = (a – b)/2

Hence, -(a + b)/2 and (a – b)/2 are the roots of the given equation.

53. Solve x2 + 5x - (a2 + a – 6) = 0

Solution

x2 + 5x – (a2 + a – 6) = 0

⇒ x2 + 5x – (a + 3)(a – 2) = 0

⇒ x2 + [(a + 3) – (a – 2)]x – (a + 3)(a – 2) = 0

⇒ x2 + (a + 3)x – (a – 2)x – (a + 3)(a – 2) = 0

⇒ x[x + (a + 3)] – (a – 2)[x + (a + 3)] = 0

⇒ [x + (a + 3)][x – (a – 2)] = 0

⇒ x + (a + 3) = 0 or x – (a – 2) = 0

⇒ x = - (a + 3) or x = (a – 2)

54. Solve x2 + 6x – (a2 + 2a – 8) = 0

Solution

x2 + 6x – (a2 – 2a – 8) = 0

⇒ x2 + 6x – (a + 4)(a – 2) = 0

⇒ x2 + [(a + 4) – (a – 2)]x – (a + 4)(a – 2) = 0

⇒ x2 + (a + 4)x – (a – 2)x – (a + 4)(a – 2) = 0

⇒ x[x + (a + 4)] – (a – 2)[x + (a + 4)] = 0

⇒ [x + (a + 4)][x – (a – 2)] = 0

⇒ x + (a + 4) = 0 x – (a – 2) = 0

⇒ x = - (a + 4) or x = (a – 2)

Hence, -(a + 4) and (a – 2) are the roots of the given equation.

55. Solve x2 – 4ax + 4a2 – b2 = 0

Solution

x2 – 4ax + 4a2 – b2 = 0

⇒ x2 – 4ax + (2a + b)(2a – b) = 0

⇒ x2 – [(2a + b) + (2a – b)]x + (2a + b)(2a – b) = 0

⇒ x2 – (2a + b)x – (2a – b)x + (2a + b)(2a – b) = 0

⇒ x[x – (2a + b)] – (2a – b)[x – (2a + b)] = 0

⇒ [x – (2a + b)] [x – (2a – b)] =0

⇒ x – (2a + b) = 0 or x – (2a – b) = 0

⇒ x = (2a + b) or x = (2a – b)

Hence, (2a + b) and (2a – b) are the roots of the given equation.