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 10A
  • Exercise 10B
  • Exercise 10C
  • Exercise 10D
  • Exercise 10E

Related Study

NCERT Solutions for Class 10 Maths

Quadratic Equation MCQ Solutions

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.

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