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 |
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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
(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.
