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 

Related Study  NCERT Solutions for Class 10 Maths 
Quadratic Equation MCQ Solutions
1. Which of the following is a quadratic equation?
(a) x^{3}  3√x + 2 = 0
(b) x + 1/x = x^{2 }
(c) x^{2} + 1/x^{2} = 5
(d) 2x^{2} – 5x = (x – 1)^{2}
Solution
(d) 2x^{2} – 5x = (x – 1)^{2}
A quadratic equation is the equation with degree 2.
∵ 2x^{2} – 5x = (x – 1)^{2}
⇒ 2x^{2} – 5x = x^{2} – 2x + 1
⇒ 2x^{2} – 5x – x^{2} + 2x – 1 = 0
⇒ x^{2} – 3x – 1 = 0, which is a quadratic equation.
2. Which of the following is a quadratic equation?
(a) (x^{2} + 1) = (2 – x)^{2} + 3
(b) (x^{3} – x^{2}) = (x – 1)^{3}
(c) 2x^{2} + 3 = (5 + x)(2x – 3)
(d) None of these
Solution
(b) x^{3} – x^{2} = (x – 1)^{3}
∵ x^{3 }– x^{2} = (x – 1)^{3}
⇒ x^{3} – x^{2} = x^{3} – 3x^{2} + 3x – 1
⇒ 2x^{2} – 3x + 1 = 0, which is a quadratic equation
3. Which of the following is not a quadratic equation?
(a) 3x – x^{2} = x^{2} + 5
(b) (x + 2) = 2(x^{2} – 5)
(c) (√2x + 3)^{2} = 2x^{2} + 6
(d) (x – 1)^{2} = 3x^{2} + x – 2
Solution
(c) (√2x + 3)^{2} = 2x^{2 }+ 6
∵ (√2x + 3)^{2} = 2x^{2} + 6
⇒ 2x^{2} + 9 + 6√2x = 2x^{2 }+ 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 3x^{2} + (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 2x^{2} + 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 2x^{2} + ax + 6 = 0 is 2.
∴ 2 × 2^{2} + a × 2 + 6 = 0
⇒ 2a + 14 = 0
⇒ a =  7
6. The sum of the roots of the equation x^{2} – 6x + 2 = 0 is
(a) 2
(b) 2
(c) 6
(d) – 6
Solution
(b) – 2
Sum of roots of the equation x^{2} – 6x + 2 = 0 is
Î± + Î² = b/a =  (6)/1 = 6, where Î± and Î² are the roots of the equation.
7. If the product of the roots of the equation x^{2} – 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 x^{2} – 3x + k = 10 is 2.
The equation can be rewritten as:
x^{2} – 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 7x^{2} – 12x + 18 = 0 is
(a) 7:12
(b) 7:18
(c) 3:2
(d) 2:3
Solution
(d) 2:3
Given:
7x^{2 }– 12 + 18 = 0
∴ Î± + Î² = 12/7 and Î² = 18/7, where Î± and Î² 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 3x^{2} – 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:
3x^{2} – 10x + 3 = 0
One root of the equation is 1/3.
Let the other root be Î±
Product of the roots = c/a
⇒ 1/3 × Î± = 3/3
⇒ Î± = 3
10. If one root of 5x^{2} + 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 Î± and 1/Î±.
∴ Product of the roots = c/a
⇒ Î± × 1/Î± = k/5
⇒ Î± = k/5
⇒ k = 5
11. If the sum of the roots of the equation kx^{2} + 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:
kx^{2} + 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) x^{2} – 3x + 10 = 0
(b) x^{2} – 3x – 10 = 0
(c) x^{2} + 3x – 10 = 0
(d) x^{2} + 3x + 10 = 0
Solution
(b) x^{2} – 3x – 10 = 0
It is given that the roots of the quadratic equation are 5 and 2.
Then, the equation is:
x^{2} – (5 – 2)x + 5 × (2) = 0
⇒ x^{2} – 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) x^{2} – 6x + 6 = 0
(b) x^{2} + 6x + 6 = 0
(c) x^{2} – 6x – 6 =
(d) x^{2} + 6x + 6 = 0
Solution
(a) x^{2} – 6x + 6 = 0
Given:
Sum of roots = 6
Product of roots = 6
Thus, the equation is:
x^{2} – 6x + 6 = 0
14. If Î± and Î² are the roots of the equation 3x^{2} + 8x + 2 = 0 then (1/Î± + 1/Î²) = ?
(a) 3/8
(b) 2/3
(c) 4
(d) 4
Solution
(c) 4
It is given that Î± and Î² are the roots of the equation 3x^{2} + 8x + 2 = 0
∴ Î± + Î² = 8/3 and Î±Î² = 2/3
1/Î± + 1/Î² = (Î± + Î²)/Î±Î² = (8/3)/(2/3)
=  4
15. The roots of the equation ax^{2 }+ 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 (ax^{2} + bx + c = 0) be Î± and 1/Î±.
∴ Product of the roots = Î± × 1/Î± = 1
⇒ c/a = 1
⇒ c = a
16. If the roots of the equation ax^{2} + bx + c = 0 are equal then c = ?
(a) b/2a
(b) b/2a
(c) b^{2}/4a
(d) b^{2}/4a
Solution
(d) b^{2}/4a
It is given that the roots of the equation (ax^{2} + bx + c = 0) are equal.
∴ (b^{2} – 4ac) = 0
⇒ b^{2 }= 4ac
⇒ c = b^{2}/4a
17. If the equation 9x^{2} + 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 (9x^{2} + 6kx + 4 = 0) are equal.
∴ (b^{2} – 4ac) = 0
⇒ (6k)^{2} – 4 × 9 × 4 = 0
⇒ 36k^{2} = 144
⇒ k^{2} = 4
⇒ k = ±2
18. If the equation x^{2} + 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 (x^{2} + 2(k + 2)x + 9k = 0) are equal.
∴ (b^{2} – 4ac) = 0
⇒ {2(k + 2)}^{2} – 4 × 1 × 9k = 0
⇒ 4(k^{2} + 4k + 4) – 36k = 0
⇒ 4k^{2} + 16k + 16 – 36k = 0
⇒ 4k^{2} – 20 + 16 = 0
⇒ k^{2} – 5k + 4 = 0
⇒ k^{2} – 4k – k + 4 = 0
⇒ k(k – 4) – (k – 4) = 0
⇒ (k – 4)(k – 1) = 0
⇒ k = 4 or k = 1
19. If the equation 4x^{2} – 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 (4x^{2} – 3kx + 1 = 0) are equal.
∴ (b^{2} – 4ac) = 0
⇒ (3k)^{2} – 4 × 4 × 1 = 0
⇒ 9k^{2 }= 16
⇒ k^{2} = 16/9
⇒ k = ± 4/3
20. The roots of ax^{2} + bx + c = 0, a ≠ 0 are real and unequal, if (b^{2} – 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 (b^{2 }– 4ac) > 0.
21. In the equation ax^{2} + bx + c = 0, it is given that D = (b^{2} – 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 2x^{2} – 6x + 7 = 0 are
(a) real, unequal and rational
(b) real, unequal and irrational
(c) real and equal
(d) imaginary
Solution
(d) imaginary
∵ D = (b^{2} – 4ac)
= (6)^{2}  4×2×7
= 36 – 56
=  20 < 0
Thus, the roots of the equation are imaginary
23. The roots of the equation 2x^{2} – 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 = (b^{2} – 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 5x^{2 }– 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 (5x^{2} – k + 1 = 0) are real and distinct.
∴ (b^{2} – 4ac) > 0
⇒ (k)^{2} – 4 × 5 × 1 > 0
⇒ k^{2 }– 20 > 0
⇒ k^{2 }> 20
25. If the equation x^{2 }+ 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 (x^{2 }+ 5kx + 16 = 0) has no real roots.
∴ (b^{2} – 4ac) < 0
⇒ (5k)^{2} – 4 × 1 × 16 < 0
⇒ 25k^{2} – 64 < 0
⇒ k^{2} < 64/25
⇒ 8/5 < k < 8/5
26. If the equation x^{2} – 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 x^{2 }– kx + 1 = 0 has no real roots.
∴ (b^{2} – 4ac) < 0
⇒ (k)^{2} – 4 × 1 × 1 < 0
⇒ k2 < 4
⇒ 2 < k < 2
27. For what value of k, the equation kx^{2} – 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 (kx^{2} – 6x – 2 = 0) are real.
∴ D ≥ 0
⇒ (b^{2} – 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
⇒ (x^{2 }+ 1)/x = 41/20
⇒ 20x^{2} – 41x + 20 = 0
⇒ 20x^{2 }– 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 m^{2}. 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 m^{2}
⇒ l × b = 400 m^{2}
⇒ (41 – b)b = 400 (using (i))
⇒ 41b – b^{2} = 400
⇒ b^{2} – 41b + 400 = 0
⇒ b^{2} – 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 m^{2}. 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 m^{2} (Given)
∴ (x + 8) × x = 20 (Area = Length × Breadth)
⇒ x^{2} + 8x – 240 = 0
⇒ x^{2} + 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 2x^{2} – 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 2x^{2} – x – 6 = 0
2x^{2} – x – 6 = 0
⇒ 2x^{2} – 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 – x^{2} = 15
⇒ x^{2} – 8x + 15 = 0
⇒ x^{2} – 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 x^{2} + 6x + 9 = 0
Solution
The given equation is x^{2} + 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 3x^{2} + 13x + 14 = 0
Solution
The given equation is 3x^{2} + 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 3x^{2} + 2kx – 3 = 0. Find the value of k.
Solution
It is given that x = 1/2 is a solution of the quadratic equation 3x^{2 }+ 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 2x^{2} – x – 6 = 0.
Solution
The given quadratic equation is 2x^{2} – x – 6 = 0.
2x^{2} – x – 6 = 0
⇒ 2x^{2} – 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√3x^{2} + 10x + √3 = 0
Solution
The given quadratic equation is √3x^{2} + 10x + √3 = 0
3√3x^{2} + 10x + √3 = 0
⇒ 3√3x^{2} + 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 2x^{2} + 8x + k = 0 are equal then find the value of k.
Solution
It is given that the roots of the quadratic equation 2x^{2} + 8x + k = 0 are equal.
∴ D = 0
⇒ 8^{2} – 4 × 2 × k = 0
⇒ 64 – 8k = 0
⇒ k = 8
Hence, the value of k is 8.
39. If the quadratic px^{2}  2√5px + 15 = 0 has two equal roots then find the value of p.
Solution
It is given that the quadratic equation px^{2}  2√5px + 15 = 0 has two equal roots.
∴ D = 0
⇒ (2√p)^{2} – 4 × p × 15 = 0
⇒ 20p^{2} – 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 ay^{2} + ay + 3 = 0 and y^{2} + y + b = 0 then find the value of ab.
Solution
It is given that y = 1 is a root of the equation ay^{2} + 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 x^{2} – 4x + 1 is (2 + √3), write the other zero.
Solution
Let the other zero of the given polynomial be Î±.
Now,
Sum of zeros of the given polynomial = (4)/1 = 4
∴ Î± + (2 + √3) = 4
⇒ Î± = 4 – 2  √3
= 2  √3
Hence, the other zero of the given polynomial is (2  √3).
42. If one root of the quadratic equation 3x^{2} – 10x + k = 0 is reciprocal of the other, find the value of k.
Solution
Let Î± and Î² be the roots of the equation 3x^{2} + 10x + k = 0.
∴ Î± = 1/Î² (Given)
⇒ Î±Î² = 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
⇒ 4p^{2} – 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 x^{2} – 4kx + k = 0 has equal roots.
Solution
It is given that the quadratic equation x^{2} – 4kx + k = 0 has equal roots.
∴ D = 0
⇒ (4k)^{2} – 4 × 1 × k = 0
⇒ 16k^{2} – 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 9x^{2} – 3kx + k = 0 has equal roots.
Solution
It is given that the quadratic equation 9x^{2} – 3kx + k = 0 has equal roots.
∴ D = 0
⇒ (3k)^{2} – 4 × 9 × k = 0
⇒ 9k^{2} – 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 x^{2} – (√3 + 1)x + √3 = 0
Solution
x^{2} – (√3 + 1)x + √3 = 0
⇒ x^{2}  √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 2x^{2} + ax – a^{2} = 0
Solution
2x^{2} + ax – a^{2} = 0
⇒ 2x^{2} + 2ax – ax – a^{2} = 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: 3x^{2} + 5√5x – 10 = 0
Solution
3x^{2} + 5√5x – 10 = 0
⇒ 3x^{2} + 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 √3x^{2} + 10x  8√3 = 0
Solution
√3x^{2} + 10x  8√3 = 0
⇒ 3x^{2} + 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: √3x^{2 } 2√2x  2√3 = 0
Solution
√3x^{2}  2√2x  2√3 = 0
⇒ √3x^{2}  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√3x^{2} + 5x  2√3 = 0
Solution
4√3x^{2} + 5x  2√3 = 0
⇒ 4√3x^{2} + 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: 4x^{2}+ 4bx – (a^{2} – b^{2}) = 0
Solution
4x^{2} + 4bx – (a^{2}  b^{2}) = 0
⇒ 4x^{2} + 4bx – (a – b)(a + b) = 0
⇒ 4x^{2 }+ 2[(a + b) – (a – b)]x – (a – b)(a + b) = 0
⇒ 4x^{2 }+ 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 x^{2} + 5x  (a^{2} + a – 6) = 0
Solution
x^{2 }+ 5x – (a^{2} + a – 6) = 0
⇒ x^{2} + 5x – (a + 3)(a – 2) = 0
⇒ x^{2} + [(a + 3) – (a – 2)]x – (a + 3)(a – 2) = 0
⇒ x^{2} + (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 x^{2} + 6x – (a^{2} + 2a – 8) = 0
Solution
x^{2 }+ 6x – (a^{2} – 2a – 8) = 0
⇒ x^{2} + 6x – (a + 4)(a – 2) = 0
⇒ x^{2} + [(a + 4) – (a – 2)]x – (a + 4)(a – 2) = 0
⇒ x^{2} + (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 x^{2} – 4ax + 4a^{2} – b^{2} = 0
Solution
x^{2} – 4ax + 4a^{2} – b^{2 }= 0
⇒ x^{2} – 4ax + (2a + b)(2a – b) = 0
⇒ x^{2 }– [(2a + b) + (2a – b)]x + (2a + b)(2a – b) = 0
⇒ x^{2} – (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.