Class 10 Maths MCQ – Quadratic Equations – Determination of Types of Roots

This set of Class 10 Maths Chapter 4 Multiple Choice Questions & Answers (MCQs) focuses on “Quadratic Equations – Determination of Types of Roots”.

1. For the equation x2 + 5x – 1, which of the following statements is correct?
a) The roots of the equation are equal
b) The discriminant of the equation is negative
c) The roots of the equation are real, distinct and irrational
d) The discriminant is equal to zero
View Answer

Answer: c
Explanation: Roots are real. ∴ b2 – 4ac ≥ 0
52 – 4(1)(1)
25 – 4 = 21 which is greater than 0. Hence, the discriminant of the equation is greater than zero, so roots are real.

2. If the roots of the equation ax2 + bx + c are real and equal, what will be the relation between a, b, c?
a) b = ±\(\sqrt {ac}\)
b) b = ±\(\sqrt {4c}\)
c) b = ±\(\sqrt {-4ac}\)
d) b = ±\(\sqrt {4ac}\)
View Answer

Answer: d
Explanation: Roots are real and equal. ∴ b2 – 4ac = 0
b2 = 4ac
b = ±\(\sqrt {4ac}\)

3. What will be the value of k, so that the roots of the equation are x2 + 2kx + 9 are imaginary?
a) -5 < k < 5
b) -3 < k < 3
c) 3 < k < -3
d) -5 < k < 3
View Answer

Answer: b
Explanation: Roots are imaginary. ∴ b2 – 4ac < 0
(2k)2 – 4(9)(1) < 0
4k2 – 36 < 0
k2 – 9 < 0
k2 < 9
k < ±3
-3 < k < 3
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4. What will be the nature of the roots of the quadratic equation 5x2 – 11x + 13?
a) Imaginary
b) Real
c) Irrational
d) Equal
View Answer

Answer: a
Explanation: To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
Discriminant = b2 – 4ac = – 112 – 4 × 5 × 13 = 121 – 260 = – 139
Since discriminant is less than zero, the roots of the equation are imaginary.

5. What will be the nature of the roots of the quadratic equation x2 + 10x + 25?
a) Imaginary
b) Real
c) Irrational
d) Equal
View Answer

Answer: d
Explanation: To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
Discriminant = b2 – 4ac = 102 – 4 × 25 × 1 = 100 – 100 = 0
Since discriminant is equal to zero, the roots of the equation are equal.
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6. What will be the nature of the roots of the quadratic equation 2x2 + 10x + 9?
a) Imaginary
b) Real
c) Irrational
d) Equal
View Answer

Answer: b
Explanation: To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
Discriminant = b2 – 4ac = 102 – 4 × 2 × 9 = 100 – 72 = 28
Since discriminant is greater than zero, the roots of the equation are real and distinct.

7. The equation 9x2 – 2x + 5 is not true for any real value of x.
a) False
b) True
View Answer

Answer: b
Explanation: To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
Discriminant = b2 – 4ac = -22 – 4 × 9 × 5 = 4 – 180 = -176
Since discriminant is less than zero, the roots of the equation are imaginary. Hence, for any real value of x the equation is not true.
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8. The value of p for which the equation 8x2 + 9px + 15 has equal roots is \(\frac {4\sqrt {30}}{9}\).
a) True
b) False
View Answer

Answer: a
Explanation: Roots are equal. ∴ b2 – 4ac = 0
(9p)2 – 4(8)(15) = 0
81p2 – 480 = 0
81p2 = 480
p = ± \(\sqrt {\frac {480}{81}}\) = ± \(\frac {4\sqrt {30}}{9}\)

9. What will be the value of a, for which the equation 5x2 + ax + 5 and x2 – 12x + a will have real roots?
a) a = 37
b) 10 < a < 36
c) 36 < a < 10
d) a = 9
View Answer

Answer: b
Explanation: The roots of both the equations are real.
Discriminant of 5x2 + ax + 5 : b2 – 4ac = a2 – 4 × 5 × 5 = a2 – 100
Since, roots are real; discriminant will be greater than 0.
a2 ≥ 100
a ≥ ±10
Now, discriminant of x2 – 12x + a : b2 – 4ac = -122 – 4 × 1 × a = 144 – 4a
Since, roots are real; discriminant will be greater than 0.
144 ≥ 4a
a ≤ \(\frac {144}{4}\) = 36
For both the equations to have real roots the value of a must lie between 36 and 10.
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10. What will be the value of k, if the roots of the equation (k – 4)x2 – 2kx + (k + 5) = 0 are equal?
a) 18
b) 19
c) 20
d) 21
View Answer

Answer: c
Explanation: Roots are equal. ∴ b2 – 4ac = 0
-(2k)2 – 4(k – 4)(k + 5) = 0
4k2 – 4(k2 – 4k + 5k – 20) = 0
4k2 – 4(k2 + k – 20) = 0
4k2 – 4k2 – 4k + 80 = 0
-4k = – 80
k = \(\frac {-80}{-4}\) = 20

Sanfoundry Global Education & Learning Series – Mathematics – Class 10.

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