Nature of roots — Class 10 CBSE Mathematics MCQs with Solutions
Free Class 10 CBSE Mathematics Nature of roots MCQs with step-by-step solutions (8 questions). Part of Quadratic Equations. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Nature of roots · easy · theory
The discriminant of the quadratic equation $2x^2 - 4x + 3 = 0$ is:
A. $8$
B. $-8$ ✓ Correct
C. $40$
D. $-40$
Solution: The discriminant is $D = b^2 - 4ac$ with $a = 2$, $b = -4$, $c = 3$. So $D = (-4)^2 - 4(2)(3) = 16 - 24 = -8$. Since $D < 0$, this equation has no real roots; writing $16 + 24 = 40$ is the common sign error.
Q2 — Nature of roots · easy · theory
If the discriminant $b^2 - 4ac > 0$ for a quadratic equation, then the equation has:
A. two distinct real roots ✓ Correct
B. two equal real roots
C. no real roots
D. exactly one real root
Solution: When $D = b^2 - 4ac > 0$, the square root $\sqrt{D}$ is a positive real number, so the formula $x = \frac{-b \pm \sqrt{D}}{2a}$ gives two different real values. Equal roots occur only when $D = 0$, and no real roots when $D < 0$. A quadratic equation can never have exactly one real root unless the two roots coincide (the $D = 0$ case, which counts as two equal roots).
Q3 — Nature of roots · medium · theory
The values of $k$ for which the equation $2x^2 + kx + 3 = 0$ has two equal real roots are:
A. $k = 2\sqrt{6}$ only
B. $k = \pm\sqrt{6}$
C. $k = \pm 12$
D. $k = \pm 2\sqrt{6}$ ✓ Correct
Solution: For equal roots, the discriminant must be zero: $k^2 - 4(2)(3) = 0$, so $k^2 = 24$ and $k = \pm\sqrt{24} = \pm 2\sqrt{6}$. Both values are valid — dropping the negative value (writing $2\sqrt{6}$ only) loses half the answer.
Q4 — Nature of roots · medium · theory
The roots of the equation $3x^2 - 4\sqrt{3}x + 4 = 0$ are:
A. real and distinct
B. not real
C. real and equal ✓ Correct
D. real, distinct and irrational
Solution: The discriminant is $D = (-4\sqrt{3})^2 - 4(3)(4) = 48 - 48 = 0$. Since $D = 0$, the equation has two equal real roots, each equal to $\frac{-b}{2a} = \frac{4\sqrt{3}}{6} = \frac{2}{\sqrt{3}}$. Note that $(-4\sqrt{3})^2 = 16 \times 3 = 48$, not $-48$.
Q5 — Nature of roots · medium · theory
The nature of the roots of the quadratic equation $2x^2 - 6x + 3 = 0$ is:
A. two distinct real roots ✓ Correct
B. two equal real roots
C. no real roots
D. cannot be determined without solving
Solution: The discriminant is $D = (-6)^2 - 4(2)(3) = 36 - 24 = 12$. Since $D > 0$, the equation has two distinct real roots (in fact $x = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}$). The discriminant alone decides the nature of the roots — no need to solve the equation fully.
Q6 — Nature of roots · hard · theory
The value of $k$ for which the equation $kx^2 - 2kx + 6 = 0$ has two equal real roots is:
A. $k = 0$ or $k = 6$
B. $k = 6$ only ✓ Correct
C. $k = -6$
D. $k = 3$
Solution: For equal roots, $D = (-2k)^2 - 4(k)(6) = 4k^2 - 24k = 0$, so $4k(k - 6) = 0$, giving $k = 0$ or $k = 6$. But if $k = 0$ the equation becomes $6 = 0$, which is not a quadratic equation at all, so $k = 0$ must be rejected. Hence $k = 6$; check: $6x^2 - 12x + 6 = 6(x-1)^2$, which has the repeated root $x = 1$.
Q7 — Nature of roots · hard · theory
Which of the following equations has no real roots?
A. $x^2 + 4x - 3\sqrt{2} = 0$
B. $x^2 - 4x - 3\sqrt{2} = 0$
C. $3x^2 + 4\sqrt{3}x + 4 = 0$
D. $x^2 - 4x + 3\sqrt{2} = 0$ ✓ Correct
Solution: For $x^2 - 4x + 3\sqrt{2} = 0$, the discriminant is $16 - 12\sqrt{2} \approx 16 - 16.97 < 0$, so it has no real roots. In the first two options the constant term is negative, making $D = 16 + 12\sqrt{2} > 0$ (real roots exist), and for $3x^2 + 4\sqrt{3}x + 4 = 0$ we get $D = 48 - 48 = 0$ (real equal roots). The key comparison is $12\sqrt{2} > 16$ since $\sqrt{2} \approx 1.414$.
Q8 — Nature of roots · hard · theory
The equation $2x^2 + kx + 2 = 0$ has real roots only when:
A. $-4 < k < 4$
B. $k \geq 4$ only
C. $k \leq -4$ or $k \geq 4$ ✓ Correct
D. $k = \pm 4$ only
Solution: For real roots, the discriminant must be non-negative: $k^2 - 4(2)(2) \geq 0$, i.e. $k^2 \geq 16$. This means $|k| \geq 4$, so $k \leq -4$ or $k \geq 4$ — remember that $k^2 \geq 16$ includes large negative values of $k$ too. For $-4 < k < 4$ the discriminant is negative and there are no real roots, while $k = \pm 4$ gives exactly equal roots (a special case, not the whole answer).