Nature of roots — Class 10 CBSE Mathematics MCQs with Solutions
Free Class 10 CBSE Mathematics Nature of roots MCQs with step-by-step solutions (10 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
If the discriminant $\Delta = b^2 - 4ac < 0$ for a quadratic equation, what can we conclude about its roots?
A. The equation has two distinct real roots
B. The equation has two equal real roots
C. The equation has no real roots; roots are complex ✓ Correct
D. The equation has exactly one root
Solution: The discriminant $\Delta = b^2 - 4ac$ determines the nature of roots. When $\Delta < 0$, the term $\sqrt{b^2-4ac}$ is not real, so the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ yields complex roots. Geometrically, this means the parabola does not intersect the x-axis.
Q2 — Nature of roots · easy · theory
What does it mean for a quadratic equation when $\Delta = b^2 - 4ac = 0$?
A. The equation has no real roots
B. The equation has two distinct real roots
C. The equation has two equal real roots (a repeated root) ✓ Correct
D. The equation is not quadratic
Solution: When $\Delta = 0$, the quadratic formula gives $x = \frac{-b \pm 0}{2a} = \frac{-b}{2a}$, a single repeated root. Graphically, the parabola just touches the x-axis at one point, confirming the root has multiplicity 2.
Q3 — Nature of roots · medium · theory
A quadratic $x^2 + px + q = 0$ has equal roots. What must be the relationship between $p$ and $q$?
A. $p^2 = 4q$ ✓ Correct
B. $p = 2q$
C. $p^2 + q^2 = 0$
D. $q = p^2$
Solution: Equal roots occur when the discriminant $\Delta = 0$. For $x^2 + px + q = 0$, we have $\Delta = p^2 - 4(1)(q) = 0$, which gives $p^2 = 4q$. This is the condition connecting the coefficients when roots coincide. Geometrically, the parabola is tangent to the x-axis.
Q4 — Nature of roots · medium · theory
Why does the parabola corresponding to $y = x^2 - 6x + 9$ touch the x-axis at exactly one point?
A. Because the discriminant is zero, meaning the quadratic has a double (repeated) root ✓ Correct
B. Because the coefficient of $x$ is negative
C. Because the constant term is positive
D. Because the parabola opens upward
Solution: The discriminant is $(-6)^2 - 4(1)(9) = 36 - 36 = 0$. A zero discriminant means a repeated root ($x = 3$ with multiplicity 2). Geometrically, this is a parabola tangent to the x-axis—touching at one point without crossing. The equation $x^2 - 6x + 9 = (x-3)^2 = 0$ confirms this.
Q5 — Nature of roots · medium · theory
Consider the statement: 'If the discriminant of a quadratic is positive, the parabola must cross the x-axis at two points.' Is this true, and why?
A. Yes, because a positive discriminant guarantees two distinct real roots ✓ Correct
B. No, because a positive discriminant means the roots are irrational
C. Yes, but only if the leading coefficient is positive
D. No, because the parabola might not intersect the x-axis
Solution: $\Delta > 0$ implies $\sqrt{b^2-4ac}$ is real and non-zero, giving $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ as two distinct real values. These correspond to the two x-intercepts of the parabola. Whether the parabola opens up or down (sign of $a$) does not affect the existence of two intersection points.
Q6 — Nature of roots · hard · theory
For a quadratic $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, why does $\Delta = b^2 - 4ac = a^2(\alpha - \beta)^2$ hold geometrically?
A. Because the discriminant measures the square of the difference between the roots scaled by $a^2$, determining how far apart the two x-intercepts are ✓ Correct
B. Because the discriminant is always positive when roots are real
C. Because $\alpha - \beta$ is always equal to $b/a$
D. Because the parabola's vertex lies on the x-axis
Solution: From Vieta: $\alpha - \beta = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = \sqrt{\frac{b^2}{a^2} - \frac{4c}{a}} = \frac{\sqrt{b^2-4ac}}{|a|}$. Thus $a^2(\alpha - \beta)^2 = b^2 - 4ac = \Delta$. Geometrically, $|\alpha - \beta|$ is the distance between the x-intercepts. When $\Delta = 0$, the roots coincide (one point); when $\Delta > 0$, they separate (two points).
Q7 — Nature of roots · hard · theory
If $p$ is a positive integer and the quadratic $x^2 - px + p = 0$ has real roots, what can be concluded about the range of $p$?
A. $p \geq 4$ (because the discriminant $p^2 - 4p \geq 0$ requires $p(p-4) \geq 0$, and for positive $p$ this means $p \geq 4$) ✓ Correct
B. $p \geq 1$ (any positive integer works)
C. $p < 4$ (the discriminant must be negative for real roots)
D. $p = 4$ (only one value works)
Solution: For real roots, we need $\Delta \geq 0$: $(-p)^2 - 4(1)(p) \geq 0$ gives $p^2 - 4p \geq 0$, so $p(p-4) \geq 0$. Since $p > 0$ (positive integer), we need $p - 4 \geq 0$, thus $p \geq 4$. This shows how discriminant conditions constrain parameters.
Q8 — Nature of roots · hard · theory
A quadratic $ax^2 + bx + c = 0$ with $a > 0$ has no real roots. What does this imply about the parabola $y = ax^2 + bx + c$, and why does the sign of $a$ matter?
A. The parabola opens upward ($a > 0$) and lies entirely above the x-axis (never touches or crosses it), consistent with the absence of real roots. ✓ Correct
B. The parabola opens downward and touches the x-axis at one point.
C. The parabola opens upward but must cross the x-axis.
D. The sign of $a$ does not affect where the parabola is positioned relative to the x-axis.
Solution: When $a > 0$, the parabola opens upward. If $\Delta < 0$ (no real roots), the quadratic $ax^2 + bx + c$ has no x-intercepts. Since the parabola opens upward and has a minimum at the vertex, if this minimum value is positive, the entire curve lies above the x-axis. Conversely, if $a < 0$ with $\Delta < 0$, the parabola opens downward and lies entirely below the x-axis.
Q9 — Nature of roots · hard · theory
For a quadratic $f(x) = ax^2 + bx + c$, the vertex is at $x = -\frac{b}{2a}$, and the minimum/maximum value is $f(-\frac{b}{2a}) = -\frac{\Delta}{4a}$. How does this formula explain why $\Delta < 0$ guarantees no real roots when $a > 0$?
A. When $\Delta < 0$ and $a > 0$, the term $-\frac{\Delta}{4a}$ becomes positive, so the minimum value is positive, meaning the entire parabola stays above the x-axis. ✓ Correct
B. The vertex formula only applies when the discriminant is positive.
C. If $\Delta < 0$, the parabola has no vertex.
D. The minimum value is always negative when $\Delta < 0$.
Solution: The parabola $y = ax^2 + bx + c$ with $a > 0$ reaches its minimum at the vertex. This minimum value is $-\frac{\Delta}{4a} = -\frac{b^2-4ac}{4a}$. When $\Delta < 0$, we have $b^2 < 4ac$, so $-\frac{\Delta}{4a} = \frac{4ac-b^2}{4a} > 0$. Thus, the minimum $y$-value is positive, and since the parabola opens upward, it never touches the x-axis—confirming no real roots. This connects the discriminant directly to the parabola's position.
Q10 — Nature of roots · hard · theory
Given that the quadratic $3x^2 - kx + 12 = 0$ has two equal roots, determine the value(s) of $k$ and explain the geometric significance of equal roots.
A. $k = \pm 12$ (from $\Delta = 0$: $k^2 - 4(3)(12) = 0$ gives $k^2 = 144$). Equal roots mean the parabola is tangent to the x-axis. ✓ Correct
B. $k = 12$ only; equal roots mean the parabola opens upward.
C. $k = -12$ only; equal roots mean the two roots coincide.
D. $k = 6$; equal roots have no special geometric meaning.
Solution: For equal roots, $\Delta = 0$: $(-k)^2 - 4(3)(12) = 0$ gives $k^2 = 144$, so $k = 12$ or $k = -12$. Both values make the quadratic a perfect square: when $k=12$, we get $3x^2 - 12x + 12 = 3(x-2)^2$; when $k=-12$, we get $3x^2 + 12x + 12 = 3(x+2)^2$. Geometrically, the parabola touches the x-axis at exactly one point (at $x=2$ or $x=-2$ respectively)—tangent, not intersecting.