Quadratic Equations — Class 10 CBSE Mathematics MCQs with Solutions

Solution: A quadratic equation has the form $ax^2 + bx + c = 0$ where $a \neq 0$ and the highest power of $x$ is 2. In $x^2 - 5x + 6 = 0$, we have $a=1$, $b=-5$, $c=6$, making it quadratic. The other equations are linear, cubic, or rational—not quadratic.

Solution: If $a = 0$, the equation $0 \cdot x^2 + bx + c = 0$ becomes $bx + c = 0$, which is linear, not quadratic. Additionally, the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ requires $a \neq 0$ in the denominator. The condition $a \neq 0$ is what distinguishes a quadratic equation from a linear one.

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.

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.

Solution: By Vieta's formulas, for roots $\alpha$ and $\beta$ of $ax^2 + bx + c = 0$, we have $\alpha + \beta = -\frac{b}{a}$. This relationship connects the sum of roots directly to the coefficients, without needing to compute the roots themselves.

Solution: By Vieta's formulas, the product of roots $\alpha \cdot \beta = \frac{c}{a}$. In $2x^2 - 8x + 6 = 0$, we have $a=2$ and $c=6$, so the product is $\frac{6}{2} = 3$. Note: the sum of roots would be $-\frac{-8}{2} = 4$, not the product.

Solution: Factorisation works when the quadratic can be written as $(px+q)(rx+s) = 0$, which happens only when the roots are real. Here, $\Delta = 1^2 - 4(1)(1) = -3 < 0$, so the roots are complex. Without real roots, factorisation into real linear factors is impossible.

Solution: When we complete the square: $a(x^2 + \frac{b}{a}x) + c = 0$ becomes $a(x + \frac{b}{2a})^2 = \frac{b^2-4ac}{4a}$, so $(x + \frac{b}{2a})^2 = \frac{b^2-4ac}{4a^2}$. Taking the square root: $x + \frac{b}{2a} = \pm\sqrt{\frac{b^2-4ac}{4a^2}}$, giving $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. The $\pm$ reflects the fundamental fact that a square root has two values (positive and negative).

Solution: Expanding $(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$, we get $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. Using Vieta's formulas: $\alpha^2 + \beta^2 = (-\frac{b}{a})^2 - 2(\frac{c}{a}) = \frac{b^2}{a^2} - \frac{2c}{a} = \frac{b^2-2ac}{a^2}$. This expresses a property of roots using only coefficients—a powerful technique in algebra.

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).

Solution: For a monic quadratic $x^2 + px + q = 0$ with roots whose sum is $S$ and product is $P$, we have $p = -S$ and $q = P$. Here, $p = -7$ and $q = 10$, giving $x^2 - 7x + 10 = 0$. The discriminant is $(-7)^2 - 4(1)(10) = 49 - 40 = 9 > 0$, confirming two distinct real roots.

Solution: A perfect-square quadratic $(x-k)^2 = x^2 - 2kx + k^2$ has the form where the discriminant $\Delta = (-2k)^2 - 4(1)(k^2) = 4k^2 - 4k^2 = 0$. This zero discriminant corresponds to the repeated root $x = k$ with multiplicity 2. Perfect squares are the unique case where a quadratic factors into an identical linear factor twice.

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.

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.

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.

Solution: If $a > 0$, $b > 0$, $c > 0$, and $x > 0$, then $ax^2 > 0$, $bx > 0$, and $c > 0$, so $ax^2 + bx + c > 0$. The quadratic is always positive for positive $x$, meaning the parabola never crosses the x-axis on the right side. Any real roots must be non-positive. This illustrates Descartes' Rule of Signs: the number of positive roots equals the number of sign changes in the coefficient sequence.

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.

Solution: Factorisation reduces a quadratic to the product form $(px+q)(rx+s) = 0$. By the zero-product property, this is satisfied only when $px+q=0$ or $rx+s=0$, giving us two linear equations to solve. Here, $2x+1=0 \Rightarrow x=-\frac{1}{2}$ and $x-3=0 \Rightarrow x=3$. This is the conceptual basis for why factorisation works.

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.

Solution: The sum of roots is $\alpha + \beta = 2 + (-3) = -1$, and the product is $\alpha \beta = 2 \cdot (-3) = -6$. By Vieta's formulas (reversed), the equation is $x^2 - (\text{sum})x + (\text{product}) = 0$, giving $x^2 - (-1)x + (-6) = 0$ or $x^2 + x - 6 = 0$. This demonstrates reconstructing a quadratic from root properties.

Solution: Factorisation is strategic. Here, we can factor out 3: $3(x^2 - 4x + 3) = 0$, simplifying to $x^2 - 4x + 3 = 0$, which factors as $(x-1)(x-3)=0$ by inspection. This avoids fractions and is faster than applying the quadratic formula. The choice between factorisation and formula depends on recognizing factorable patterns.

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.

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.

Solution: Factorisation over the reals works only when the quadratic has real (typically rational) roots, so it can be written as $(px+q)(rx+s)=0$. The quadratic formula, however, applies universally—it produces real roots when $\Delta \geq 0$, complex roots when $\Delta < 0$. This makes the formula the universal tool for solving any quadratic.