Quadratic Equations: 19 MCQs with answers

$ax^2 + bx + c = 0$

A quadratic equation has the form $ax^2 + bx + c = 0$ with $a \neq 0$. The highest power of the variable must be 2.

$a \neq 0$

If $a = 0$, the $x^2$ term vanishes and the equation becomes linear. The condition $a \neq 0$ is essential.

$b^2 - 4ac$

The discriminant is $D = b^2 - 4ac$. It is the expression under the square root in the quadratic formula.

Two distinct real roots

When $D > 0$, the square root is a positive real number, giving two different real values for $x$.

Two equal real roots

When $D = 0$, the $\pm$ part adds and subtracts zero, so both roots are the same: $x = -b/(2a)$.

No real roots

When $D < 0$, the square root of a negative number is not real, so the equation has no real roots (the roots are complex conjugates).

Upward

The sign of $a$ determines the direction. If $a > 0$, the parabola opens upward (like a U). If $a < 0$, it opens downward.

True

The quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ solves every quadratic equation with $a \neq 0$.

$-\dfrac{b}{a}$

The sum of the roots is $\alpha + \beta = -\dfrac{b}{a}$. This comes from expanding $(x - \alpha)(x - \beta) = 0$.

$\dfrac{c}{a}$

The product of the roots is $\alpha\beta = \dfrac{c}{a}$. This is a standard result from comparing coefficients.

Quadratic equation

The height depends on $t^2$, making it a quadratic equation. Projectile motion is a classic real-world application of quadratics.

$x = -\dfrac{b}{2a}$

The vertex (turning point) of the parabola is at $x = -\dfrac{b}{2a}$, which is also the axis of symmetry.

Factorisation

Factorisation is fastest when roots are simple integers. Here $(x-2)(x-3)=0$ gives roots 2 and 3 immediately.

The $x$-intercepts

The roots are the values of $x$ where $y = 0$, i.e., where the parabola crosses the $x$-axis. These are the $x$-intercepts.

Forgot to include the $\pm$ and missed the second root

The quadratic formula has a $\pm$, meaning there are two roots. The student only computed the $+$ case and missed $x = \frac{-3 - 7}{4} = -\frac{5}{2}$.

$x^2 - 5x + 6 = 0$

If roots are $\alpha$ and $\beta$, the equation is $x^2 - (\alpha+\beta)x + \alpha\beta = 0$. Substituting sum $=5$ and product $=6$ gives $x^2 - 5x + 6 = 0$.

8

The roots are equal when the discriminant \(D = b^2 - 4ac = 0\). Here, \(D = (-4)^2 - 4(2)(k) = 16 - 8k\). Setting this to zero gives \(16 - 8k = 0\) or \(k = 2\).

6

The sum of the roots for a quadratic \(ax^2 + bx + c = 0\) is given by \(-rac{b}{a}\). Here, \(b = -6\) and \(a = 1\), thus the sum of the roots is \(-rac{-6}{1} = 6$.

12

The product of the roots is given by \(r_1 r_2 = rac{c}{3}\). If \(r_1 r_2 = 4\), then \(4 = rac{c}{3}\) leads to \(c = 12\).