Quadratic Equations: 15 MCQs with answers

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

A quadratic equation is of the form $ax^2 + bx + c = 0$ with $a \neq 0$. Only $x^2 - 5x + 6 = 0$ fits this form.

$ax^2 + bx + c = 0$

The standard form of a quadratic equation in variable $x$ is $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

$b^2 - 4ac$

The discriminant, denoted by $D$ or $\Delta$, is $b^2 - 4ac$. It determines the nature of the roots.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, derived by completing the square.

Substitution method

Substitution method is used for solving systems of linear equations, not a standard method for a single quadratic equation.

$3$ and $-3$

$x^2 - 9 = (x-3)(x+3) = 0$, so $x = 3$ or $x = -3$.

True

It is of the form $ax^2 + bx + c = 0$ with $a = 3 \neq 0$, so it is a quadratic equation.

True

A quadratic equation is of degree 2, so it can have at most two roots (real or complex).

$3$ and $4$

Factorising: $x^2 - 7x + 12 = (x-3)(x-4) = 0$, so $x = 3$ or $x = 4$.

Real and equal

When $D = 0$, the quadratic equation has two equal real roots, given by $x = -\frac{b}{2a}$.

$-\frac{b}{a}$

If $\alpha$ and $\beta$ are the roots, then $\alpha + \beta = -\frac{b}{a}$.

$\frac{c}{a}$

If $\alpha$ and $\beta$ are the roots, then $\alpha \beta = \frac{c}{a}$.

Two distinct real roots

When $D = b^2 - 4ac > 0$, the equation has two distinct real roots.

$-2, -2$

$x^2 + 4x + 4 = (x+2)^2 = 0$, so $x = -2$ (twice). The roots are real and equal.

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

If roots are $2$ and $3$, sum $= 5$, product $= 6$. The equation is $x^2 - (\text{sum})x + (\text{product}) = 0 \Rightarrow x^2 - 5x + 6 = 0$.