Quadratic Equations

What you will learn

• Solve quadratics by factorising, the formula and completing the square.
• Use the discriminant to count solutions.
• Link roots to where the parabola crosses the x-axis.

Before you start: you should be able to factorise expressions and rearrange formulas. See Factorising with a Common Factor.

The idea

A quadratic equation contains a squared unknown and no higher power. Written in standard form it looks like

$ax^2 + bx + c = 0$

where $a$, $b$ and $c$ are numbers and $a$ is not zero.

The graph of $y = ax^2 + bx + c$ is a parabola, a symmetrical U-shaped curve (upside-down if $a$ is negative). Solving the equation means finding where that curve crosses the $x$-axis. A parabola can cross twice, touch once, or miss altogether, which is why a quadratic has two, one or zero real solutions.

Three methods

1. Factorising: quickest when the numbers are friendly.
2. Completing the square: reveals the turning point of the parabola.
3. The quadratic formula: always works.

All three rely on the same fact: if two numbers multiply to give zero, at least one of them must be zero.

Worked examples

Example 1: Factorising

Solve $x^2 - x - 12 = 0$.

1. Find two numbers that multiply to $-12$ and add to $-1$. They are $-4$ and $+3$.
2. Write the factorised form: $(x - 4)(x + 3) = 0$.
3. Set each bracket equal to zero: $x - 4 = 0$ or $x + 3 = 0$.
4. Check $x = 4$: $16 - 4 - 12 = 0$. ✓

Answer: $x = 4$ or $x = -3$

Example 2: The quadratic formula

Solve $2x^2 + 3x - 4 = 0$, giving answers to 2 decimal places.

1. Identify $a = 2$, $b = 3$, $c = -4$.
2. Find the discriminant: $b^2 - 4ac = 9 - 4(2)(-4) = 9 + 32 = 41$.
3. It is positive, so there are two roots: $x = \dfrac{-3 \pm \sqrt{41}}{4}$.
4. Evaluate: $\sqrt{41} \approx 6.403$, so $x \approx \dfrac{3.403}{4}$ or $x \approx \dfrac{-9.403}{4}$.

Answer: $x \approx 0.85$ or $x \approx -2.35$

Example 3: Completing the square

Write $x^2 + 6x + 2$ in completed-square form and find the turning point.

1. Halve the coefficient of $x$: $6 \div 2 = 3$.
2. $(x + 3)^2 = x^2 + 6x + 9$, which is $7$ too big, so $x^2 + 6x + 2 = (x + 3)^2 - 7$.
3. The squared bracket is smallest (zero) when $x = -3$, giving $y = -7$.

Answer: $(x + 3)^2 - 7$, with minimum point $(-3, -7)$.

Before using the formula, rearrange the equation so one side is zero. $x^2 = 5x$ becomes $x^2 - 5x = 0$. Do not divide by $x$, or you will lose the solution $x = 0$.

Why it matters

Anything thrown or launched under gravity follows a parabola, so quadratics predict where a basketball lands or how high a fountain jet rises. Satellite dishes and car headlights use parabolic reflectors because of a special focusing property of the curve.

Quadratics also appear in area problems (a garden $3$ m longer than it is wide with area $40$ m² gives $x(x + 3) = 40$) and in business, where profit often rises then falls as price increases, with the best price at the turning point.