What is sign errors in the formula?

With c negative, -4ac becomes positive. For 2x^2 + 3x - 4, the term is -4(2)(-4) = +32.

SingaporeMathsSyllabus dot point

How do we solve a quadratic equation, where the unknown appears squared, by factorising or using the formula?

Solve quadratic equations of the form ax^2 + bx + c = 0 by factorisation and by the quadratic formula

A focused answer to the N(A)-Level Mathematics outcome on quadratic equations. The zero product rule, solving by factorisation, the quadratic formula, and recognising when each is appropriate.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to solve quadratic equations - equations of the form $ax^2 + bx + c = 0$ where the unknown appears squared - by factorising when possible and by the quadratic formula when it does not factorise neatly. A quadratic usually has two solutions (also called roots), and finding both is part of the answer.

The answer

The standard form and the zero product rule

A quadratic equation is first written with everything on one side, equal to zero: $ax^2 + bx + c = 0$ . The key idea is the zero product rule: if two things multiply to give zero, at least one of them must be zero. So once the left side is factorised into two brackets, each bracket can be set to zero in turn.

Solving by factorisation

When the quadratic factorises into two brackets, solving is quick.

Make sure the equation equals zero.
Factorise the quadratic into two brackets.
Set each bracket equal to zero and solve.

For $x^2 + 5x + 6 = 0$ , factorise to $(x + 2)(x + 3) = 0$ , so $x = -2$ or $x = -3$ . To factorise, find two numbers that multiply to the constant term and add to the coefficient of $x$ .

A common factor or a difference of two squares

Some quadratics factorise specially. If every term shares a factor, take it out first: $2x^2 - 8x = 0$ becomes $2x(x - 4) = 0$ , so $x = 0$ or $x = 4$ . A difference of two squares factorises as $x^2 - 9 = (x - 3)(x + 3)$ , giving $x = 3$ or $x = -3$ .

The quadratic formula

When a quadratic does not factorise neatly, use the formula:

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

Read off $a$ , $b$ and $c$ from $ax^2 + bx + c = 0$ , substitute carefully (watching signs), and the $\pm$ gives the two solutions. The quantity $b^2 - 4ac$ under the root is called the discriminant.

Checking and rounding

Substitute a solution back to confirm it satisfies the equation. When the formula gives a surd, round to the accuracy the question asks for, usually $2$ decimal places or $3$ significant figures.

Worked example

Solve $x^2 + 2x - 15 = 0$ by factorisation.

Step 1: Confirm it equals zero

The equation is already $x^2 + 2x - 15 = 0$ , so no rearranging is needed.

Step 2: Find two numbers

We need two numbers that multiply to $-15$ and add to $+2$ . The pair $+5$ and $-3$ works, since $5 \times (-3) = -15$ and $5 + (-3) = 2$ .

Step 3: Factorise

x^2 + 2x - 15 = (x + 5)(x - 3) = 0.

Step 4: Solve each bracket

x + 5 = 0 \Rightarrow x = -5, \qquad x - 3 = 0 \Rightarrow x = 3.

So $x = -5$ or $x = 3$ . Check: $(-5)^2 + 2(-5) - 15 = 25 - 10 - 15 = 0$ , correct.

Examples in context

Example 1. Area of a rectangle. A rectangle has length $x + 2$ and width $x$ , and an area of $24$ . Then $x(x + 2) = 24$ , so $x^2 + 2x - 24 = 0$ , which factorises to $(x + 6)(x - 4) = 0$ . Since a length cannot be negative, $x = 4$ , giving dimensions $4$ and $6$ . Rejecting the negative root is part of a sensible real-world answer.

Example 2. Where a curve crosses the x-axis. The curve $y = x^2 - 4x + 3$ meets the $x$ -axis where $y = 0$ , that is $x^2 - 4x + 3 = 0$ . Factorising gives $(x - 1)(x - 3) = 0$ , so the curve crosses at $x = 1$ and $x = 3$ . The roots of the equation are exactly the $x$ -intercepts of the graph, which connects this topic to quadratic graphs.

Try this

Cue. Solve $x^2 - 9 = 0$ . This is a difference of two squares: $(x - 3)(x + 3) = 0$ , so $x = 3$ or $x = -3$ .
Cue. Solve $x^2 + 6x + 8 = 0$ . Two numbers multiply to $8$ and add to $6$ : $2$ and $4$ , so $(x + 2)(x + 4) = 0$ and $x = -2$ or $x = -4$ .
Cue. Solve $x^2 - 5x = 0$ . Take out $x$ : $x(x - 5) = 0$ , so $x = 0$ or $x = 5$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksSolve the equation

x^2 - 7x + 12 = 0

by factorisation.

Show worked answer →

Find two numbers that multiply to $12$ and add to $-7$ . These are $-3$ and $-4$ .

Factorise: $x^2 - 7x + 12 = (x - 3)(x - 4) = 0$ .

For a product to be zero, one factor must be zero:

$x - 3 = 0$ gives $x = 3$ , and $x - 4 = 0$ gives $x = 4$ .

So $x = 3$ or $x = 4$ .

What markers reward: the correct pair of numbers, a correct factorisation, using the zero product rule, and both solutions. Giving only one solution loses a mark.

Original4 marksSolve

2x^2 + 3x - 4 = 0

, giving your answers correct to

2

decimal places.

Show worked answer →

This does not factorise neatly, so use the quadratic formula with $a = 2$ , $b = 3$ , $c = -4$ .

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-3 \pm \sqrt{9 - 4(2)(-4)}}{2(2)} = \dfrac{-3 \pm \sqrt{41}}{4}$ .

$\sqrt{41} \approx 6.403$ , so $x = \dfrac{-3 + 6.403}{4} = 0.85$ or $x = \dfrac{-3 - 6.403}{4} = -2.35$ .

What markers reward: correct substitution into the formula (especially the signs, with $-4ac$ becoming $+32$ ), the discriminant $41$ , and both answers rounded as asked. A negative under the root with a sign slip is the usual mistake.