Solve quadratic equations by factorisation and interpret the solutions, including equations that must first be rearranged into standard form

What this dot point is asking

SEAB wants you to solve quadratic equations by factorisation, including ones you must first rearrange into standard form, and to interpret what the solutions mean. Factorisation is the quickest method when the quadratic factorises with whole numbers, and it builds directly on the factorising skills from the algebra strand.

The answer

Standard form

A quadratic equation is one whose highest power of the unknown is $2$. The standard form is:

You must rearrange any quadratic into this form, with one side equal to zero, before solving by factorisation.

The zero product property

Once one side is zero and the other is factorised, use the fact that if a product of two factors is zero then at least one factor is zero:

This is why setting one side to zero is essential, the property only works against zero.

Rearranging before solving

If the equation is not already equal to zero, expand and move every term to one side. For $x(x + 3) = 18$, expand to $x^2 + 3x = 18$ then subtract $18$ to get $x^2 + 3x - 18 = 0$, which factorises.

Interpreting the solutions

The solutions are the values of the unknown that satisfy the equation, also called the roots. A quadratic usually has two solutions, but it can have one repeated solution (a perfect square such as $x^2 - 6x + 9 = 0$) when both factors are the same.

Factorising when the leading coefficient is not 1

When the coefficient of $x^2$ is greater than $1$, the reliable factorising method is to "split the middle term". Find two numbers that multiply to $a \times c$ and add to $b$, use them to break the middle term into two, then factorise the four terms in pairs. For $2x^2 - 5x - 3$, the product $ac = -6$ and the numbers $-6$ and $1$ add to $-5$, so $2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)$. Checking that the chosen pair multiplies to $ac$ and adds to $b$ before splitting keeps this method dependable on the harder quadratics E-Maths sets.

Rejecting a root that does not fit the context

In a worded problem both algebraic roots are correct, but often only one is physically sensible, and stating which you reject earns a method mark. A length, a time, or a count cannot be negative, so a negative root is discarded. For the rectangle with $x(x + 3) = 40$, the roots are $x = 5$ and $x = -8$; only $x = 5$ is kept because a width cannot be negative. Always solve the quadratic fully first and then apply the context to choose the valid root, rather than discarding a root before solving, which could hide an arithmetic error.

Examples in context

Example 1. A rectangle's dimensions. A rectangle has area $40\ \text{cm}^2$ and its length is $3\ \text{cm}$ more than its width. Letting the width be $x$, the equation $x(x + 3) = 40$ rearranges to a quadratic whose positive root gives the width. The negative root is rejected as a length cannot be negative.

Example 2. Where a curve meets the axis. The graph of $y = x^2 - 7x + 10$ crosses the $x$-axis where $y = 0$, that is at the roots $x = 2$ and $x = 5$. Solving the quadratic locates these intercepts, linking algebra to graphs.

Try this

Q1. Solve $x^2 - 9 = 0$. [2 marks]

• Cue. Difference of two squares: $(x - 3)(x + 3) = 0$, so $x = 3$ or $x = -3$.

Q2. Solve $x^2 + 4x = 0$. [2 marks]

• Cue. Factorise: $x(x + 4) = 0$, so $x = 0$ or $x = -4$.

Q3. Solve $x^2 - 5x - 14 = 0$. [2 marks]

• Cue. Factors of $-14$ adding to $-5$ are $-7$ and $2$, so $(x - 7)(x + 2) = 0$ and $x = 7$ or $x = -2$.