Sec 2 Quadratic Equations: Factorisation & Word Problems

Secondary 2 G3 Mathematics · Study guide

Make one side zero, use factorisation to find every root and decide which solutions make sense in a word problem.

Concept → worked examples → practice12 min read · practice linked below

A quadratic equation can have more than one answer

A quadratic equation can be written as $ax^2+bx+c=0$ , where $a\ne0$ . The squared term makes it different from a linear equation: there may be two distinct real solutions, one repeated real solution, or no real solution. A solution is also called a root.

Before you begin: be able to expand brackets, factorise quadratic expressions and solve a linear equation. This guide uses factorisation and the meaning of a square, rather than a new solving formula.

The zero-product rule: if $AB=0$ , then $A=0$ or $B=0$ . A product of two non-zero numbers cannot be zero. The rule needs zero on one side; from $AB=6$ , you cannot conclude that $A=6$ or $B=6$ .

The reliable sequence is: make one side zero, factorise completely, set each factor equal to zero, and check every root.

Keep the possibility that the unknown is zero

Worked example 1 — do not divide away a solution

Solve $x^2=5x$ .

Subtract $5x$ from both sides and take out the common factor $x$ :

\begin{aligned}x^2-5x&=0\\x(x-5)&=0.\end{aligned}

Either $x=0$ or $x-5=0$ , so the solutions are $\boxed{x=0\text{ or }x=5}$ .

Check: $0^2=5(0)=0$ and $5^2=5(5)=25$ . If we had divided the original equation by $x$ , that operation would have assumed $x\ne0$ , losing the valid zero root.

An equation may need expanding before factorising

Worked example 2 — the displayed brackets are not yet zero factors

Solve $(2x-1)(x+3)=9$ .

The product equals 9, so setting either displayed bracket equal to zero would not solve the equation. Expand, then subtract 9 from both sides:

\begin{aligned}2x^2+6x-x-3&=9\\2x^2+5x-12&=0.\end{aligned}

To factorise, try $2x$ and $x$ , with constants $-3$ and $4$ . The cross-products are $8x$ and $-3x$ , which add to $5x$ :

\begin{aligned}(2x-3)(x+4)&=0\\2x-3=0&\quad\text{or}\quad x+4=0\\x=\frac32&\quad\text{or}\quad x=-4.\end{aligned}

Check in the original equation: at $x=\frac32$ , $(3-1)(\frac32+3)=2\times\frac92=9$ . At $x=-4$ , $(-8-1)(-4+3)=(-9)(-1)=9$ . Both roots are valid.

If an equation has numerical fractions, first multiply every term by a common denominator. If a denominator contains the unknown, record where it is zero before multiplying. See algebraic fractions and formulae for a complete example.

A given root is a value that makes the equation true

Worked example 3 — find a coefficient, then the other root

The equation $2x^2+kx-6=0$ has a root $x=-2$ . Find $k$ and the other root.

Since $-2$ is a root, substitute it for $x$ in the original equation. Use brackets when squaring a negative number:

\begin{aligned}2(-2)^2+k(-2)-6&=0\\8-2k-6&=0\\2-2k&=0\\k&=1.\end{aligned}

Now solve the completed equation:

\begin{aligned}2x^2+x-6&=0\\(2x-3)(x+2)&=0\\x=\frac32&\quad\text{or}\quad x=-2.\end{aligned}

The other root is $\frac32$ . Check it: $2(\frac32)^2+\frac32-6=\frac92+\frac32-6=0$ . The given root $-2$ also gives $8-2-6=0$ .

Keep the jobs separate: substituting the known root finds the coefficient; factorising the resulting equation finds all roots.

Form an equation and interpret its roots

Define the unknown with its units, write related quantities in terms of it, then translate the condition into an equation. An algebraic root may still be unsuitable for the situation.

Worked example 4 — a rectangular garden

A rectangular garden has area $84\text{ m}^2$ . Its length is 5 m greater than its width. Find its dimensions.

Let the width be $x$ metres. Then the length is $(x+5)$ metres, with $x>0$ .

Area equals length multiplied by width. Use metres for both dimensions so their product is in square metres.

\begin{aligned}x(x+5)&=84\\x^2+5x-84&=0\\(x+12)(x-7)&=0\\x=-12&\quad\text{or}\quad x=7.\end{aligned}

The numbers 12 and $-7$ multiply to $-84$ and add to 5, giving the factorisation. Reject $x=-12$ because a width must be positive. With $x=7$ , the length is $7+5=12$ .

Answer: the garden is 7 m wide and 12 m long. Check: $7\times12=84\text{ m}^2$ , and $12-7=5\text{ m}$ . Both the area and the difference condition hold.

What a repeated root means

If $(x-3)^2=0$ , both factors give the same value, $x=3$ . This is one distinct solution, called a repeated root. On the graph of $y=(x-3)^2$ , the curve touches the x-axis at $(3,0)$ .

A real square cannot be negative. For example, $(x-3)^2+4=0$ would require $(x-3)^2=-4$ , so it has no real solution. The graph stays above the x-axis. Failure to find integer factors alone is not proof of no real roots: $x^2=2$ has the real roots $\pm\sqrt2$ .

Explore roots as zero factors and x-intercepts

For $y=(x-r)(x-s)$ , either factor becomes zero at its matching root. Change the roots and compare two crossings with a repeated root. This explorer uses factorised quadratics, so every setting has at least one real root.

The default graph is $y=(x+2)(x-3)=x^2-x-6$ , crossing the x-axis at $(-2,0)$ and $(3,0)$ .

The horizontal and vertical axes use different scales. Only the part within the displayed window is shown; the curve continues beyond it.

(x + 2)(x − 3) = 0 gives x = −2 or x = 3. The expanded equation is x² − x − 6 = 0. The graph has two distinct x-intercepts.

Common mistakes and final checks

Using the zero-product rule too early: move all terms to one side first.
Dividing by the unknown: factor out a common $x$ instead, so a zero root remains possible.
Giving the bracket constant as the root: $2x-3=0$ gives $x=\frac32$ , not $x=3$ or $-3$ .
Discarding every negative root: use the actual restrictions. A negative algebraic answer can be valid; a negative garden width cannot.
Stopping at the unknown: if the question asks for dimensions, ages or a total time, calculate that requested quantity and state its units.

Check each candidate in the original equation, then check the context. Give exact fractions or surds when appropriate; use decimal rounding only if the question requests it.

Practise solving and modelling

Try the Sec 2 quadratic equations practice bank, moving from factorisation to known-root questions and word problems. Revisit expansion and factorisation if identifying the factors is the main obstacle.