Learn9709Paper 1 · Pure Mathematics 1QuadraticsSolving quadratic equations

Quadratics · Subtopic 2 of 5

Solving Quadratic Equations

Three ways to solve $ax^2+bx+c=0$ (factorising, completing the square and the formula), choosing the quickest, giving exact answers, and spotting quadratics in disguise such as $x^4-5x^2+4=0$ .

Cambridge 9709 Paper 1Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2AQA 7357 Paper 3≈ 45 minRevision 5 · updated 2026-10-27

Section 1 of 15

Learning objectives

Rearrange an equation into the form $ax^2+bx+c=0$ and solve it by factorising.
Solve by completing the square or by the quadratic formula, giving exact answers with surds.
Choose the quickest method for a given equation.
Solve equations that are quadratic in $x^2$ , $\sqrt x$ , $a^x$ or a bracket, and reject impossible values.

Pause and think

If $A\times B=0$ , what can you say about $A$ and $B$ ? Is the same true if $A\times B=6$ ?

Section 2 of 15

Core idea

Start hereThe garden puzzle

A rectangular vegetable patch is $3$ m longer than it is wide, and its area must be exactly $40$ m². How wide should it be? Call the width $x$ . Then the length is $x+3$ and the area is $x(x+3)=40$ . You can't just "divide to get $x$ ", because $x$ appears twice ( $x^2+3x$ ). Trial and error might find $5$ , but would you be sure there isn't another answer, or that $5$ isn't really $4.98$ ? This lesson gives three methods that find every solution, with certainty.

The big idea

To solve a quadratic equation, first rearrange it to $ax^2+bx+c=0$ . Then use one of three methods: factorise (fast when it works), complete the square, or use the quadratic formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (always works). The solutions are called roots; on a graph they are where the curve meets the $x$ -axis.

1. What "solving" means

A root (or solution) of an equation is a value of $x$ that makes it true. For $x^2+3x-40=0$ , $x=5$ is a root because $25+15-40=0$ . A quadratic equation has at most two real roots. Graphically, the roots of $ax^2+bx+c=0$ are the $x$ -coordinates where $y=ax^2+bx+c$ crosses the $x$ -axis (where $y=0$ ).

The roots are where the curve meets the x-axis

On the

x

-axis,

y=0

. So the points where the graph crosses the axis are exactly the solutions of

x^2-5x+6=0

Key takeawayAlways get zero on one side first Every method below needs the form

ax^2+bx+c=0

. For the garden:

x(x+3)=40

becomes

x^2+3x=40

, then

x^2+3x-40=0

(subtract

40

from both sides).

2. The zero-product property

Factorising works because of one simple fact about multiplication:

AB=0

, then

A=0

B=0

$AB=0$
Suppose two numbers multiply to zero.
$\text{If }A\neq0:\quad\frac{AB}{A}=\frac0A$
We are allowed to divide by $A$ because it isn't zero.
$B=0$
Simplify both sides.

So either

A=0

already, or (if not)

B

must be

0

. Zero is the only number with this property: if

AB=6

A

and

B

could be anything.

Common misconception $(x-2)(x+5)=6$ does not mean $x-2=6$ The zero-product property only works with zero. Expand and rearrange first:

x^2+3x-10=6\Rightarrow x^2+3x-16=0

3. Method 1: factorising (when a = 1)

Expanding $(x+m)(x+n)$ shows what to look for:

The pattern behind factorising

$(x+m)(x+n)=x^2+nx+mx+mn$
Multiply out: four products.
$=x^2+(m+n)x+mn$
Collect the two $x$ terms.

To factorise

x^2+bx+c

, find two numbers

m

and

n

that multiply to $c$ and add to $b$ .

Worked exampleThe garden: $x^2+3x-40=0$

Solve $x^2+3x-40=0$ and find the garden's width.

Goal: Find all roots, then choose the one that makes sense.
Strategy: Look for two numbers with product $-40$ and sum $3$ .

Factor pairs of $-40$ : $(1,-40),(-1,40),(2,-20),(-2,20),(4,-10),(-4,10),(5,-8),(-5,8)$ . Only $-5+8=3$ .
One number must be negative since the product is negative.
$(x-5)(x+8)=0$
$m=-5$ , $n=8$ .
$x-5=0\quad\text{or}\quad x+8=0$
Zero-product property.
$x=5\quad\text{or}\quad x=-8$
Solve each linear equation.

AnswerRoots

5

and

-8

. A width can't be negative, so the garden is

5

m wide (and

8

m long).

✓ Check

5\times8=40

✓. Also

(-8)^2+3(-8)-40=64-24-40=0

✓, so

-8

is a genuine root, just not a sensible width.

4. Factorising when a ≠ 1: splitting the middle term

For $ax^2+bx+c$ , multiply $a\times c$ . Find two numbers that multiply to $ac$ and add to $b$ ; use them to split $bx$ into two terms, then factorise in pairs. Why does this work? $(px+q)(rx+s)=prx^2+(ps+qr)x+qs$ , and the two middle parts $ps$ and $qr$ multiply to $pqrs=(pr)(qs)=ac$ .

Worked example $2x^2+7x+3=0$

Solve $2x^2+7x+3=0$ by factorising.

Goal: Write the left side as two brackets, then use the zero-product property.
Strategy: $ac=6$ ; find a pair with product $6$ and sum $7$ .

$6$ and $1$ : $6\times1=6$ , $6+1=7$ ✓.
$2x^2+6x+x+3=0$
Split $7x$ into $6x+x$ .
$2x(x+3)+1(x+3)=0$
Factor each pair; the brackets must match.
$(2x+1)(x+3)=0$
Take out the common bracket $(x+3)$ .
$x=-\tfrac12\quad\text{or}\quad x=-3$
Set each bracket to zero.

Answer

x=-\frac12

x=-3

✓ Check

x=-\frac12

2\cdot\frac14-\frac72+3=\frac12-\frac72+3=0

✓.

5. Method 2: completing the square

This works even when factorising fails. Write the quadratic as $(x+p)^2+q$ , move $q$ across, and square-root.

Worked example $x^2+6x-2=0$

Solve $x^2+6x-2=0$ , exactly.

Goal: Roots in surd form.
Strategy: Integer factor pairs of $-2$ don't add to $6$ , so complete the square.

$(x+3)^2-9-2=0$
Halve $6$ , square, take away.
$(x+3)^2=11$
Add $11$ to both sides.
$x+3=\pm\sqrt{11}$
Square root, both signs.
$x=-3\pm\sqrt{11}$
Subtract $3$ .

Answer

x=-3+\sqrt{11}\approx0.317

x=-3-\sqrt{11}\approx-6.317

✓ CheckSum of roots

=-6=-b

and product

=9-11=-2=c

✓ (for

a=1

the roots always add to

-b

and multiply to

c

, from section 3).

6. Method 3: the quadratic formula

Quadratic formula

ax^2+bx+c=0\;\Longrightarrow\;x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

$\pm$: "plus or minus": there are two answers, one with $+$ and one with $-$ .
$b^2-4ac$: the discriminant; it must be $\ge0$ for real roots (next lesson).

It comes from completing the square on $ax^2+bx+c=0$ in general (proved fully in the completing-the-square lesson), so it always works.

Worked example $3x^2-5x-1=0$

Solve $3x^2-5x-1=0$ , giving answers to 3 significant figures.

Goal: Two decimal roots.
Strategy: Identify $a,b,c$ carefully (with signs) and substitute.

$a=3$ , $b=-5$ , $c=-1$ .
$b^2-4ac=(-5)^2-4(3)(-1)=25+12=37$
Work out the discriminant first; brackets protect the signs.
$x=\frac{-(-5)\pm\sqrt{37}}{2\times3}=\frac{5\pm\sqrt{37}}6$
$-b=+5$ .
$x=\frac{5+6.083}{6}=1.85\quad\text{or}\quad x=\frac{5-6.083}6=-0.180$
$\sqrt{37}\approx6.083$ .

Answer

x=1.85

x=-0.180

✓ Check

3(1.847)^2-5(1.847)-1=10.23-9.24-1\approx0

✓.

Common misconceptionSign slips with negative b With

b=-5

-b=+5

and

b^2=+25

. Writing

-5^2

on a calculator gives

-25

; always use brackets:

(-5)^2

y = x² − 5x + c as c increases

c = 2.00

Two roots, then one (the curve just touches), then none. The number under the square root in the formula,

25-4c

, decides which.

7. Choosing a method

Which method should I use?

Drawing diagram…

All three methods give the same roots. Pick the quickest that works; the formula is the universal fallback.

Common misconceptionNever divide both sides by $x$ From

x^2=5x

, dividing by

x

gives only

x=5

and loses

x=0

. Instead rearrange:

x^2-5x=0\Rightarrow x(x-5)=0\Rightarrow x=0

5

8. Disguised quadratics

Worked example $x^4-5x^2+4=0$

Solve $x^4-5x^2+4=0$ .

Goal: All real $x$ .
Strategy: $x^4=(x^2)^2$ , so substitute $u=x^2$ to get an ordinary quadratic.

$u^2-5u+4=0$
Let $u=x^2$ , so $x^4=u^2$ .
$(u-1)(u-4)=0$
Product $4$ , sum $-5$ : $-1,-4$ .
$u=1\quad\text{or}\quad u=4$
Zero-product property.
$x^2=1\Rightarrow x=\pm1;\qquad x^2=4\Rightarrow x=\pm2$
Undo the substitution; remember both square roots.

Answer

x=-2,-1,1,2

(four roots: a quartic can have up to four).

✓ Check

x=2

16-20+4=0

✓;

x=-1

1-5+4=0

✓.

Pause and think

Solve $x-5\sqrt x+6=0$ .

Summary

Rearrange to $=0$ first.
Zero-product property: $AB=0\Rightarrow A=0$ or $B=0$ .
Factorise: numbers multiplying to $ac$ and adding to $b$ .
Complete the square or use the formula when factorising fails.
Substitute $u$ for disguised quadratics; never divide by $x$ .

Root: A value that makes the equation true; where the graph meets the x-axis.
Factorise: Write as a product of brackets.
Zero-product property: If $AB=0$ then $A=0$ or $B=0$ .
Discriminant: $b^2-4ac$ .

HistoryA rule in verse. In 628 CE the Indian mathematician Brahmagupta described, in words, a rule for solving quadratic equations that is equivalent to the formula we use today. Mathematicians in that era often wrote their rules as poetry so that students could memorise them.

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