Mathematics 9709 · Pure Mathematics 1

Solving by factorising

A quadratic is $ax^2+bx+c$ .
To solve it, find the $x$ -values that make it zero. These are called its roots.

Where the curve crosses zero

Solving $ax^2+bx+c=0$ means finding where the curve crosses the line $y=0$ . The curve is a parabola (a U-shape). It can cross in two places, so you can get up to two answers. Here is $x^2+5x+6=0$ :

y = x² + 5x + 6 crosses zero at x = −3 and x = −2. These are its two answers.

Keep both crossings. If you solve $x^2=5x$ by dividing by $x$ , you get $x=5$ but you lose $x=0$ .
A calculator gives you both answers at once.

The fast method: your calculator

Get the equation to $ax^2+bx+c=0$ . Type $a,\ b,\ c$ into your calculator's equation solver. Read both roots from the screen.

Rearrange so one side is $0$ : $ax^2+bx+c=0$ .
Open the polynomial solver — on a Casio fx-991EX, Menu → Equation → Polynomial, degree $2$ .
Type $a,\ b,\ c$ and read off both roots.

Newer fx-991CW (no Menu key): Home → Equation → Polynomial.

Then work backwards to the brackets. Each root is where a bracket equals zero. So $x=-\tfrac12$ comes from $(2x+1)$ , and $x=-3$ comes from $(x+3)$ .
The bracket line earns the method mark. The roots earn the answer mark.

Worked example

Solve $2x^2+7x+3=0$

Already $=0$ ; the calculator gives $x=-\tfrac12$ and $x=-3$ .
Work backwards to the brackets: $(2x+1)(x+3)=0$ . (method mark)

Answer

x=-\tfrac{1}{2}\ \text{ or }\ x=-3

Common mistake

A calculator answer on its own can lose the method mark. Always write the two lines:

ax^2+bx+c=0

, then

(x+\dots)(x+\dots)=0

. For an exact answer (with a surd), use the formula (§1.3).

Your turn— tap to reveal the worked answer (your turn: 6x² − 17x + 5 = 0)

Calculator → $x=\tfrac52,\ \tfrac13$ . Now work backwards to the brackets:

Answer

(2x-5)(3x-1)=0\ \Rightarrow\ x=\tfrac{5}{2}\ \text{ or }\ x=\tfrac{1}{3}

Factorising by hand

You still need this to factorise an expression, to show working, or when no calculator is allowed.

For $x^2+bx+c$ , find two numbers that multiply to $c$ and add to $b$ .

Quadratic	Two numbers	Factors
$x^2+5x+6$	2, 3	$(x+2)(x+3)$
$x^2+2x-15$	5, −3	$(x+5)(x-3)$

When a ≠ 1: the split-and-group method

Multiply $a\times c$ . For $2x^2+7x+3$ that is $2\times3=6$ .
Find two numbers that multiply to $6$ and add to $b=7$ : that is $6$ and $1$ .
Split the middle term and group: $2x^2+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3)$ .

Getting to a clean = 0

First get to a clean $=0$ . Rearrange the equation. Clear any fractions by multiplying every term by the bottom. Then divide out any common factor.

Common mistake

Sometimes the question limits

x

, for example if

x

is a length. Then reject any root that breaks the rule. Examiners put that root there on purpose. (9709/11 Nov 2020 Q11)

Where it shows up

You rarely solve a quadratic on its own.
It is usually the last step in a bigger question, in series, in trigonometry, or where a line meets a curve.

Worked example

Real question: 9709/12 Nov 2022 Q2

An arithmetic progression starts $a,\ 2a,\ a^2$ ( $a>0$ ). Equal spacing gives $a^2=3a$ , so:

$a^2-3a=0\ \Rightarrow\ a(a-3)=0\ \Rightarrow\ a=3$ ( $a=0$ rejected, $a>0$ )

Answer

a=3

Your turn— tap to reveal the worked answer (your turn — x(10x − 13) = 3)

To zero: $10x^2-13x-3=0$ . Calculator (or factorise to $(5x+1)(2x-3)=0$ ):

Answer

x=-\tfrac{1}{5}\ \text{ or }\ x=\tfrac{3}{2}

These same roots come back as the points where the graph crosses the x-axis, its x-intercepts (§1.4). They are also what the discriminant counts (§1.5).

NextCompleting the square

Solving by factorising

A quadratic is $ax^2+bx+c$ .
To solve it, find the $x$ -values that make it zero. These are called its roots.

Where the curve crosses zero

y = x² + 5x + 6 crosses zero at x = −3 and x = −2. These are its two answers.

Keep both crossings. If you solve $x^2=5x$ by dividing by $x$ , you get $x=5$ but you lose $x=0$ .
A calculator gives you both answers at once.

The fast method: your calculator

Get the equation to $ax^2+bx+c=0$ . Type $a,\ b,\ c$ into your calculator's equation solver. Read both roots from the screen.

Rearrange so one side is $0$ : $ax^2+bx+c=0$ .
Open the polynomial solver — on a Casio fx-991EX, Menu → Equation → Polynomial, degree $2$ .
Type $a,\ b,\ c$ and read off both roots.

Newer fx-991CW (no Menu key): Home → Equation → Polynomial.

Then work backwards to the brackets. Each root is where a bracket equals zero. So $x=-\tfrac12$ comes from $(2x+1)$ , and $x=-3$ comes from $(x+3)$ .
The bracket line earns the method mark. The roots earn the answer mark.

Worked example

Solve $2x^2+7x+3=0$

Already $=0$ ; the calculator gives $x=-\tfrac12$ and $x=-3$ .
Work backwards to the brackets: $(2x+1)(x+3)=0$ . (method mark)

Answer

x=-\tfrac{1}{2}\ \text{ or }\ x=-3

Common mistake

A calculator answer on its own can lose the method mark. Always write the two lines:

ax^2+bx+c=0

, then

(x+\dots)(x+\dots)=0

. For an exact answer (with a surd), use the formula (§1.3).

Your turn— tap to reveal the worked answer (your turn: 6x² − 17x + 5 = 0)

Calculator → $x=\tfrac52,\ \tfrac13$ . Now work backwards to the brackets:

Answer

(2x-5)(3x-1)=0\ \Rightarrow\ x=\tfrac{5}{2}\ \text{ or }\ x=\tfrac{1}{3}

Factorising by hand

You still need this to factorise an expression, to show working, or when no calculator is allowed.

For $x^2+bx+c$ , find two numbers that multiply to $c$ and add to $b$ .

Quadratic	Two numbers	Factors
$x^2+5x+6$	2, 3	$(x+2)(x+3)$
$x^2+2x-15$	5, −3	$(x+5)(x-3)$

When a ≠ 1: the split-and-group method

Multiply $a\times c$ . For $2x^2+7x+3$ that is $2\times3=6$ .
Find two numbers that multiply to $6$ and add to $b=7$ : that is $6$ and $1$ .
Split the middle term and group: $2x^2+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3)$ .

Getting to a clean = 0

First get to a clean $=0$ . Rearrange the equation. Clear any fractions by multiplying every term by the bottom. Then divide out any common factor.

Common mistake

Sometimes the question limits

x

, for example if

x

is a length. Then reject any root that breaks the rule. Examiners put that root there on purpose. (9709/11 Nov 2020 Q11)

Where it shows up

You rarely solve a quadratic on its own.
It is usually the last step in a bigger question, in series, in trigonometry, or where a line meets a curve.

Worked example

Real question: 9709/12 Nov 2022 Q2

An arithmetic progression starts $a,\ 2a,\ a^2$ ( $a>0$ ). Equal spacing gives $a^2=3a$ , so:

$a^2-3a=0\ \Rightarrow\ a(a-3)=0\ \Rightarrow\ a=3$ ( $a=0$ rejected, $a>0$ )

Answer

a=3

Your turn— tap to reveal the worked answer (your turn — x(10x − 13) = 3)

To zero: $10x^2-13x-3=0$ . Calculator (or factorise to $(5x+1)(2x-3)=0$ ):

Answer

x=-\tfrac{1}{5}\ \text{ or }\ x=\tfrac{3}{2}

These same roots come back as the points where the graph crosses the x-axis, its x-intercepts (§1.4). They are also what the discriminant counts (§1.5).