Solving quadratic equations | Year 9

01 / LEARN

The idea, step by step

An equation can have two roots

A root is a solution of an equation. Both $5^{2}$ and ${(- 5)}^{2}$ equal $25$ , so $x^{2} = 25$ has two roots.

Write both roots unless the context rules one out. A length must be positive, but an equation on its own can have a negative root.

x^{2} = 25 \Rightarrow x = 5 or x = - 5

Make one side zero

The zero-product rule says that if two numbers multiply to zero, at least one of them is zero. It applies when the equation has a product on one side and zero on the other.

For $x^{2} + x - 6 = 0$ , factorise first: $(x + 3) (x - 2) = 0$ . Then set each factor equal to zero.

(x + 3) (x - 2) = 0 \Rightarrow x = - 3 or x = 2

Do not lose zero or a repeated root

In $x^{2} - 4 x = 0$ , taking out $x$ gives $x (x - 4) = 0$ . The roots are $0$ and $4$ . Dividing by $x$ would lose the root $0$ .

In ${(x - 3)}^{2} = 0$ , both factors give the same answer. There is one distinct root, $x = 3$ , called a repeated root. Check each proposed root in the original equation.

x (x - 4) = 0 \Rightarrow x = 0 or x = 4

A common trap

Dividing $x^{2} - 7 x = 0$ by $x$ and giving only $x = 7$ .

Instead: Factorise instead: $x (x - 7) = 0$ . Then $x = 0$ or $x = 7$ . Both work in the original equation.

02 / WORKED EXAMPLES

See the thinking

Example 1

Take square roots

Solve $x^{2} = 36$ .

Both $6$ and $- 6$ have square $36$ .
Check: $6^{2} = 36$ and ${(- 6)}^{2} = 36$ .

x = 6

x = - 6

Example 2

Factorise, then solve

Solve $x^{2} - 5 x + 6 = 0$ .

Find two numbers with sum $- 5$ and product $6$ : $- 2, - 3$ .
Factorise: $(x - 2) (x - 3) = 0$ . Set each factor to zero.
Check: $2^{2} - 5 (2) + 6 = 0$ and $3^{2} - 5 (3) + 6 = 0$ .

x = 2

x = 3

Example 3

Choose the possible length

A rectangle is $x$ cm wide and $(x + 3)$ cm long. Its area is $40$ cm $2$ . Find its width.

Write $x (x + 3) = 40$ , so $x^{2} + 3 x - 40 = 0$ .
Factorise: $(x + 8) (x - 5) = 0$ . The roots are $- 8$ and $5$ .
A width cannot be negative, so choose $5$ . Check: $5 \times 8 = 40$ .

The width is

5

cm.

Check your understanding

Solve $x^{2} + 4 x = 0$ .

Show answer

$x = 0$ or $x = - 4$

Write $x (x + 4) = 0$ . Either $x = 0$ , or $x + 4 = 0$ .

03 / WORKSHEET

Your turn

6 questions · 17 marks

Worksheet PDF Worked solutions

Read the questions online

Solve $x^{2} = 49$ .
2 marks
Solve $x^{2} - 9 x = 0$ .
3 marks
Solve $x^{2} + 7 x + 10 = 0$ .
3 marks
Solve $x^{2} - 2 x - 15 = 0$ .
3 marks
Solve $x^{2} - 8 x + 16 = 0$ . How many different roots are there?
3 marks
A square has area $64$ cm $2$ . Its side length is $s$ cm. Write and solve an equation to find $s$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Solve $x^{2} = 49$ .
2 marks
Give $7$ [1] and $- 7$ [1].
$x = 7$ or $x = - 7$ .
Solve $x^{2} - 9 x = 0$ .
3 marks
Factorise to $x (x - 9) = 0$ [1].
Use the zero-product rule to give $x = 0$ [1] and $x = 9$ [1].
$x = 0$ or $x = 9$ .
Solve $x^{2} + 7 x + 10 = 0$ .
3 marks
Factorise to $(x + 5) (x + 2) = 0$ [1].
Give $x = - 5$ [1] and $x = - 2$ [1].
$x = - 5$ or $x = - 2$ .
Solve $x^{2} - 2 x - 15 = 0$ .
3 marks
Factorise to $(x - 5) (x + 3) = 0$ [1].
Give $x = 5$ [1] and $x = - 3$ [1].
$x = 5$ or $x = - 3$ .
Solve $x^{2} - 8 x + 16 = 0$ . How many different roots are there?
3 marks
Factorise to ${(x - 4)}^{2} = 0$ [1].
Give $x = 4$ [1] and explain that both factors give the same root [1].
$x = 4$ . There is one distinct root.
A square has area $64$ cm $2$ . Its side length is $s$ cm. Write and solve an equation to find $s$ .
3 marks
Write $s^{2} = 64$ [1].
Find the positive square root $s = 8$ , or give the algebraic roots $s = 8$ and $s = - 8$ [1].
Choose the positive length and state $8$ cm [1].
$s^{2} = 64$ , so the side length is $8$ cm.

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Curriculum reference & lesson scope

AC9M9A04: Solve monic quadratic equations with integer roots

Hand methods for integer-root monic quadratic equations: square roots, a common factor, factorisation and the zero-product rule. Graphical connections are developed in the following lesson. No quadratic formula or completing-the-square method is required.

Official curriculum source ↗

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