Quadratic formula | Year 10

01 / LEARN

The idea, step by step

Put the equation in standard form

The coefficients in the formula belong to ax squared plus bx plus c equal to zero. Move every term to one side first. The coefficient b includes its sign.

Substitute with brackets, particularly when b or c is negative. The denominator is twice a, and it divides the whole numerator.

a x^{2} + b x + c = 0, x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}, a \neq 0

Exact answers and useful decimals

If the square root is not an integer, an exact surd answer often communicates the result best. Simplify common factors only after treating each whole numerator correctly.

A calculator approximation is useful for a practical length or for checking a graph. Retain more digits during the calculation and round the final result to the requested precision.

A common trap

Using $b = 4$ when the equation contains $- 4 x$ .

Instead: The sign is part of the coefficient; bracket the signed value whenever it is substituted.

02 / WORKED EXAMPLES

See the thinking

Example 1

Negative b and c

Solve $3 x^{2} - 4 x - 2 = 0$ exactly.

$a = 3, b = - 4, c = - 2$ .
$x = \frac{4 \pm \sqrt{{(- 4)}^{2} - 4 (3) (- 2)}}{6} = \frac{4 \pm \sqrt{40}}{6}$ .
Use $\sqrt{40} = 2 \sqrt{10}$ and divide numerator and denominator by $2$ .

x = \frac{2 \pm \sqrt{10}}{3}

Example 2

Rearrange before substituting

Solve $(x + 1) (x - 2) = 5$ .

Expand: $x^{2} - x - 2 = 5$ , so $x^{2} - x - 7 = 0$ .
The coefficients are $a = 1, b = - 1, c = - 7$ .
$x = \frac{1 \pm \sqrt{1 + 28}}{2}$ .

x = \frac{1 \pm \sqrt{29}}{2}

Check your understanding

What are a, b and c in $2 x^{2} + 3 x - 5 = 0$ ?

Show answer

$a = 2, b = 3, c = - 5$

Each coefficient includes its preceding sign.

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Use the formula to solve $2 x^{2} + 3 x - 1 = 0$ exactly.
3 marks
Solve $3 x^{2} + 4 x - 2 = 0$ exactly.
3 marks
Solve $x^{2} - 5 x + 3 = 0$ exactly.
3 marks
Solve $4 x^{2} - 4 x - 1 = 0$ exactly.
3 marks
Solve $x^{2} - 6 x + 2 = 0$ . Give both answers to three decimal places.
3 marks
A student writes the roots of $x^{2} + 5 x - 2 = 0$ as $(5 \pm \sqrt{33}) / 2$ . Correct the answer and explain the error.
3 marks

Read the worked solutions

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

Use the formula to solve $2 x^{2} + 3 x - 1 = 0$ exactly.
3 marks
$a = 2, b = 3, c = - 1$ , so $b^{2} - 4 a c = 17$ .
$x = \frac{- 3 \pm \sqrt{17}}{4}$
Solve $3 x^{2} + 4 x - 2 = 0$ exactly.
3 marks
$x = \frac{- 4 \pm \sqrt{40}}{6} = \frac{- 2 \pm \sqrt{10}}{3}$ .
$x = \frac{- 2 \pm \sqrt{10}}{3}$
Solve $x^{2} - 5 x + 3 = 0$ exactly.
3 marks
$b^{2} - 4 a c = 25 - 12 = 13$ .
$x = \frac{5 \pm \sqrt{13}}{2}$
Solve $4 x^{2} - 4 x - 1 = 0$ exactly.
3 marks
$x = \frac{4 \pm \sqrt{32}}{8} = \frac{4 \pm 4 \sqrt{2}}{8}$ .
$x = \frac{1 \pm \sqrt{2}}{2}$
Solve $x^{2} - 6 x + 2 = 0$ . Give both answers to three decimal places.
3 marks
$x = 3 \pm \sqrt{7}$ .
Using $\sqrt{7} \approx 2.645751$ , round each final root.
$x \approx 0.354$ or $x \approx 5.646$
A student writes the roots of $x^{2} + 5 x - 2 = 0$ as $(5 \pm \sqrt{33}) / 2$ . Correct the answer and explain the error.
3 marks
Here $b = 5$ , so $- b = - 5$ .
The discriminant $25 + 8 = 33$ is correct.
$x = \frac{- 5 \pm \sqrt{33}}{2}$ ; the numerator starts with $- b$ .

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Curriculum reference & lesson scope

Australian Curriculum v9.0 and South Australian Curriculum

Year 10 quadratics, including the fuller school-course algebra requested for Ascent: non-monic factorisation, completing the square and the quadratic formula. These lessons consolidate Year 9 monic quadratics and extend the available solving methods.

Official curriculum source ↗

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