Quadratic formula | Year 10

01 / LEARN

Understand the idea.

Put the equation in standard form

The quadratic formula solves $a x^{2} + b x + c = 0$ when $a \neq 0$ , even when integer factorisation is unavailable. Expand and rearrange first so the coefficients belong to an equation with zero on one side.

Include the sign in every coefficient. Substituting a negative value with brackets protects both $- b$ and $b^{2}$ . The denominator $2 a$ divides the entire numerator, including the square-root term.

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

Exact answers and useful decimals

Keep a square root exact when the question asks for an exact answer. Simplify its square factors, then cancel a common factor only when it divides every term of the numerator and the denominator.

For an approximate practical answer, evaluate both branches before applying the model’s restrictions. Retain calculator digits until the final rounding; rounding the square root early can change the last reported digit.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Negative b and c

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

Worked solution

1
Identify the signed coefficients.
$a = 3, b = - 4, c = - 2$
The equation already has zero on the right-hand side.
2
Substitute with brackets.
$x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (3) (- 2)}}{2 (3)}$
Brackets distinguish squaring a negative coefficient from negating a square.
3
Simplify the discriminant and radical.
$\begin{array}{c} {(- 4)}^{2} - 4 (3) (- 2) = 16 + 24 = 40 \\ \sqrt{40} = 2 \sqrt{10} \\ x = \frac{4 \pm 2 \sqrt{10}}{6} \end{array}$
The square factor $4$ comes out of the radical as $2$ .
4
Cancel the common factor.
$x = \frac{2 \pm \sqrt{10}}{3}$
Both numerator terms and the denominator share a factor of $2$ ; both signs remain.

Answer

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

EXAMPLE 02

Rearrange before substituting

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

Worked solution

1
Expand and rearrange.
$\begin{aligned} x^{2} - x - 2 & = 5 \\ x^{2} - x - 7 & = 0 \end{aligned}$
The formula uses coefficients from the equation after everything is moved to one side.
2
Read the coefficients.
$a = 1, b = - 1, c = - 7$
The constant is $- 7$ , rather than the $- 2$ from the initial expansion.
3
Apply the quadratic formula.
$x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (1) (- 7)}}{2 (1)}$
The formula gives both possible roots of the rearranged equation.
4
Simplify without rounding.
$\begin{aligned} x & = \frac{1 \pm \sqrt{1 + 28}}{2} \\ = \frac{1 \pm \sqrt{29}}{2} \end{aligned}$
No nontrivial square factor can be removed from $29$ , so this is an exact simplified answer.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

The coefficient of $x^{2}$ is $2$ , the coefficient of $x$ is $3$ , and the constant term is $- 5$ . Therefore $a = 2$ , $b = 3$ and $c = - 5$ .

Keep practising

6 worksheet questions · 18 marks

Open worksheet 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 $\frac{5 \pm \sqrt{33}}{2}$ . 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 $\frac{5 \pm \sqrt{33}}{2}$ . 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 ↗

In this lesson

Understand the idea.

Put the equation in standard form

Exact answers and useful decimals

See the method in action.

Negative b and c

Identify the signed coefficients.

Substitute with brackets.

Simplify the discriminant and radical.

Cancel the common factor.

Rearrange before substituting

Expand and rearrange.

Read the coefficients.

Apply the quadratic formula.

Simplify without rounding.

Try it yourself.

Check your understanding

Keep practising