Quadratic formula and exact surd answers | Year 11

01 / LEARN

Understand the idea.

A method for every quadratic

For $a x^{2} + b x + c = 0$ with $a \neq 0$ , the quadratic formula gives $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ . It comes from completing the square and works even when simple factorisation is unavailable.

Put every term on one side before identifying the signed coefficients. Write a negative coefficient in brackets when squaring or multiplying it. The complete numerator, including both terms, is divided by $2 a$ .

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Exact before approximate

An exact surd retains the square root. Extract square factors, such as $\sqrt{20} = 2 \sqrt{5}$ , then cancel a factor only when it is common to every numerator term and the denominator. A term of a sum cannot be cancelled alone.

Keep exact values or full calculator precision during calculations, then round once at the end if requested. The two signs represent two candidate roots. Reject a root only for a stated mathematical or contextual restriction, rather than simply because it is negative.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Irrational roots

Solve $x^{2} - 4 x - 1 = 0$ exactly.

Worked solution

1
Identify the signed coefficients and discriminant.
$a = 1, b = - 4, c = - 1, Δ = {(- 4)}^{2} - 4 (1) (- 1) = 20$
The equation is already in standard form, and both negative signs affect the substitution.
2
Substitute into the quadratic formula.
$x = \frac{- (- 4) \pm \sqrt{20}}{2 (1)} = \frac{4 \pm \sqrt{20}}{2}$
The leading coefficient belongs in the denominator beneath the whole numerator.
3
Simplify the surd and both numerator terms.
$x = \frac{4 \pm 2 \sqrt{5}}{2} = 2 \pm \sqrt{5}$
Twenty contains the square factor four, and every numerator term has a common factor two.
4
Check the two branches in a completed square.
$x^{2} - 4 x - 1 = {(x - 2)}^{2} - 5, {(\pm \sqrt{5})}^{2} - 5 = 0$
Either sign makes the original quadratic zero, confirming both exact roots.

Answer

$x = 2 \pm \sqrt{5}$

EXAMPLE 02

A non-unit denominator

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

Worked solution

1
Calculate the discriminant.
$a = 3, b = 2, c = - 2, Δ = 2^{2} - 4 (3) (- 2) = 28$
The constant coefficient is negative, so its contribution increases the discriminant.
2
Use the complete denominator.
$x = \frac{- 2 \pm \sqrt{28}}{6}$
The denominator is twice the leading coefficient, hence six.
3
Extract the square factor and reduce.
$\begin{aligned} x & = \frac{- 2 \pm 2 \sqrt{7}}{6} \\ = \frac{- 1 \pm \sqrt{7}}{3} \end{aligned}$
The common factor two divides both numerator terms and the denominator.
4
Retain both real roots.
$x = \frac{- 1 + \sqrt{7}}{3} or x = \frac{- 1 - \sqrt{7}}{3}$
The positive discriminant gives two distinct real roots, and the question imposes no contextual restriction.

Answer

$x = (- 1 \pm \sqrt{7}) / 3$

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$x = - 1 \pm \sqrt{3}$

Here $Δ = 2^{2} - 4 (1) (- 2) = 12$ . The formula gives $x = \frac{- 2 \pm 2 \sqrt{3}}{2} = - 1 \pm \sqrt{3}$ , retaining both exact branches.

Keep practising

6 worksheet questions · 17 marks

Open worksheet Worked solutions

Read the questions online

Solve $x^{2} - 6 x + 1 = 0$ exactly.
3 marks
Solve $2 x^{2} + 4 x - 3 = 0$ exactly.
3 marks
Solve $3 x^{2} - 5 x + 1 = 0$ exactly.
3 marks
Solve $x^{2} = 3 x + 2$ exactly.
3 marks
A positive length $x$ satisfies $x^{2} + 2 x = 8$ . Find $x$ .
2 marks
Give both roots of $x^{2} - 2 x - 4 = 0$ to 3 decimal places.
3 marks

Read the worked solutions

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

Solve $x^{2} - 6 x + 1 = 0$ exactly.
3 marks
The discriminant is 32.
$x = \frac{6 \pm 4 \sqrt{2}}{2}$ .
$x = 3 \pm 2 \sqrt{2}$
Solve $2 x^{2} + 4 x - 3 = 0$ exactly.
3 marks
The discriminant is 40.
$x = \frac{- 4 \pm 2 \sqrt{10}}{4}$ .
$x = \frac{- 2 \pm \sqrt{10}}{2}$
Solve $3 x^{2} - 5 x + 1 = 0$ exactly.
3 marks
Use $a = 3, b = - 5, c = 1$ .
The discriminant is 13.
$x = \frac{5 \pm \sqrt{13}}{6}$
Solve $x^{2} = 3 x + 2$ exactly.
3 marks
Rearrange $x^{2} - 3 x - 2 = 0$ .
Use the quadratic formula.
$x = \frac{3 \pm \sqrt{17}}{2}$
A positive length $x$ satisfies $x^{2} + 2 x = 8$ . Find $x$ .
2 marks
The equation is $x^{2} + 2 x - 8 = 0$ .
Roots are 2 and -4; only 2 is positive.
$x = 2$
Give both roots of $x^{2} - 2 x - 4 = 0$ to 3 decimal places.
3 marks
Exact roots are $1 \pm \sqrt{5}$ .
Evaluate both branches and round the final values.
$x \approx - 1.236, 3.236$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topics 2.1–2.2: quadratic relationships, cubic and quartic polynomials. Original Ascent teaching sequence.

Official curriculum source ↗

In this lesson

Understand the idea.

A method for every quadratic

Exact before approximate

See the method in action.

Irrational roots

Identify the signed coefficients and discriminant.

Substitute into the quadratic formula.

Simplify the surd and both numerator terms.

Check the two branches in a completed square.

A non-unit denominator

Calculate the discriminant.

Use the complete denominator.

Extract the square factor and reduce.

Retain both real roots.

Try it yourself.

Check your understanding

Keep practising