Quadratic equations with non-real roots | Year 11

01 / LEARN

The idea, step by step

A negative discriminant is not the end

For $a z^{2} + b z + c = 0$ with real coefficients and $a \neq 0$ , the quadratic formula still applies over the complex numbers. If $Δ = b^{2} - 4 a c < 0$ , rewrite $\sqrt{Δ} = i \sqrt{- Δ}$ .

Give both roots and simplify the real and imaginary parts. The signs in front of the imaginary part usually distinguish the two answers; a negative discriminant means no real roots, not no complex roots.

Completing the square and checking

A form such as ${(z - h)}^{2} = - k$ with $k > 0$ immediately gives $z = h \pm i \sqrt{k}$ . Completing the square can be quicker than the formula when coefficients are simple.

A complex solution must satisfy the original equation. Substitute it and use $i^{2} = - 1$ ; the real and imaginary components of the result must both be zero. Exact forms make that check more reliable than rounded decimals.

A common trap

Reporting 'no solutions' whenever the discriminant is negative.

Instead: Say 'no real solutions'. Over the complex numbers the quadratic has two roots, counted with multiplicity.

02 / WORKED EXAMPLES

See the thinking

Example 1

Complete the square

Solve $z^{2} - 6 z + 13 = 0$ .

$z^{2} - 6 z + 13 = {(z - 3)}^{2} + 4$ .
Thus ${(z - 3)}^{2} = - 4$ .
$z - 3 = \pm 2 i$ , so $z = 3 \pm 2 i$ .

z = 3 \pm 2 i

Example 2

A non-monic quadratic

Solve $2 z^{2} + 4 z + 10 = 0$ .

Divide by $2$ : $z^{2} + 2 z + 5 = 0$ .
The discriminant is $2^{2} - 4 (1) (5) = - 16$ .
$z = (- 2 \pm 4 i) / 2 = - 1 \pm 2 i$ .

z = - 1 \pm 2 i

Check your understanding

Solve $z^{2} + 2 z + 5 = 0$ .

Show answer

$z = - 1 \pm 2 i$ .

Complete the square to ${(z + 1)}^{2} = - 4$ .

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $z^{2} + 25 = 0$ .
2 marks
Solve $z^{2} - 4 z + 8 = 0$ .
3 marks
Solve $3 z^{2} + 12 = 0$ .
2 marks
Solve $z^{2} + 6 z + 11 = 0$ exactly.
3 marks
Check directly that $1 + 3 i$ solves $z^{2} - 2 z + 10 = 0$ .
3 marks
For real $k$ , determine when $z^{2} + 4 z + k = 0$ has non-real roots.
3 marks

Read the worked solutions

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

Solve $z^{2} + 25 = 0$ .
2 marks
$z^{2} = - 25$ , whose two roots are $\pm 5 i$ .
$z = \pm 5 i$ .
Solve $z^{2} - 4 z + 8 = 0$ .
3 marks
Complete the square: ${(z - 2)}^{2} = - 4$ .
$z = 2 \pm 2 i$ .
Solve $3 z^{2} + 12 = 0$ .
2 marks
Divide by $3$ to obtain $z^{2} = - 4$ .
$z = \pm 2 i$ .
Solve $z^{2} + 6 z + 11 = 0$ exactly.
3 marks
${(z + 3)}^{2} = - 2$ .
Take both roots and subtract $3$ .
$z = - 3 \pm i \sqrt{2}$ .
Check directly that $1 + 3 i$ solves $z^{2} - 2 z + 10 = 0$ .
3 marks
${(1 + 3 i)}^{2} = - 8 + 6 i$ .
$(- 8 + 6 i) - 2 (1 + 3 i) + 10 = 0 + 0 i$ .
The substitution gives zero.
For real $k$ , determine when $z^{2} + 4 z + k = 0$ has non-real roots.
3 marks
The discriminant is $16 - 4 k$ .
It is negative exactly when $16 - 4 k < 0$ , or $k > 4$ .
$k > 4$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, real and complex numbers: Cartesian arithmetic, conjugates, quadratic roots and Argand geometry. Polar form and roots of unity are developed in Year 12.

Official curriculum source ↗

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