Conjugate roots of real quadratics | Year 11

01 / LEARN

The idea, step by step

Why the other root is a conjugate

Let $p (z) = a z^{2} + b z + c$ with real $a, b, c$ . Conjugating $p (z) = 0$ gives $a {\bar{z}}^{2} + b \bar{z} + c = 0$ , because conjugation leaves real coefficients unchanged and respects addition and multiplication.

So if $p (a_{0} + b_{0} i) = 0$ , then $p (a_{0} - b_{0} i) = 0$ . For a non-real root these values differ. The real-coefficient condition matters: it is not a rule for every polynomial with complex coefficients.

Recovering a quadratic

A conjugate pair $α = a + b i$ and $\bar{α} = a - b i$ produces the real factor pair $(z - α) (z - \bar{α}) = z^{2} - 2 a z + (a^{2} + b^{2})$ .

This gives a monic quadratic. Any nonzero real multiple has the same roots. State whether the question requires a monic equation before assuming the leading coefficient is one.

A common trap

Assuming a complex polynomial must always have conjugate roots.

Instead: The guarantee applies to polynomials with real coefficients. For example, $z - (1 + i)$ has only root $1 + i$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Find a monic quadratic

A monic real quadratic has root $2 + 3 i$ . Find the other root and its equation.

Real coefficients require the conjugate root $2 - 3 i$ .
The sum of roots is $4$ and product is $(2 + 3 i) (2 - 3 i) = 13$ .
The monic equation is $z^{2} - 4 z + 13 = 0$ .

Other root

2 - 3 i

; equation

z^{2} - 4 z + 13 = 0

Example 2

Find two real coefficients

Given that $- 1 + 2 i$ solves $z^{2} + p z + q = 0$ , with real $p, q$ , find $p, q$ .

The other root is $- 1 - 2 i$ .
The sum is $- 2 = - p$ , so $p = 2$ .
The product is ${(- 1)}^{2} + 2^{2} = 5 = q$ .
Check: ${(z + 1)}^{2} + 4 = z^{2} + 2 z + 5$ .

p = 2

q = 5

Check your understanding

A real quadratic has root $4 - 2 i$ . What is its other root?

Show answer

$4 + 2 i$ .

Conjugate the given non-real root because the coefficients are real.

03 / WORKSHEET

Your turn

6 questions · 17 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

A real quadratic has root $- 3 + 5 i$ . State the other root.
2 marks
Find the monic real quadratic with roots $1 \pm 4 i$ .
3 marks
Find the monic real quadratic with roots $- 2 \pm i$ .
3 marks
One root of $2 z^{2} + b z + c = 0$ is $3 + i$ . Find real $b, c$ .
3 marks
A student claims a real quadratic has exactly the roots $2 + i$ and $3 - i$ . Explain why this is impossible.
3 marks
For $p (z) = z - (1 + i)$ , test whether the conjugate of its root is also a root. Explain the result.
3 marks

Read the worked solutions

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

A real quadratic has root $- 3 + 5 i$ . State the other root.
2 marks
Use the conjugate-root property.
$- 3 - 5 i$ .
Find the monic real quadratic with roots $1 \pm 4 i$ .
3 marks
The sum is $2$ and product is $1^{2} + 4^{2} = 17$ .
$z^{2} - 2 z + 17 = 0$ .
Find the monic real quadratic with roots $- 2 \pm i$ .
3 marks
Expand $(z + 2 - i) (z + 2 + i) = {(z + 2)}^{2} + 1$ .
$z^{2} + 4 z + 5 = 0$ .
One root of $2 z^{2} + b z + c = 0$ is $3 + i$ . Find real $b, c$ .
3 marks
The monic quadratic is $z^{2} - 6 z + 10$ .
Multiply by the required leading coefficient $2$ .
$b = - 12$ , $c = 20$ .
A student claims a real quadratic has exactly the roots $2 + i$ and $3 - i$ . Explain why this is impossible.
3 marks
$(2 + i) (3 - i) = 7 + i$ .
A real quadratic's product of roots must be real.
The two roots are not conjugates; their sum is $5$ , but their product is $7 + i$ , which is not real.
For $p (z) = z - (1 + i)$ , test whether the conjugate of its root is also a root. Explain the result.
3 marks
The coefficient $- (1 + i)$ is non-real.
The real-coefficient condition is not satisfied.
$1 - i$ is not a root because $p (1 - i) = - 2 i$ .

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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.

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