Conjugate roots of real quadratics | Year 11

01 / LEARN

Understand the idea.

Why the other root is a conjugate

For a polynomial with real coefficients, conjugating a zero equation leaves its coefficients unchanged. Since conjugation respects sums and products, $p (α) = 0$ implies $p (\bar{α}) = 0$ . Thus a non-real root brings its distinct conjugate with it.

A real quadratic with one non-real root therefore has that conjugate pair as its two roots. The real-coefficient condition is essential: conjugating a polynomial with complex coefficients generally changes the polynomial itself.

Recovering a quadratic

For roots $a + b i$ and $a - b i$ , the sum is $2 a$ and the product is $a^{2} + b^{2}$ . Their monic quadratic is ${(z - a)}^{2} + b^{2} = z^{2} - 2 a z + (a^{2} + b^{2})$ . The imaginary contributions cancel in its coefficients.

Monic means the leading coefficient is one. Without that requirement, any nonzero real multiple of this quadratic has the same roots. Keep the leading coefficient separate when recovering an equation from roots.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find a monic quadratic

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

Worked solution

1
Use the real-coefficient condition.
$α = 2 + 3 i, \bar{α} = 2 - 3 i$
A non-real root of a real quadratic requires the conjugate as the other root.
2
Find the root sum and product.
$\begin{aligned} α + \bar{α} & = 4 \\ α \bar{α} & = 2^{2} + 3^{2} = 13 \end{aligned}$
The imaginary parts cancel in the sum and conjugate product.
3
Form the monic equation.
$z^{2} - 4 z + 13 = 0$
A monic quadratic is $z^{2} - (root sum) z + root product$ .
4
Check the factor pair.
$(z - 2 - 3 i) (z - 2 + 3 i) = {(z - 2)}^{2} + 9 = z^{2} - 4 z + 13$
Expanding the conjugate linear factors reproduces every coefficient.

Answer

Other root $2 - 3 i$ ; equation $z^{2} - 4 z + 13 = 0$ .

EXAMPLE 02

Find two real coefficients

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

Worked solution

1
Identify the conjugate root.
$β = - 1 - 2 i$
The coefficients p and q are real, so the given non-real root has a conjugate partner.
2
Match the sum of roots.
$(- 1 + 2 i) + (- 1 - 2 i) = - 2 = - p ⟹ p = 2$
In the monic equation the root sum is the negative of the linear coefficient.
3
Match the product of roots.
$q = (- 1 + 2 i) (- 1 - 2 i) = {(- 1)}^{2} + 2^{2} = 5$
The constant term is the product of the two roots.
4
Check the resulting quadratic.
$(z + 1 - 2 i) (z + 1 + 2 i) = {(z + 1)}^{2} + 4 = z^{2} + 2 z + 5$
The expanded factors confirm p equals two and q equals five.

Answer

$p = 2$ , $q = 5$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$4 + 2 i$ .

The coefficients are real, so conjugate the given root by changing only its imaginary sign. The other root is $4 + 2 i$ .

Keep practising

6 worksheet questions · 17 marks

Open worksheet 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.

Official curriculum source ↗

In this lesson

Understand the idea.

Why the other root is a conjugate

Recovering a quadratic

See the method in action.

Find a monic quadratic

Use the real-coefficient condition.

Find the root sum and product.

Form the monic equation.

Check the factor pair.

Find two real coefficients

Identify the conjugate root.

Match the sum of roots.

Match the product of roots.

Check the resulting quadratic.

Try it yourself.

Check your understanding

Keep practising