Quadratic equations with complex solutions

Warm-up

Three quick questions from earlier lessons.

Discover

$x^2 + 9 = 0$ has no real solution. With $i$ , which numbers solve it?

$x = 3i$ and $x = -3i$
$x = 3$ and $x = -3$

Key idea

Every quadratic has solutions among the complex numbers.

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

. If the discriminant

b^2 - 4ac < 0

, the solutions are two non-real complex conjugates

p \pm qi

; if

= 0

, one real solution; if

> 0

, two real solutions.

Worked example

$x^2 - 4x + 13 = 0$ .

Discriminant

= -36

16 − 52.

x = \dfrac{4 \pm 6i}{2}

√(−36) = 6i.

x = 2 \pm 3i

Divide both parts by 2.

Your turn, guided

Your turn: the discriminant of $x^2 + 2x + 5 = 0$ .

-16

: two non-real solutions

4 − 20.

Practice ★

$x^2 + 36 = 0$ . Enter the solution with positive imaginary part.
$x^2 + 81 = 0$ . Enter the solution with positive imaginary part.

Find the discriminant of $1x^2 + 7x + 4 = 0$ .

Watch out!

Practice ★★

$x^2 - 4x + 20 = 0$ . Enter the solution with positive imaginary part.
$x^2 - 6x + 34 = 0$ . Enter the solution with positive imaginary part.

Boss problem. Solve $x^2 - 10x + 29 = 0$ with the quadratic formula.

Exit ticket

Three questions to check you have got it.

$x^2 + 36 = 0$ . Enter the solution with positive imaginary part.
$x^2 + 81 = 0$ . Enter the solution with positive imaginary part.
$x^2 + 25 = 0$ . Enter the solution with positive imaginary part.

Summary card

Complex solutions

Discriminant < 0 → $x = p \pm qi$ (conjugates).

Watch out: divide the whole numerator by $2a$ .

End of the lesson

Next lesson: Simplifying, multiplying and dividing rational expressions Back to the chapter

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