MATHSLIVE .ie
COMPLEX NUMBERS · HLComplex Numbers 6 — Quadratics
COMPLEX NUMBERS · HL

Complex Numbers 6 — Quadratics

Proving roots and using the complex conjugate root.

Section 1 of 2

Prove a root

Prove $x = 3$ is a root of $x^2 - 5x + 6 = 0$.

Method
1.Sub in $x = 3$ and answer should equal $0$.
$3^2 - 5(3) + 6$
$9 - 15 + 6$
$0 = 0$   is a root

Prove $x = 5$ is a root of $x^2 - 7x + 10 = 0$.

Sub in $x = 5$
$5^2 - 7(5) + 10 = 0$
$25 - 35 + 10 = 0$
$0 = 0$   is a root
Section 2 of 2

Complex roots and the conjugate

Prove $z = 1 + i$ is a root of $z^2 - 2z + 2 = 0$ and state the other root.

Method
1.Sub in $1 + i$.
$(1 + i)^2 - 2(1 + i) + 2 = 0$
$(1 + i)(1 + i) - 2 - 2i + 2 = 0$
$1(1 + i) + i(1 + i) - 2i = 0$
$1 + i + i + i^2 - 2i = 0$
$1 + 2i - 1 - 2i = 0$
$0 = 0$
The other root
1.Other root is always the complex conjugate.
Ans:   $1 - i$
SUM

The lot in one box

Quadratics toolkit
1.To prove a value is a root: sub it in, answer should equal $0$.
2.$0 = 0$ confirms it is a root.
3.The other root is always the complex conjugate.

End of lesson

Complex Numbers 6 — Quadratics · HL · Mathslive.ie

Tap NEXT to reveal the first line
0%0 / 0