Check the solution to "Complex Quadraticz2−2(1−2i)z−8i=0"

High School
Algebra
Algebra I
Quadratic function and equation

London Douglas

Answered question

2022-03-28

Complex Quadratic

z^{2} - 2 (1 - 2 i) z - 8 i = 0

Answer & Explanation

okusen8m1a

Beginner2022-03-29Added 8 answers

$\begin{aligned} x_{1, 2} & = \frac{2 - 4 i \pm \sqrt{(4 i - 2)^{2} + 4 \cdot 8 i}}{2} \\ = \frac{2 - 4 i \pm \sqrt{- 16 + 4 - 16 i + 32 i}}{2} \\ = \frac{2 - 4 i \pm \sqrt{- 12 + 16 i}}{2} \\ = \frac{2 - 4 i \pm \sqrt{4 (- 3 + 4 i)}}{2} \\ = \frac{2 - 4 i \pm 2 \sqrt{(4 i - 3)}}{2} \\ = 1 - 2 i \pm \sqrt{4 i - 3} \end{aligned}$
Let $a + b i = \sqrt{4 i - 3}$
$a^{2} - b^{2} + a b i = 4 i - 3$
$a^{2} - b^{2} = - 3, a b = 4$
$b = \frac{4}{a}$
$a^{2} - {(\frac{4}{a})}^{2} = - 3$
$a^{4} + 3 a^{2} - 16 = 0$
$t^{2} + 3 t - 16 = 0$
$t_{1, 2} = \frac{- 3 \pm \sqrt{9 + 4 \cdot 16}}{2}$
Answers should be $z = 2, z = - 4 i$ , I don't know what I am doing wrong, I can get it right when substituting $z = a + b i$ but I don't when using this method.

haiguetenteme7zyu

Beginner2022-03-30Added 13 answers

The mistake is when you square

a + b i

, you should get

a^{2} - b^{2} + 2 a b i

and not

a^{2} - b^{2} + a b i

a^{4} + 3 a^{2} - 4 = 0

(a^{2} + 4) (a^{2} - 1) = 0

a = \pm 1, b = \pm 2

Hence,

x_{1, 2} = 1 - 2 i \pm (1 + 2 i)

which gives you the desired solution of 2 and -4i.

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