What is the answer to -x^2+6ix+8=0 ?

The answer to -x^2+6ix+8=0 is x=4i,x=2i

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-x^2+6ix+8=0

- x^{2} + 6 i x + 8 = 0

x = 4 i, x = 2 i

Solution steps

- x^{2} + 6 i x + 8 = 0

Substitute

x = a + bi

- (a + bi)^{2} + 6 i (a + bi) + 8 = 0

Expand

- (a + bi)^{2} + 6 i (a + bi) + 8 : (- a^{2} + b^{2} - 6 b + 8) + i (6 a - 2 ab)

(- a^{2} + b^{2} - 6 b + 8) + i (6 a - 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(- a^{2} + b^{2} - 6 b + 8) + i (6 a - 2 ab) = 0 + 0 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[- a^{2} + b^{2} - 6 b + 8 = 0 6 a - 2 ab = 0]

[- a^{2} + b^{2} - 6 b + 8 = 0 6 a - 2 ab = 0] : (a = 0, a = 0, b = 4 b = 2)

Substitute back

x = a + bi

x = 4 i, x = 2 i

2 x + 3 y i = 10 + 18 i

(x + y i)^{2} = 3 + 4 i

3 x + 5 y - 7 = - 2 i y + i x + 5 i

r = 6 e^{\frac{π}{3} i}

(x + i y)^{2} + 1 = 1