What is the answer to x^2+(2-2i)x+3-6i=0 ?

The answer to x^2+(2-2i)x+3-6i=0 is x=3i,x=-2-i

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x^2+(2-2i)x+3-6i=0

x^{2} + (2 - 2 i) x + 3 - 6 i = 0

x = 3 i, x = - 2 - i

Solution steps

x^{2} + (2 - 2 i) x + 3 - 6 i = 0

Substitute

x = a + bi

(a + bi)^{2} + (2 - 2 i) (a + bi) + 3 - 6 i = 0

Expand

(a + bi)^{2} + (2 - 2 i) (a + bi) + 3 - 6 i : (a^{2} - b^{2} + 2 a + 2 b + 3) + i (- 6 - 2 a + 2 b + 2 ab)

(a^{2} - b^{2} + 2 a + 2 b + 3) + i (- 6 - 2 a + 2 b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + 2 a + 2 b + 3) + i (- 6 - 2 a + 2 b + 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} + 2 a + 2 b + 3 = 0 - 6 - 2 a + 2 b + 2 ab = 0]

[a^{2} - b^{2} + 2 a + 2 b + 3 = 0 - 6 - 2 a + 2 b + 2 ab = 0] : (a = 0, a = - 2, b = 3 b = - 1)

Substitute back

x = a + bi

x = 3 i, x = - 2 - i

a^{2} - b^{2} - a - bi = 1

(2 x - 3 y i) (1 - 2 i) = 1 - 4 i

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

(1 + 2 i) (x + y i) - (1 + 2 i) (x - y i) = 16

z^{2} = - 4 i

What is the answer to x^2+(2-2i)x+3-6i=0 ?
The answer to x^2+(2-2i)x+3-6i=0 is x=3i,x=-2-i