What is the answer to x^2-(2-4i)x-(3+6i)=0 ?

The answer to x^2-(2-4i)x-(3+6i)=0 is x=-3i,x=2-i

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x^2-(2-4i)x-(3+6i)=0

x^{2} - (2 - 4 i) x - (3 + 6 i) = 0

x = - 3 i, x = 2 - i

Solution steps

x^{2} - (2 - 4 i) x - (3 + 6 i) = 0

Substitute

x = a + bi

(a + bi)^{2} - (2 - 4 i) (a + bi) - (3 + 6 i) = 0

Expand

(a + bi)^{2} - (2 - 4 i) (a + bi) - (3 + 6 i) : (a^{2} - b^{2} - 2 a - 4 b - 3) + i (- 6 - 2 b + 4 a + 2 ab)

(a^{2} - b^{2} - 2 a - 4 b - 3) + i (- 6 - 2 b + 4 a + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} - 2 a - 4 b - 3) + i (- 6 - 2 b + 4 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} - 2 a - 4 b - 3 = 0 - 6 - 2 b + 4 a + 2 ab = 0]

[a^{2} - b^{2} - 2 a - 4 b - 3 = 0 - 6 - 2 b + 4 a + 2 ab = 0] : (a = 0, a = 2, b = - 3 b = - 1)

Substitute back

x = a + bi

x = - 3 i, x = 2 - i

y - 8 = 3 (x + 1) in

x, R s i R (x) = 8 - 2 x + x^{2}

3 x - 200 s i \cdot = - 10

i (x) = - x^{2} + 60 x

m, (m - n)^{2} = - 2 i

What is the answer to x^2-(2-4i)x-(3+6i)=0 ?
The answer to x^2-(2-4i)x-(3+6i)=0 is x=-3i,x=2-i