What is the answer to x^2+6(3-i)-26=0 ?

The answer to x^2+6(3-i)-26=0 is x=3+i,x=-3-i

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x^2+6(3-i)-26=0

x^{2} + 6 (3 - i) - 26 = 0

x = 3 + i, x = - 3 - i

Solution steps

x^{2} + 6 (3 - i) - 26 = 0

Substitute

x = a + bi

(a + bi)^{2} + 6 (3 - i) - 26 = 0

Expand

(a + bi)^{2} + 6 (3 - i) - 26 : (a^{2} - b^{2} - 8) + i (- 6 + 2 ab)

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

Rewrite

0

in standard complex form:

0 + 0 i

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

[a^{2} - b^{2} - 8 = 0 - 6 + 2 ab = 0] : (a = 3, a = - 3, b = 1 b = - 1)

Substitute back

x = a + bi

x = 3 + i, x = - 3 - i

i z^{2} + 4 z + 3 = 0

20 - 12 i = 5 l + (4 m) i

x^{2} - 4 x i + 5 = 0

2 z^{2} - 5 i z - 2 z = 0

(2 x + y) + (3 i - 5 x i) = 1 - 7 i

What is the answer to x^2+6(3-i)-26=0 ?
The answer to x^2+6(3-i)-26=0 is x=3+i,x=-3-i