What is the answer to x^2+(6+2i)x+(11+10i)=0 ?

The answer to x^2+(6+2i)x+(11+10i)=0 is x=-2-3i,x=-4+i

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x^2+(6+2i)x+(11+10i)=0

x^{2} + (6 + 2 i) x + (11 + 10 i) = 0

x = - 2 - 3 i, x = - 4 + i

Solution steps

x^{2} + (6 + 2 i) x + (11 + 10 i) = 0

Substitute

x = a + bi

(a + bi)^{2} + (6 + 2 i) (a + bi) + 11 + 10 i = 0

Expand

(a + bi)^{2} + (6 + 2 i) (a + bi) + 11 + 10 i : (a^{2} - b^{2} + 6 a - 2 b + 11) + i (10 + 2 a + 6 b + 2 ab)

(a^{2} - b^{2} + 6 a - 2 b + 11) + i (10 + 2 a + 6 b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + 6 a - 2 b + 11) + i (10 + 2 a + 6 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} + 6 a - 2 b + 11 = 0 10 + 2 a + 6 b + 2 ab = 0]

[a^{2} - b^{2} + 6 a - 2 b + 11 = 0 10 + 2 a + 6 b + 2 ab = 0] : (a = - 2, a = - 4, b = - 3 b = 1)

Substitute back

x = a + bi

x = - 2 - 3 i, x = - 4 + i

(3 - i) z + (1 + 2 i) = 7 + i es

3 + ai = b + 5 i

2 x + 3 i x - 3 i y + 4 i y + x - y = - 400

10 + 6 Yi = 5 X + 24 i

z^{2} + 3 + i = 0

What is the answer to x^2+(6+2i)x+(11+10i)=0 ?
The answer to x^2+(6+2i)x+(11+10i)=0 is x=-2-3i,x=-4+i