What is the answer to x^2+2(1+i)x-5(1+2i)=0 ?

The answer to x^2+2(1+i)x-5(1+2i)=0 is x=2+i,x=-4-3i

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x^2+2(1+i)x-5(1+2i)=0

x^{2} + 2 (1 + i) x - 5 (1 + 2 i) = 0

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

Solution steps

x^{2} + 2 (1 + i) x - 5 (1 + 2 i) = 0

Substitute

x = a + bi

(a + bi)^{2} + 2 (1 + i) (a + bi) - 5 (1 + 2 i) = 0

Expand

(a + bi)^{2} + 2 (1 + i) (a + bi) - 5 (1 + 2 i) : (a^{2} - b^{2} + 2 a - 2 b - 5) + i (- 10 + 2 a + 2 b + 2 ab)

(a^{2} - b^{2} + 2 a - 2 b - 5) + i (- 10 + 2 a + 2 b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + 2 a - 2 b - 5) + i (- 10 + 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 - 5 = 0 - 10 + 2 a + 2 b + 2 ab = 0]

[a^{2} - b^{2} + 2 a - 2 b - 5 = 0 - 10 + 2 a + 2 b + 2 ab = 0] : (a = 2, a = - 4, b = 1 b = - 3)

Substitute back

x = a + bi

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

i (x) = \frac{640000}{4 - 4}

s i f (x) = 2 - 3 x^{2}

a \cdot (a + bi) = 4 + 6 i

x^{2} - 9 i + 11 = 3 i

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

What is the answer to x^2+2(1+i)x-5(1+2i)=0 ?
The answer to x^2+2(1+i)x-5(1+2i)=0 is x=2+i,x=-4-3i