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

The answer to x^2+(4-2i)x-(1+i)=0 is x=0.12132…+0.29289…i,x=-4.12132…+1.70710…i

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x^2+(4-2i)x-(1+i)=0

x^{2} + (4 - 2 i) x - (1 + i) = 0

x = 0.12132 \dots + 0.29289 \dots i, x = - 4.12132 \dots + 1.70710 \dots i

Solution steps

x^{2} + (4 - 2 i) x - (1 + i) = 0

Substitute

x = a + bi

(a + bi)^{2} + (4 - 2 i) (a + bi) - (1 + i) = 0

Expand

(a + bi)^{2} + (4 - 2 i) (a + bi) - (1 + i) : (a^{2} - b^{2} + 4 a + 2 b - 1) + i (- 1 - 2 a + 4 b + 2 ab)

(a^{2} - b^{2} + 4 a + 2 b - 1) + i (- 1 - 2 a + 4 b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + 4 a + 2 b - 1) + i (- 1 - 2 a + 4 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} + 4 a + 2 b - 1 = 0 - 1 - 2 a + 4 b + 2 ab = 0]

[a^{2} - b^{2} + 4 a + 2 b - 1 = 0 - 1 - 2 a + 4 b + 2 ab = 0] : (a = 0.12132 \dots, a = - 4.12132 \dots, b = 0.29289 \dots b = 1.70710 \dots)

Substitute back

x = a + bi

x = 0.12132 \dots + 0.29289 \dots i, x = - 4.12132 \dots + 1.70710 \dots i

x^{2} - 4 x + 40 = 2 - 6 i

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

3 x + 8 i = 12 + 12

(x + y) + (x - y) \cdot i = 5 + 5 \cdot i

5 x + 6 i = - 8 + 6

What is the answer to x^2+(4-2i)x-(1+i)=0 ?
The answer to x^2+(4-2i)x-(1+i)=0 is x=0.12132…+0.29289…i,x=-4.12132…+1.70710…i