What is the answer to x^2+3x+2-i=0 ?

The answer to x^2+3x+2-i=0 is x=-0.69975…+0.62481…i,x=-2.30024…-0.62481…i

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x^2+3x+2-i=0

x^{2} + 3 x + 2 - i = 0

x = - 0.69975 \dots + 0.62481 \dots i, x = - 2.30024 \dots - 0.62481 \dots i

Solution steps

x^{2} + 3 x + 2 - i = 0

Substitute

x = a + bi

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

Expand

(a + bi)^{2} + 3 (a + bi) + 2 - i : (a^{2} - b^{2} + 3 a + 2) + i (- 1 + 3 b + 2 ab)

(a^{2} - b^{2} + 3 a + 2) + i (- 1 + 3 b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + 3 a + 2) + i (- 1 + 3 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} + 3 a + 2 = 0 - 1 + 3 b + 2 ab = 0]

[a^{2} - b^{2} + 3 a + 2 = 0 - 1 + 3 b + 2 ab = 0] : (a = - 0.69975 \dots, a = - 2.30024 \dots, b = 0.62481 \dots b = - 0.62481 \dots)

Substitute back

x = a + bi

x = - 0.69975 \dots + 0.62481 \dots i, x = - 2.30024 \dots - 0.62481 \dots i

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

- 8.4 x i - 3.15 = - x^{2}

i x = x^{2} - 2 x + 2

x^{2} + 2 x i + 1 = 0

(2 + i) 2 = c + d i

What is the answer to x^2+3x+2-i=0 ?
The answer to x^2+3x+2-i=0 is x=-0.69975…+0.62481…i,x=-2.30024…-0.62481…i