What is the answer to x^2-(2+i)x+(-1+7i)=0 ?

The answer to x^2-(2+i)x+(-1+7i)=0 is x=-1+2i,x=3-i

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x^2-(2+i)x+(-1+7i)=0

x^{2} - (2 + i) x + (- 1 + 7 i) = 0

x = - 1 + 2 i, x = 3 - i

Solution steps

x^{2} - (2 + i) x + (- 1 + 7 i) = 0

Substitute

x = a + bi

(a + bi)^{2} - (2 + i) (a + bi) - 1 + 7 i = 0

Expand

(a + bi)^{2} - (2 + i) (a + bi) - 1 + 7 i : (a^{2} - b^{2} - 2 a + b - 1) + i (7 - a - 2 b + 2 ab)

(a^{2} - b^{2} - 2 a + b - 1) + i (7 - a - 2 b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} - 2 a + b - 1) + i (7 - 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 + b - 1 = 0 7 - a - 2 b + 2 ab = 0]

[a^{2} - b^{2} - 2 a + b - 1 = 0 7 - a - 2 b + 2 ab = 0] : (a = - 1, a = 3, b = 2 b = - 1)

Substitute back

x = a + bi

x = - 1 + 2 i, x = 3 - i

i (x) = \frac{1}{x + 1}

\frac{z + 1}{z - 1} = 2 + i

z^{2} - z - 1 - 3 i = 0

\frac{3 - a + 2 b}{5} - 3 i + a + bi = 4 - i

What is the answer to x^2-(2+i)x+(-1+7i)=0 ?
The answer to x^2-(2+i)x+(-1+7i)=0 is x=-1+2i,x=3-i