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

The answer to x^2-(2-i)x+(3-i)=0 is No Solution

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

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

No Solution

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

[a^{2} - b^{2} - 2 a - b + 3 = 0 - 1 + a - 2 b + 2 ab = 0] : No Solution

Substitute back

x = a + bi

No Solution

(x + y) - (3 x + 2 y) i = 5 + 2 i

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

(2 x - 5 i y) (1 + i)^{2} = 10 (2 + i)

z^{2} = 3 + i

z^{2} + (2 i - 3) + 5 - i = 0

What is the answer to x^2-(2-i)x+(3-i)=0 ?
The answer to x^2-(2-i)x+(3-i)=0 is No Solution