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

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

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

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

No Solution

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

x = a + bi

No Solution

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

2 x - 4 y i = 10 - 5 i

\frac{( z - i )}{z} = 2 + i

(3 + bi) (3 + d i) = 11 - 3 i

(x + y i) + (4 + 6 i) = 7 - 2 i

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