What is the answer to ix^2-2(i+1)x+10=0 ?

The answer to ix^2-2(i+1)x+10=0 is x=-1-3i,x=3+i

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ix^2-2(i+1)x+10=0

i x^{2} - 2 (i + 1) x + 10 = 0

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

Solution steps

i x^{2} - 2 (i + 1) x + 10 = 0

Substitute

x = a + bi

i (a + bi)^{2} - 2 (i + 1) (a + bi) + 10 = 0

Expand

i (a + bi)^{2} - 2 (i + 1) (a + bi) + 10 : (- 2 ab + 2 b - 2 a + 10) + i (a^{2} - b^{2} - 2 a - 2 b)

(- 2 ab + 2 b - 2 a + 10) + i (a^{2} - b^{2} - 2 a - 2 b) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(- 2 ab + 2 b - 2 a + 10) + i (a^{2} - b^{2} - 2 a - 2 b) = 0 + 0 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[- 2 ab + 2 b - 2 a + 10 = 0 a^{2} - b^{2} - 2 a - 2 b = 0]

[- 2 ab + 2 b - 2 a + 10 = 0 a^{2} - b^{2} - 2 a - 2 b = 0] : (a = - 1, a = 3, b = - 3 b = 1)

Substitute back

x = a + bi

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

s i y = 33 y + 1

\frac{1 + 4 i}{3 - 6 i} = a + bi

z^{2} + (1 - 2 i) z + 1 + 5 i = 0

a^{2} + 2 abi - b^{2} = 2 i

a + (b + 1) \cdot i = 4

What is the answer to ix^2-2(i+1)x+10=0 ?
The answer to ix^2-2(i+1)x+10=0 is x=-1-3i,x=3+i