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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ix^2+(1-2i)x+3i-1=0

i x^{2} + (1 - 2 i) x + 3 i - 1 = 0

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

Solution steps

i x^{2} + (1 - 2 i) x + 3 i - 1 = 0

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

[- 2 ab + a + 2 b - 1 = 0 3 + a^{2} + b - b^{2} - 2 a = 0]

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

Substitute back

x = a + bi

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

x - 2 i = 2 + y i

z - \frac{1}{z} = 4 + 4 i

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

z^{2} - (5 + 4 i) z - (3 - 5 i) = 0

s i f (x) = x^{2} - x, f (x + 1) = f (- x)

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