What is the answer to x^2+3(1-i)x-5i=0 ?

The answer to x^2+3(1-i)x-5i=0 is x=-1+2i,x=-2+i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+3(1-i)x-5i=0

x^{2} + 3 (1 - i) x - 5 i = 0

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

Solution steps

x^{2} + 3 (1 - i) x - 5 i = 0

Substitute

x = a + bi

(a + bi)^{2} + 3 (1 - i) (a + bi) - 5 i = 0

Expand

(a + bi)^{2} + 3 (1 - i) (a + bi) - 5 i : (a^{2} - b^{2} + 3 a + 3 b) + i (- 5 - 3 a + 3 b + 2 ab)

(a^{2} - b^{2} + 3 a + 3 b) + i (- 5 - 3 a + 3 b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + 3 a + 3 b) + i (- 5 - 3 a + 3 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} + 3 a + 3 b = 0 - 5 - 3 a + 3 b + 2 ab = 0]

[a^{2} - b^{2} + 3 a + 3 b = 0 - 5 - 3 a + 3 b + 2 ab = 0] : (a = - 1, a = - 2, b = 2 b = 1)

Substitute back

x = a + bi

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

(2 + 3 i) (x + i y) = (2 + i)^{2}

(9 - 8 i) - 6 = x - (2 + y i)

(x + i y) (4 i) = 8

\frac{1285}{8} - 60 i = 1 0^{6} M^{2}

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

What is the answer to x^2+3(1-i)x-5i=0 ?
The answer to x^2+3(1-i)x-5i=0 is x=-1+2i,x=-2+i