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

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

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

x^{2} + x + 3 i = 0

x = \frac{1}{2} - \frac{3}{2} i, x = - \frac{3}{2} + \frac{3}{2} i

Solution steps

x^{2} + x + 3 i = 0

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

x = a + bi

x = \frac{1}{2} - \frac{3}{2} i, x = - \frac{3}{2} + \frac{3}{2} i

\frac{1}{0} = a + bi

(x^{2} - 2 x - 15) + (3 y - 9) i = 3 i

3 ai - b - 3 i = a - 5 + bi

∣ z - 1 + 3 i ∣ = 2

∣ x + bi + 1 + i ∣ = 1

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