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

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

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

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

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

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

x = a + bi

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

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

x^{2} + (4 - 2 i) x - (1 + i) = 0

x^{2} - 4 x + 40 = 2 - 6 i

(x + i y) (2 + 1) = 3 - i

3 x + 8 i = 12 + 12

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