What is the answer to 2ix^2+x+3i=0 ?

The answer to 2ix^2+x+3i=0 is x=-i,x= 3/2 i

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2ix^2+x+3i=0

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

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

Solution steps

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

Substitute

x = a + bi

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

Expand

2 i (a + bi)^{2} + a + bi + 3 i : (- 4 ab + a) + i (3 + b + 2 a^{2} - 2 b^{2})

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

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

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

Substitute back

x = a + bi

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

w^{2} - i (w - 2) = (w - 2)

x^{2} - y^{2} + i 2 x y = y + i x

\frac{- 4 + x i}{2 - 3 i} = y - 2 i

3 x - 2 i y - i x + 5 y = 7 + 5 i

x + i y = ∣ x - i y ∣

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