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

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

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x^2+ix+2=0

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

x = - 2 i, x = i

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

x = a + bi

x = - 2 i, x = i

3 (x + 5 i) = 7 y i - 6 + 8

x - 5 i + 2 = 4 - 3 i + y i

(x + y i) + (4 + 9 i) = 9 - 4 i

x - i y = 3 + 4 i

5 i - 6 x = (10 y) i + 2