# Solve: x^2 + i = 0show complete solution and explain the answer.

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You need to move i to the right side such that:

`x^2 = -i`

You need to consider a complex number `x=a+bi` . The square of this number yields `-i` such that:

`(a+bi)^2 = -i`

You need to expand binomial such that:

`a^2 + 2abi + (bi)^2 = -i`

You need to use that `i^2=-1` , hence:

`a^2 + 2abi - b^2 = -i`

You need to equate real parts both sides and imaginary parts both sides such that:

`a^2 - b^2 ` = 0

`2ab = -1 =gt ab = -1/2 =gt a=-1/(2b)`

`1/(4b^2) - b^2 = 0 =gt (1/(2b) - b)(1/(2b)+ b) = 0`

`1/(2b) - b = 0 =gt 1-2b^2 = 0 =gt 2b^2=1 =gt b^2=1/2`

`b_(1,2) = +-sqrt2/2`

`a_(1,2) = - + sqrt2/ 2`

**Hence, the complex numbers that squared yield -i are `x_1 = -sqrt2/2 + sqrt2/2*i` and `x_2 = sqrt2/2 - sqrt2/2*i.` **