# Algebra II

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Please help me with this problem.

Simplify.

√(-24 - 10i)

• Algebra II -

oh, and also, if it is not too much too ask, please include steps and explanations. ty

• Algebra II -

√(-24 - 10i)
If you want to find a perfect square which equals -24-10i, it will have to include -5i, so

(a-5i)^2 = -24-10i
a^2 - 10ai - 25 = -24-10i
Hmmm. Looks like a=1

(1-5i)^2 = 1 - 10i - 25 = -24-10i
So,
√(-24 - 10i) = 1-5i or -1+5i

Or, if you use polar form, you can see that
r=26
tanθ = 5/12
the square root has
r = √26
tanθ = (1-(-12/13))/(-5/13) = -5
so, we get the same result.

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