# Thread: nth roots of a complex number

1. ## nth roots of a complex number

Find all solutions of the equation

x^3+125=0

2. ## x^3+125=0

Wouldn't that be -5 ?

If you must have it in complex number form, a + bi, that could be -5 + 0i.

A trivial answer to a trivial question, I hope.

But if I am wrong, perhaps more mysteries will be revealed by the next respondent.

3. Originally Posted by The Math Man
Find all solutions of the equation

x^3+125=0
$a^3+b^3=(a+b)(a^2-ab+b^2)$

a=x, b=5 After that you can use the quadratic formula to get the two complex roots.

Hope that helps.

4. ## more roots

Thanks, Jen.

Would it be too much for you to show the steps to the roots?

I don't recall ever being asked to go beyond factoring an expressions of this sort and it would be enlightening to see an example.

Thanks, Jen.

Would it be too much for you to show the steps to the roots?

I don't recall ever being asked to go beyond factoring an expressions of this sort and it would be enlightening to see an example.
$x^3+125=x^3+5^3=(x+5)(x^2-5x+25)$

So one root is x=-5.

Now using the quadratic formula we get

$x=\frac{-(-5) \pm \sqrt{(-5)^2-4(1)(25)}}{2(1)}=\frac{5 \pm \sqrt{-75}}{2}=\frac{5 \pm 5i\sqrt{3}}{2}$

Since these are zero's $x-\frac{5 \pm 5i\sqrt{3}}{2}$ are factors.

$\left( x+5 \right) \left(x- \frac{5 + 5i\sqrt{3}}{2}\right) \left( x- \frac{5 - 5i\sqrt{3}}{2}\right)$