Substition vs completing the square on an irreducible polynimial of degree 4

Find the all the roots of the polynomial x^{4} + x^{2} + 1.

The above is from an example in a text book. I attempted the question and solved it (I think) by

letting u = x^{2}

then solving the rewritten polynomial u^{2} + u + 1 using the quadratic formula before solving for x.

I get the roots, $\displaystyle \sqrt[]{\frac {-1 + \sqrt[]{3} i} {2}}, -\sqrt[]{\frac {-1 + \sqrt[]{3} i} {2}}, \sqrt[]{\frac {-1 - \sqrt[]{3} i} {2}}, -\sqrt[]{\frac {-1 - \sqrt[]{3} i} {2}} $

But the book's author use completing the square to obtain the factors (x ^{2} + x + 1)(x^{2} - x + 1)

Which he then proceeds to solve using the quadratic formula to obtain $\displaystyle \frac {1 + \sqrt[]{3} i} {2}, \frac {-1 + \sqrt[]{3} i} {2}, \frac {1 - \sqrt[]{3} i} {2}, \frac {-1 - \sqrt[]{3} i} {2}$

My question is, why the different answers? Am I doing it wrong? That would be ironic because I learned that substitution trick from the very same book. Or is there something am missing?

Re: Substition vs completing the square on an irreducible polynimial of degree 4

You might be interested to know that $\displaystyle \sqrt{\frac{-1 - \sqrt{3} i }{2}} = \frac{1 - \sqrt{3} i }{2}$

So those answers you got are the same as the ones in the book.

Re: Substition vs completing the square on an irreducible polynimial of degree 4

To the OP: You are expected to write your solutions in the form a + bi, where a and b are real numbers. To find the square root of a complex number, it's easiest to convert them to polar form first.

Re: Substition vs completing the square on an irreducible polynimial of degree 4

It's good to know the answers are the same. I was going crazy trying to figure out why the answers were different. But, if its not too much trouble, could you explain why they are the same? Assuming of course that the explanation itself does not require familiarity with an advanced mathematical concept.

Re: Substition vs completing the square on an irreducible polynimial of degree 4

The polar form looks like it utilizes trigonometric functions. I am studying the book linearly and I haven't yet gotten to the trigonometry section which is where they explain the polar form of a complex number. I guess, till then, I'll just have to take Educated's word for it that the answers I got are the same as the ones in the book. But you are encouraged to offer a 'dumbed-down' explanation as to why that is.

Re: Substition vs completing the square on an irreducible polynimial of degree 4

The easiest way to check if $\displaystyle \displaystyle \begin{align*} \sqrt{\frac{-1 - \sqrt{3} \, i}{2}} = \frac{ 1 - \sqrt{3}\,i}{2} \end{align*}$ is to check if $\displaystyle \displaystyle \begin{align*} \left( \frac{ 1 - \sqrt{3}\, i}{2} \right)^2 = \frac{-1 - \sqrt{3} \, i}{2} \end{align*}$...