# Solve this simple equation

1. Apr 10, 2014

### skrat

1. The problem statement, all variables and given/known data

$z^4-6z^2+1=0$

2. Relevant equations

3. The attempt at a solution

$z^{2}=\frac{6\pm \sqrt{36-4}}{2}=3\pm 2\sqrt{2}$. Now this has to be wrong already yet I don't know why..

Wolfram gives me some solutions $1+\sqrt{2}$, $1-\sqrt{2}$, $-1+\sqrt{2}$, $-1-\sqrt{2}$...

Now how can I find them without wolfram alpha?

2. Apr 10, 2014

### jbunniii

No, it's fine. Note that you have solved for $z^2$, not $z$. If the goal is to solve for $z$, you need to do some more work. Start with one of your solutions, say $z^2 = 3 + 2 \sqrt{2}$. There are two values of $z$ that satisfy this equation, namely $z = \pm \sqrt{3 + 2 \sqrt{2}}$. Note that this expression can be simplified. Indeed, $(1 + \sqrt{2})^2 = 3 + 2 \sqrt{2}$, so $\sqrt{3 + 2 \sqrt{2}} = 1 + \sqrt{2}$ is one solution. See if you can find the other solutions. There should be four since the original polynomial $z^4 -6z^2 + 1$ has degree 4.

3. Apr 10, 2014

### skrat

See, I couldn't see that. Now It is obvious. Thanks!

4. Apr 10, 2014

### LCKurtz

Your original question asked how to get the roots without Wolfram. Once you know that you have $$z=\pm\sqrt{3+2\sqrt 2}$$it is easy enough to verify that $z=1+2\sqrt 2$ works, but it's another thing to find that form in the first place. You might find it instructive to set$$(a+b\sqrt 2)^2 = 3+2\sqrt 2$$square both sides, and solve for $a$ and $b$.