# Math Help - Solving an equation

1. ## Solving an equation

I'm not getting this one.

$3z^4 - 5z^2 +1 = 0$

If I substitute variable u in place of $z^2$, I get

$3u^2 - 5u + 1 = 0$

solving with the Quadratic Equation gives me

$u= 5 \pm \frac{\sqrt13}{6}$

and since u = $z^2$, the solutions should be

z = $\pm \sqrt{5 \pm \frac{\sqrt13}{6}}$

But, the book says the solution is

$z = \pm \frac{ \sqrt{30 \pm 6\sqrt13}} {6}$

What have I done wrong?

$u=\frac{5\pm\sqrt{13}}{6}$.
Then $z=\pm\sqrt{\frac{5\pm\sqrt{13}}{6}}=\pm\frac{\sqrt {30\pm 6\sqrt{13}}}{6}$