# Solving Quartics

• April 24th 2009, 08:52 PM
xwrathbringerx
Solving Quartics
Hi

$b^4 - 25b^2 + 37.20 = 0$

Solve for max. height b.

Could someone please show me how to solve b?
• April 24th 2009, 08:55 PM
TheEmptySet
Quote:

Originally Posted by xwrathbringerx
Hi

$b^4 - 25b^2 + 37.20 = 0$

Solve for max. height b.

Could someone please show me how to solve b?

This is quadratic in form:

It may help you see it if you let $b^2=t \implies b^4=t^2$

So we get

$t^2-25t+37.20=0$

This should get you started.

Good luck.
• April 24th 2009, 09:02 PM
xwrathbringerx
I tried solving it as a quadratic...using b = $- \frac{b}{2a}$ as well as the quadratic formula but it doesn't seem to match the answer, which is 4.84 (Doh)
• April 24th 2009, 11:30 PM
Peleus
$\frac{-b}{2a}$ is for finding the axis of symmetry, no need for it here.

$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ is what you want, a = 1, b = 25, c = 37.20