Solving Quartics

**xwrathbringerx** · Apr 24th 2009, 07:52 PM

Hi

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

Solve for max. height b.

Could someone please show me how to solve b?

**TheEmptySet** · Apr 24th 2009, 07:55 PM

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.

**xwrathbringerx** · Apr 24th 2009, 08:02 PM

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

**Peleus** · Apr 24th 2009, 10:30 PM

$\frac{-b}{2a}$ 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

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