Thread: Solving Quartics

Hi

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

Solve for max. height b.

Could someone please show me how to solve b?

Originally Posted by xwrathbringerx

Hi

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

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 $\displaystyle b^2=t \implies b^4=t^2$

So we get

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

This should get you started.

Good luck.

I tried solving it as a quadratic...using b = $\displaystyle - \frac{b}{2a} $ as well as the quadratic formula but it doesn't seem to match the answer, which is 4.84

$\displaystyle \frac{-b}{2a} $ is for finding the axis of symmetry, no need for it here.

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