Hi $\displaystyle b^4 - 25b^2 + 37.20 = 0$ Solve for max. height b. Could someone please show me how to solve b?
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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
Last edited by xwrathbringerx; Apr 24th 2009 at 08:32 PM.
$\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
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