I have a function $f(CE) = 2 \cdot \sqrt{4^2+(3-CE)^2} + \sqrt{4^2+CE^2 }$

And it's derivative $f^′ (CE)= \dfrac{−6 + 2 CE}{\sqrt{25−6CE+CE^2}}+\dfrac{CE}{\sqrt{16+CE^2 }}$

And then I'm trying to find the minimum for this function by setting $f'(CE) = 0$

However I am completely lost on solving this equation:

$\dfrac{−6 + 2 CE}{\sqrt{25−6CE+CE^2}}+\dfrac{CE}{\sqrt{16+CE^2 }}=0$

Any help would be greatly appreciated.

EDIT: I found out that the equation shortens to a quartic equation, and after solving that equation i end up getting the result: $CE = 2.0595$

  • $\begingroup$ For the equation that you are lost on, you can try moving one term to the other side of the equation and square both sides. $\endgroup$ – zyl1024 Mar 22 '15 at 20:52
  • $\begingroup$ Right, I did that and I've shortened it to the following: $CE^4 − 6CE^3 + 25CE^2 − 128CE = −192$ And Now I am clueless as to what I am supposed to do. $\endgroup$ – Fainted Mar 22 '15 at 21:44
  • $\begingroup$ That's a quartic equation. It's not fun having to solve that, but it's your only choice unless you're interested in computing the solution numerically. $\endgroup$ – Michael Grant Mar 22 '15 at 22:07

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