1
$\begingroup$

I have a problem that involves finding $e$ such that $\left( g-e \right)^2=g$. Maxima tells me that $e=g \pm \sqrt{g}$, but I can't work on that equation to get this result. Actually, I can't go past $2ge - e^2 = g^2 - g$. Can someone show me the algebraic steps to get that result?

$\endgroup$
2
  • $\begingroup$ Hint: note that $(g -e)^2 = (-(g-e))^2.$ $\endgroup$
    – Ian Mateus
    Oct 14, 2012 at 22:30
  • 7
    $\begingroup$ Don't expand it, square root. $\endgroup$
    – wj32
    Oct 14, 2012 at 22:31

2 Answers 2

6
$\begingroup$

We have $(g-e)^2 = g$. Taking square roots, $(g-e) = \pm \sqrt{g}$. Rearranging terms gives $ e = g \pm \sqrt{g}$.

Don't expand the left side; this makes it more difficult.

$\endgroup$
3
  • $\begingroup$ Oh, I let that slip. Too much math does that to your brain. Thanks everyone. $\endgroup$
    – GuiRitter
    Oct 15, 2012 at 0:56
  • $\begingroup$ Not expanding the right side is the easy part; it's the left side that you need to guard against expanding. $\endgroup$
    – jwodder
    Oct 15, 2012 at 1:48
  • $\begingroup$ @Jwodder: corrected $\endgroup$
    – Ganesh
    Oct 15, 2012 at 4:57
4
$\begingroup$

If $(g-e)^2=g$, then either $g-e=\sqrt g$, or $g-e=-\sqrt g$. Now solve these equations for $e$ to get the two possibilities.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.