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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?

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    Don't expand it, square root.2012-10-14

2 Answers 2

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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.

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    @Jwodder: corrected2012-10-15
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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.