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?
Why $e=g \pm \sqrt{g}$ when $ \left( g-e \right)^2=g$?
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algebra-precalculus
problem-solving
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7Don't expand it, square root. – 2012-10-14
2 Answers
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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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0@Jwodder: corrected – 2012-10-15
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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.