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What is the simplest method for solving $x(x+a)=a$ or $x=a/(x+a)$ for a? I think there's a trick for solving algebra questions like this. This problem comes up when deriving the sum of a geometric series. I believe a is the sum of such a series.

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Rewrite the equation $x(x+a)=a$ as $x^2+ax=a$ and then as $x^2=a-ax=a(1-x)$; provided that $x\ne 1$, you can divide through by $1-x$ to get

$$a=\frac{x^2}{1-x}=\frac{x^2-1+1}{1-x}=\frac{(x-1)(x+1)}{1-x}+\frac1{1-x}=-x-1+\frac1{1-x}\;.$$

If $x=-1$, the original equation is $a-1=a$, or $-1=0$, which is obviously impossible anyway.

For $|x|<1$ you can expand this into an infinite series:

$$a=-x-1+\sum_{n\ge 0}x^n=\sum_{n\ge 2}x^n\;.$$

Added: You can also get this directly from $a=\frac{x^2}{1-x}$:

$$a=\frac{x^2}{1-x}=x^2\sum_{n\ge 0}x^n=\sum_{n\ge 0}x^{n+2}=\sum_{n\ge 2}x^n\;.$$

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  • $\begingroup$ I'm sorry. My question was not clear. I want to solve for a, not x. $\endgroup$ – jim smity Oct 15 '13 at 23:41
  • $\begingroup$ @jim: Okay; I’ve revised my answer to fit the question. $\endgroup$ – Brian M. Scott Oct 15 '13 at 23:45
  • $\begingroup$ Thanks. There are a few algebra tricks in there that I am not familiar with. $\endgroup$ – jim smity Oct 16 '13 at 0:03
  • $\begingroup$ @jim: Which ones? $x^2=x^2-1+1$ is clear, I think, even if it’s not something that might have occurred to you; then I factored $x^2-1$ as the difference of squares to get $(x-1)(x+1)$, rewrote $\frac{a+b}c$ as $\frac{a}c+\frac{b}c$ (where $a=(x-1)(x+1)$, $b=$, and $c=1-x$), and observed that $\frac{x-1}{1-x}=\frac{-(1-x)}{1-x}=-1$. $\endgroup$ – Brian M. Scott Oct 16 '13 at 0:09
  • $\begingroup$ It didn't occur to me to add -1+1. I've seen that done before though. I just forget to do that sort of thing. I also felt that there is a method of multiplying both numerators or denominators to immediately get a. $\endgroup$ – jim smity Oct 16 '13 at 0:16
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In either case, the equation is a quadratic in $x$, so if $a$ is a constant, you can simply solve for $x$ using the quadratic formula:

$$ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}, $$ given that $ax^2 + bx + c = 0$, which is an easy form to put these into.

If $x$ is given, then both of these equations are just linear equations in $a$, which are easy to solve.

$x(x + a) = a$ becomes $x^2 + ax = a$, or $a(x - 1) = -x^2$, so $a = \frac{-x^2}{x - 1}$, assuming that $x \ne 1$. If $x = 1$, then this equation is $1(1 + a) = a$, or $a = 1 + a$, which has no solution.

A similar method works for the other form of equation.

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  • $\begingroup$ I'm sorry. My question was not clear. I want to solve for a, not x. $\endgroup$ – jim smity Oct 15 '13 at 23:41
  • $\begingroup$ As qaphla stated: "If $x$ is given, then both of these equations are just linear equations in $a$, which are easy to solve." Are you now able to solve for $a$ by seeing $x(x+a)=a \Longleftrightarrow x^2+ax=a$? $\endgroup$ – Patterns_43 Oct 16 '13 at 0:10

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