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$$\sqrt{x+a} - \sqrt{x-a} = 2\sqrt{a}$$

Squaring both sides of the equation doesn't get rid of the root.

How do I isolate $x$ from $a$?

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A general comment on such equations - remember that squaring an equation can add spurious solutions (if you were to square root the squared equation, think about the possible solutions which belong to the negative square root). This means you should always check back any solutions you find to make sure they are solutions to the original equation. –  Mark Bennet May 27 '12 at 19:48
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3 Answers

up vote 5 down vote accepted

We describe another method for isolating $x$. The case $a=0$ is uninteresting. Suppose that $a\ne 0$.

Multiply both sides by $\sqrt{x+a}+\sqrt{x-a}$. On the left, nice stuff happens, we get $2a$. On the right we get $2\sqrt{a}\left(\sqrt{x+a}+\sqrt{x-a}\right)$. Do a little cancellation. We get $$\sqrt{x+a}+\sqrt{x-a}=\sqrt{a}.$$

Adding, we find that $$2\sqrt{x+a}=3\sqrt{a},$$ and now everything is easy.

Remark: We should not necessarily be in a hurry to square, since squaring can create a mess. We were given an equation that had a nice structure. Nice structure should be preserved, and exploited. And remember that whenever $\sqrt{X}-\sqrt{Y}$ ends up in a problem, its partner $\sqrt{X}+\sqrt{Y}$ is ready to help.

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What did you do to get from $\sqrt{x+a} + \sqrt{x-a} = \sqrt{a}$ to $2\sqrt{x+a} = 3\sqrt{a}$? –  Miroslav Cetojevic May 27 '12 at 21:54
    
We also have original equation $\sqrt{x+a}-\sqrt{x-a}=2\sqrt{a}$. Add this to $\sqrt{x+a}+\sqrt{x-a}=\sqrt{a}$. The $\sqrt{x-a}$ parts cancel! –  André Nicolas May 27 '12 at 21:58
    
So THAT's how it works... –  Miroslav Cetojevic May 27 '12 at 22:18
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Square twice. The first time you obtain $2 x - 2 \sqrt{x^2-a^2} = 4 a$ so $\sqrt{x^2-a^2} = x - 2a$. The second time...

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As said, square both sides: $$2x-2\sqrt{x+a}\sqrt{x-a}=4a\Longrightarrow x-2a=\sqrt{x+a}\sqrt{x-a}\Longrightarrow$$$$\Longrightarrow x^2-4ax+4a^2=x^2-a^2\Longrightarrow 4ax=5a^2$$and now you can solve according to the different possibilities for $\,a$

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