$$\sqrt{x+a} - \sqrt{x-a} = 2\sqrt{a}$$
Squaring both sides of the equation doesn't get rid of the root.
How do I isolate $x$ from $a$?
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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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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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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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