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Let $$t = \sqrt {x^2 + a^2} - x$$

How do I isolate $x$ from this equation? I tried squaring the equation but it became pretty much complicated.

Thanks.

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    $\begingroup$ Don't square it right away; move $x$ to the LHS first. $\endgroup$ – inkievoyd Nov 11 '14 at 19:10
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Using hyperbolic trigonometry, let $x=a\sinh u$.

Then $t=a\cosh u-a\sinh u=ae^{-u}$, so that $u=-\log\frac ta$.

$$x=-a\sinh\log\frac ta=-\frac a2\left(\frac ta-\frac at\right).$$

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Hint: First, write $t+x=\sqrt{x^2+a^2}$, Then, and then only, square both sides.

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$(t+x)^2=x^2+a^2$, so $t^2+2tx=a^2\implies x=\frac{a^2-t^2}{2t}$.

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we have $(t+x)^2=x^2+a^2$ thus we get $t^2+x^2+2tx=x^2+a^2$ this will help you!

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$$ t+x=\sqrt{x^2+a^2}\\ t^2+x^2+2tx=x^2+a^2 \\ 2tx=a^2-t^2 \\ x=\frac{(a-t)(a+t)}{2t}.$$

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I hope this helps:

$$ \begin{align} t&=\sqrt{a^2 + x^2}-x\\\\ t + x&=\sqrt{a^2+x^2}\\\\ (t+x)^2&=a^2+x^2\\\\ t^2 +2tx+x^2&=a^2+x^2\\\\ t^2+2tx&=a^2\\\\ 2tx&=a^2-t^2\\\\ x&=\frac{a^2-t^2}{2t} \end{align} $$

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