I thought about this but I could not come up with a way to calculate this. Any comment?? note:$x$ is much larger than $\alpha$
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Write $\sqrt{x^2-a}$ as $\vert x \vert \sqrt{1-\dfrac{a}{x^2}}$. Now recall that $$(1-y)^{1/2} = 1-\dfrac{y}2-\dfrac{y^2}{8}-\dfrac{y^3}{16}-\dfrac{5 y^4}{128} + \mathcal{O}(y^5)$$
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Use the binomial theorem:
$$\begin{align*} \frac{\sqrt{x^2-\alpha}}x&=\left(1-\frac{\alpha}{x^2}\right)^{1/2}\\\\ &=\sum_{k\ge 0}\binom{1/2}k(-1)^k\left(\frac{\alpha}{x^2}\right)^k\\\\ &\approx1-\frac{\alpha}{2x^2}\;, \end{align*}$$
with more terms readily available if you need them. Here
$$\binom{1/2}k=\frac{(1/2)^{\underline k}}{k!}\;,$$
where $x^{\underline i}$ is a falling factorial.
Brian M. Scott
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where did you get $\frac{\sqrt{x^2-\alpha}}{x}??$ – eChung00 Oct 22 '13 at 00:06
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@eChung00: You want $x-\sqrt{x^2-\alpha}$. Obviously you can find this (by multiplying by $x$) if you can find $1-\frac{\sqrt{x^2-\alpha}}x$. – Brian M. Scott Oct 22 '13 at 00:10
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For positive $x$, multiply by $\frac{x+\sqrt{x^2-\alpha}}{x+\sqrt{x^2+\alpha}}$. We get $$\frac{\alpha}{x+\sqrt{x^2-\alpha}},$$ which can be computed to good accuracy without roundoff error issues.
André Nicolas
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In the real world, variables tend to be positive. If $x$ is negative, let $y=-x$. – André Nicolas Oct 22 '13 at 00:05