Disintegration of a of delta function A delta function has been written as 
$$\delta(x^2-a^2)$$
Do you think I can write the function as $$\delta (x+a)+ \delta (x-a)$$
?
My plan is to locate the delta function potential coordinates in x axis, and that's why I wanted to break down the quadratic term.
 A: Almost. Assuming $a\ne0$, substitute $u=x^2$ to get
$$ \int_{\mathbb{R}} \delta(x^2-a^2)f(x)\,\mathrm{d}x=\int_0^{\infty} \delta(x^2-a^2)\big[f(x)+f(-x)\big]\,\mathrm{d}x $$
$$ =\int_0^{\infty} \delta(u-a^2)\big[f(\sqrt{u})+f(-\sqrt{u})\big]\frac{\mathrm{d}u}{2\sqrt{u}}=\frac{f\big(\sqrt{a^2}\big)+f\big(-\sqrt{a^2}\big)}{2\sqrt{a^2}} $$
$$ =\frac{f(a)+f(-a)}{2|a|}=\int_{\mathbb{R}} \frac{\delta(x-a)+\delta(x+a)}{2|a|}f(x)\,\mathrm{d}x. $$
Therefore, we ought to write
$$ \delta(x^2-a^2)=\frac{\delta(x-a)+\delta(x+a)}{2|a|}. $$
We can use this trick for $\delta(g(x))$ more generally:
$$ \delta(g(x))=\sum_{g(r)=0}\frac{\delta(x-r)}{|g'(r)|}. $$
A: @runway44 beat me by 6 seconds on the conventional method, so here's a different heuristic.
In the sense of distributions, we know that for $a>0$ we have
$$\frac{2a}{x^2-a^2+\mathrm{i}0^+}=-2a\mathrm{i}\pi\delta(x^2-a^2)+\mathrm{P}\frac{2a}{x^2-a^2}\text{.}$$
But
$$\frac{2a}{x^2-a^2+\mathrm{i}0^+}=\frac{1}{x-a+\mathrm{i}0^+}- \frac{1}{x+a-\mathrm{i}0^+}$$
so
$$\frac{2a}{x^2-a^2+\mathrm{i}0^+}=-\mathrm{i}\pi\left(\delta(x-a)+\delta(-x-a)\right)+\mathrm{P}\left(\frac{1}{x-a}-\frac{1}{x+a}\right)$$
Whence
$$2a\delta(x^2-a^2)=\delta(x-a)+\delta(x+a)\text{.}$$
