Let ${x^2=p+t^2}$ so after transforming the given condition and some manipulations we arrive at ${|x|=|p-4|/\sqrt(8(2-p))}$ with condition that p<2

Jaswinder wrote: Let ${x^2=p+t^2}$ so after transforming the given condition and some manipulations we arrive at ${|x|=|p-4|/\sqrt(8(2-p))}$ with condition that p<2 Little correction : $x=\frac{|p-4|}{\sqrt{8(2-p)}}$ and not $|x|=\frac{|p-4|}{\sqrt{8(2-p)}}$ : from the original equation, we get $x>0$

In fact the solution is valid iff $0\le p\le \frac{4}{3}.$ This is problem 1 from the fifth IMO (We have $x^2\ge p$ and $2\sqrt{x^2-1}\le x$ which leads to $x^2\le \frac43$. It is no difficult to check that for these values of $p$ the solution is valid.)

You can see the 5 Solutions here^^ Post # 8, 9, 11 in http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=500396

Whew!

Because of the radicals, we will naturally have several restrictions on $x$ and $p$. The first few are $x^2\geq p$ $(1)$ since we must have $\sqrt{x^2-p}\in\mathbb{R}$, $x^2\geq1$ $(2)$ since we must have $\sqrt{x^2-1}\in\mathbb{R}$, and $x\geq0$ $(3)$ since we must have $\sqrt{x^2-p}+2\sqrt{x^2-1}\geq0$. Now, squaring the given equation, $$(x^2-p)+4(x^2-1)+4\sqrt{(x^2-p)(x^2-1)}=5x^2-p-4+4\sqrt{x^4-(p+1)x^2+p}=x^2$$or $$4\sqrt{x^4-(p+1)x^2+p}=p+4-4x^2\iff\sqrt{x^4-(p+1)x^2+p}=\frac{p}{4}+1-x^2.$$Then, we must have $\tfrac{p}{4}+1-x^2\geq0$ $(4)$, and assuming this, we can square to obtain $$x^4-(p+1)x^2+p=x^4+\left(\frac{p}{4}+1\right)^2-\left(\frac{p}{2}+2\right)x^2$$or $$\left(1-\frac{p}{2}\right)x^2=\left(\frac{p}{4}-1\right)^2\iff x^2=\frac{\left(\frac{p}{4}-1\right)^2}{1-\frac{p}{2}},$$where $p\neq2$; taking the square root, $x=|\tfrac{p}{4}-1|/\sqrt{1-\tfrac{p}{2}}=\tfrac{|p-4|}{2\sqrt{4-2p}}$, where we take the positive root because of $(3)$. Since $p<2$ in order to have $\sqrt{4-2p}\in\mathbb{R}/\{0\}$, we have $\underline{x=\frac{4-p}{2\sqrt{4-2p}}}$. Finally, we check criteria $(1)$, $(2)$, and $(4)$. For $(4)$, $$\frac{p}{4}+1-\left(\frac{4-p}{2\sqrt{4-2p}}\right)^2=\frac{p}{4}+1-\frac{(p-4)^2}{16-8p}=\frac{4p-2p^2}{16-8p}+\frac{16-8p}{16-8p}-\frac{p^2-8p+16}{16-8p}=\frac{-3p^2+4p}{16-8p}\geq0$$or $\tfrac{p(3p-4)}{2-p}\leq0\iff\underline{p\in[0,\tfrac{4}{3}]}$; for $(2)$, $$\left(\frac{4-p}{2\sqrt{4-2p}}\right)^2=\frac{p^2-8p+16}{16-8p}\geq1\iff p^2-8p+16\geq16-8p\geq p^2\geq0,$$always true because $p\in\mathbb{R}$; and for $(1)$, $$\left(\frac{4-p}{2\sqrt{4-2p}}\right)^2=\frac{p^2-8p+16}{16-8p}\geq p\iff p^2-8p+16\geq16p-8p^2\iff9p^2-24p+16=(3p-4)^2\geq0,$$again always true. Henceforth, taking the restriction from $(4)$, we have a requested answer of $$\boxed{x=\frac{4-p}{2\sqrt{4-2p}}\text{ if }p\in[0,\tfrac{4}{3}],\text{ no solutions otherwise}}.$$$\blacksquare$