Solving for $x$ in a (modified) quadratic

Question

I've come across this question which can be simplified (with a generous amount of manipulation) to a quadratic. Here it is: $$\left(\sqrt{49+20\sqrt6}\right)^{\sqrt {a \sqrt{a\sqrt{a\cdots \infty}}}}+(5-2\sqrt{6})^{x^2+x-3-\sqrt{x\sqrt{x\sqrt{x}\cdots \infty}}}=10 $$ $ \text{ where } a=x^2-3,(a\neq 0), \text{ Solve for }$ $x$.

Here's what I tried:
I first tried to simplify the powers in the given expression
$$y={\sqrt {a \sqrt{a\sqrt{a\cdots \infty}}}} \\ \Rightarrow y=\sqrt{ay} \\ \Rightarrow y=a$$
Similarly, the second power reduces to $x^2-3$ (equal to $a$). I've also condensed the first term in the expression. So our equation simplifies to: $$(5+2\sqrt{6})^{x^2-3} +(5-2\sqrt{6})^{x^2-3}=10$$ Since $5+2\sqrt{6} =\frac{1}{5-2\sqrt{6}}$, $(5+2\sqrt{6})^{x^2-3}$ can be taken as $t$ to furthur simplify the equation into: $$t^2-10t+1=0$$ Solving for $t$, we get: $$ t=5 \pm2\sqrt{6}$$ plugging back the value of $t$, we get: $$(5+2\sqrt{6})^{x^2-3}=(5+2\sqrt{6}) \text{ or } (5+2\sqrt{6})^{x^2-3}=(5+2\sqrt{6})^{-1}$$ $\Rightarrow$ $x^2-3=\pm 1$ . Solving this gives the values of $x$ to be: $\pm 2 , \pm \sqrt{2}$. $x$ cannot take the negative values as it will not satisfy the domain of the original question.
Hence the possible values of $x$ can be either $2$ or $\sqrt{2}$. But the answer tells me $x$ can only be $2$.
Where have I gone wrong or what have I overlooked?

Answer 1

There is nothing wrong but if $x=\pm\sqrt 2$ then $a=-1<0$ so $\sqrt{a\sqrt{a\dots}}$ is not well-defined. Hence, $x=\pm\sqrt 2$ must be "disqualified" as a solution.

Usually it is always good to check all solutions you got with the original equation since we usually only prove the direction that if $x$ satisfies then equation then $x$ is equal to this and this, but we actually want to prove the other direction as well.