<s>[INST] Is there a more direct way to solve in $\Bbb R$ the following equation? $\displaystyle \sqrt{x+4-4\sqrt {x}}+\sqrt{x+9-6\sqrt {x}}=1$. 
Solve in $\Bbb R$ the following equation: $\displaystyle\sqrt{x+4-4\sqrt {x}}+\sqrt{x+9-6\sqrt {x}}=1$.$\qquad\text{(1)}$

My try:
The set of Perfect squares is: $\left\{\color{blue}{0,1,4,9,16,25,36,49\ldots}\right\}$
Some guesswork: $$\begin{align}
\sqrt{\color{blue}0+4-4\sqrt {\color{blue}0}}=2 & & \sqrt{\color{blue}0+9-6\sqrt {\color{blue}0}}=3 && \text{$\boldsymbol\times$}  \\\,\\
\sqrt{\color{blue}1+4-4\sqrt {\color{blue}1}}=1 & & \sqrt{\color{blue}1+9-6\sqrt {\color{blue}1}}=2 && \text{$\boldsymbol\times$}\\\,\\
 \sqrt{\color{blue}4+4-4\sqrt {\color{blue}4}}=0 & & \sqrt{\color{blue}4+9-6\sqrt {\color{blue}4}}=1 &&\text{$\boldsymbol\checkmark$} 
\end{align}$$
So the solution is $x=4$. 
But is there a more direct way, i.e. to find $x$ without having to do guesswork?
Thanks in advance.
 [/INST] Hint:
$$x+4-4\sqrt {x}=(\sqrt{x}-2)^2$$
$$x+9-6\sqrt {x}=(\sqrt{x}-3)^2$$
And don't forget the absolute values. ;)
The simplest solution is to break the problem in $3$ cases:$\sqrt{x} \lt2$ or $2 \leqslant \sqrt{x} \leqslant 3$ or $3\lt \sqrt{x}$.
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