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.
 A: 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 $3\lt \sqrt{x}$.
A: Just to expand on @zeta's comment, because it is possible to square in order to eliminate radicals:
$$\left(\sqrt{x+4-4\sqrt{x}}\right)^2=\left(1-\sqrt{x+9-6\sqrt{x}}\right)^2$$gives, on simplifying:
$$2\sqrt{x}-6=-2\sqrt{x+9-6\sqrt x}$$
cancelling the factor $2$ and squaring gives $0=0$.
The problem is that squaring allows solutions for both signs of the square root (and therefore includes unwanted solutions, which have to be eliminated by studying cases). If the solution set consists of isolated points, this can be a way of proceeding.
It is also important to check the validity of the original formula for the solutions once they are found. Here the formula makes no sense over $\mathbb R$ for negative values of $x$, yet the squaring method does not exclude negative $x$.
[This is really a comment, but was too long for that]
