Square roots equations I had to solve this problem:
$$\sqrt{x} + \sqrt{x-36} = 2$$
So I rearranged the equation this way:
$$\sqrt{x-36} = 2 - \sqrt{x}$$
Then I squared both sides to get:
$$x-36 = 4 - 4\sqrt{x} + x$$
Then I did my simple algebra:
$$4\sqrt{x} = 40$$ 
$$\sqrt{x} = 10$$
$$x = 100$$
The problem is that when I go back and plug my $x$-value into the equation, it doesn't work.
$$\sqrt{100} + \sqrt{100-36} = 2$$
$$10+8 = 2$$
Which is obviously wrong.

Method $\#1:$
As for real $a,\sqrt a\ge0\ \ \ \ (1)$
$(\sqrt x+\sqrt{36-x})^2=36+2\sqrt{x(36-x)}\ge36$
$\implies\sqrt x+\sqrt{36-x}\ge6\ \ \ \ (2)$ or $\sqrt x+\sqrt{36-x}\le-6\ \ \ \ (3)$
Finally $(1)$ nullifies $(3)$
Method $\#2:$
WLOG let $\sqrt x=6\csc2y$   where $0<2y\le\dfrac\pi2\implies\sqrt{x-36}=+6\cot2y$
$\sqrt x+\sqrt{36-x}=6\cdot\dfrac{1+\cos2y}{\sin2y}=6\cot y$
Now $0<2y\le\dfrac\pi2\implies0<y\le\dfrac\pi4\implies\cot0>\cot y\ge\dfrac\pi4=1$ as $\cot y$ is decreasing in $\left[0,\dfrac\pi2\right]$
