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.

Surprised this hasn’t been mentioned yet. This method is helpful in general.
Let $u=\sqrt{x}$ and $v=\sqrt{x-36}$. Then $u^2-v^2=36$ and $u+v=2$. With $u^2-v^2=(u+v)(u-v)$, $u-v=18$. Thus $u=10$ and $x=100$.
