For all positive real numbers, is $f(x)=\sqrt{x}+x+2$ one to one? I understand that in order to prove this to be one to one, I need to prove $2$ numbers, $a$ and $b$, in the same set are equal. 
This is what I did:
$$\sqrt{a} + a + 2 = \sqrt{b} + b + 2$$
$$\sqrt{a} + a = \sqrt{b} + b$$
$$a + a^2 = b + b^2$$
How would I arrive at $a = b$? Is it possible?

$f(x)=\sqrt{x}+x+2$ is strictly increasing on $(0,\infty)$, so it's one-one (suppose not, use the strict monotonicity to draw a contradiction).
Or, from your second step,
$$\sqrt{a} + a = \sqrt{b} + b\iff \sqrt{a} - \sqrt{b}+ a -b=0 \iff(\sqrt{a} - \sqrt{b})(1+\sqrt{a}+\sqrt{b})=0 $$
Since $1+\sqrt{a}+\sqrt{b}>0$, we have $\sqrt{a} - \sqrt{b}=0$.
