Proving that a function is bijective I have trouble figuring out this problem:
Prove that the function $f: [0,\infty)\rightarrow[0,\infty)$ defined by $f(x)=\frac{x^2}{2x+1}$ is a bijection. 
Work: First, I tried to show that $f$ is injective. $\frac{a^2}{2a+1}=\frac{b^2}{2b+1}$
I got $a^2(2b+1)=b^2(2a+1)$. However, I get stuck here and cannot simplify the equation to get $a=b$, which would prove that the function is injective. 

Incresing: $f'(x)=\frac{2x(2x+1)-2x^2}{(2x+1)^2}=\frac{2x^2+2x}{(2x+1)^2}\gt 0$, if $x\ne0$ thus f is strictly increasing. and thus is injective, now we need to show surjectivity.
Pick $y\in[0,\infty)$, if $f(x)=y\Rightarrow \frac{x^2}{2x+1}=y\Rightarrow x^2-2xy-y=0$
Thus by the quadratic formula $x=\frac{2y\pm\sqrt{4y^2+4y}}{2}\in [0,\infty)$, thus $f$ is also surjective.
(note that we only take the "+" root)
