Finding $f(\sqrt 3 )$ 
Let $$x^2 + \left( f(x)-2 \right) x + 2\sqrt 3 - 3 - \sqrt 3 f(x) = 0$$ for all $x \in \mathbb R$ and let $f(x)$ be continuous for all $x \in \mathbb R$. Find $f(\sqrt 3)$.

When I substituted $x= \sqrt 3$, I got $0=0$. How else do I solve this problem? I don't think there are many more obvious constraints. 
The given answer is: 

 $2(1-\sqrt 3) $


If $x\ne \sqrt{3}$ then you can write $f$ explicitly: 
$$f(x)= {-x^2+2x+3-2\sqrt{3}\over x - \sqrt 3  }$$
So you can only calculate $$\lim_{x\to \sqrt{3}}f(x) =\lim_{x\to \sqrt{3}}  {-x^2+2x+3-\color{red}{2}\sqrt{3}\over x - \sqrt 3  }$$ $$= \lim_{x\to \sqrt{3}}  {(\sqrt{3}-x)(\sqrt{3}+x)+2(x-\sqrt{3})\over x - \sqrt 3  }=$$
$$= \lim_{x\to \sqrt{3}}  (-(\sqrt{3}+x)+2)=2-2\sqrt{3}$$
