Finding $f(\sqrt 3 )$

Question

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) $

Answer 1

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 $$\lim_{x\to \sqrt{3}}f(x) =\lim_{x\to \sqrt{3}} {-x^2+2x+3-\color{red}{2}\sqrt{3}\over x - \sqrt 3 }$$ $$= \lim_{x\to \sqrt{3}} {(\sqrt{3}-x)(\sqrt{3}+x)+2(x-\sqrt{3})\over x - \sqrt 3 }=$$ $$= \lim_{x\to \sqrt{3}} (-(\sqrt{3}+x)+2)=2-2\sqrt{3}$$

Answer 2

$$x^2-2x+2\sqrt3-3=f(x)(\sqrt3-x)$$

$$(x-\sqrt3)(x+\sqrt3-2)=f(x)(\sqrt3-x)$$

when $x \ne \sqrt3$, we have

$$f(x)=-(x+\sqrt3-2)$$

Now, by continuity, we have

$$f(\sqrt3)=-(\sqrt3+\sqrt3-2)=-(2\sqrt3-2)$$