<s>[INST] If $x(2+\sqrt3)=y(2-\sqrt3)$ then find the value of $\frac{1}{x+1}+\frac{1}{y+1}$ My attempt: 
Let $\frac{x}{2-\sqrt3}=\frac{y}{2+\sqrt3}=k$, 
so $x=(2-\sqrt3)k$ and $y=(2+\sqrt3)k$
Substituting these values we get: 
$\dfrac{1}{x+1}+\dfrac{1}{y+1}$
$=\dfrac{x+y+2}{xy+x+y+1}$
$=\dfrac{2(k+1)}{(k+1)^2}$
$=\dfrac{2}{k+1}$
How do I find the value of $k$?
Options for the answers are $1, \sqrt3, 2\sqrt3, 2$
If I perform backcalculation, that gives me options for k as $1, \dfrac{2-\sqrt3}{\sqrt3}$ and $\dfrac{2-2\sqrt3}{\sqrt3}$. 
If we can claim that k is an integer, then we can say $k=1$ and we are done. But can we say k is an integer, and not any rational? 
 [/INST] With a missing data, it could actually attain any value
From there you've stopped $\frac{1}{x+1}+\frac{1}{y+1} = \frac{4k+2}{k^2+4k+1}$
Now, suppose $\frac{4k+2}{k^2+4k+1} = t$ for some $t \in \mathbb{R}$, let show there is a solution for this specific t. From here $tk^2 + 4(t-1)k + (t-2) = 0$ we have $k_{1,2} = \frac{-4(t-1) \pm \sqrt{16(t-1)^2-4t(t-2)}}{2t}$.
The expression under square root is $12t^2-24t+16$ which is $12(t-1)^2 + 4$, i.e. always positive, hence there are even two values of $k$, which solves for any given $t$. (except for $t = 0$, then you get $k=-\frac{1}{2}$.
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