<s>[INST] positive root of the equation $x^2+x-3-\sqrt{3}=0$ $x^2+x-3-\sqrt{3}=0$
using the quadratic formula we get 
$$x=\frac{-1+\sqrt{13+4\sqrt{3}}}{2}$$ for the positive root
but the actual answer is simply   $x=\sqrt3$
I am unable to perform the simplification any help would we helpful
 [/INST] Let $a=\sqrt{3}$.
\begin{align*}
\text{Then}\;\;&x^2+x-3-\sqrt{3}=0\\[4pt]
\iff\;&x^2+x=3+\sqrt{3}\\[4pt]
\iff\;&x^2+x=a^2+a\\[4pt]
\end{align*}
So by inspection, we get the solution $x=a$.

By Vieta's formula, the sum of the roots is $-1$, so the other root is $-1-a$, which is negative.

Thus, $x=\sqrt{3}$ is the only positive root.

Alternatively,
\begin{align*}
&x^2+x=a^2+a\\[4pt]
\iff\;&(x^2-a^2)+(x-a)=0\\[4pt]
\iff\;&(x-a)(x+a+1)=0\\[4pt]
\iff\;&x=a\;\,\text{or}\;\,x=-1-a\\[4pt]
\iff\;&x=\sqrt{3}\;\,\text{or}\;\,x=-1-\sqrt{3}\\[4pt]
\end{align*}
so as before, $x=\sqrt{3}$ is the only positive root.
</s>