LetÂ $x=\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{...}}}}}$

$\begin{align}x^2&=\left(\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{...}}}}}\right)^2\\

&=3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{...}}}}\\

&=3+2x\end{align}$

So $x^2-2x-3=0\Rightarrow (x-3)(x+1)=0$, so $x=3$ or $x=-1$.

$x$ positive since so $x=3$.

Check: $\sqrt{3+2\times(3)}=3$

Let $x=\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{...}}}}}$

We can approximate $x$ by imagining that we are going to find that large square root as a calculation.

We can start by finding $\sqrt{3}$, and then double it, add $3$, and find the square root of that, and so on.

So if we let $a=\sqrt{3}$, and then $b=\sqrt{3+2\times a}=\sqrt{3+2\sqrt{3}}$, then $c=\sqrt{3+2\times b}=\sqrt{3+2\times\sqrt{3+2\times\sqrt{3}}}$, and $d=\sqrt{3+2\times c}=\sqrt{3+2\times\sqrt{3+2\times\sqrt{3+2\times\sqrt{3}}}}$, then the sequence $a, b, c, d, ...$ will get closer and closer to $x$.

Using a calculator, or a spreadsheet, $a=1.73205$, $b=2.54246$, $c=2.84340$, $d=2.94734$, $e=2.98239$, $f=2.99413$, ...

It looks like they are getting closer and closer to $3$, so it looks like $x=3$. If we got to $3$ in our sequence, then the next number in the sequence would be $\sqrt{3+2\times 3}$, which is $\sqrt{3+6}=\sqrt{9}=3$. So if the sequence ever gets to $3$, it will stay at $3$ forever, which strongly suggests that $x=3$.

You can find more short problems, arranged by curriculum topic, in our short problems collection.