find the real roots of

2. ## Important idea

An important idea when solving questions with infinite nested radicals.

If $\displaystyle x=\sqrt{a\sqrt{a\sqrt{a...}}}$

Then we can rewrite the equation as $\displaystyle x=\sqrt{ax}$
As it is as infinite succession, we are allowed to 'cut' the expression anywhere convenient and let it be equal to the L.H.S.

Example (from Pi in the Sky: Issue 11)

$\displaystyle x=\sqrt{3\sqrt{2\sqrt{3\sqrt{2...}}}}$. Find $\displaystyle x^3$

$\displaystyle x=\sqrt{3\sqrt{2x}}$
$\displaystyle x^2=3\sqrt{2x}$
$\displaystyle x^4=18x$
$\displaystyle x^3=18$

With this technique, you could solve your question now, I'll give another example then move on to your question.

$\displaystyle x=\sqrt{2\sqrt{2\sqrt{2...}}}$
$\displaystyle x^2=2x$
$\displaystyle x=2$
In general, if $\displaystyle x=\sqrt{a\sqrt{a\sqrt{a...}}}$
Then $\displaystyle x^2=ax$
And $\displaystyle x=a$

Now to your question
$\displaystyle \sqrt{x+2\sqrt{x+2\sqrt{x+...2\sqrt{x+2\sqrt{3x}}} }}=x$
Then using our technique, we convert to
$\displaystyle \sqrt{x+2x}=x$
$\displaystyle x^2=3x$
$\displaystyle x^2-3x=0$
$\displaystyle x(x-3)=0$

The real roots are $\displaystyle 0$ and $\displaystyle 3$.

Don't hesitate to ask anymore questions.
God bless your mathematical endeavours.