find the real roots of

2. ## Important idea

An important idea when solving questions with infinite nested radicals.

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

Then we can rewrite the equation as $x=\sqrt{ax}$
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)

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

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

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

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

$\sqrt{x+2\sqrt{x+2\sqrt{x+...2\sqrt{x+2\sqrt{3x}}} }}=x$
$\sqrt{x+2x}=x$
$x^2=3x$
$x^2-3x=0$
$x(x-3)=0$
The real roots are $0$ and $3$.