What is n?

We are given that $\underbrace{ \sqrt{ x+2\sqrt{x+2\sqrt{x+\ldots +2\sqrt{x+2\sqrt{3x}}}}}}_{\large \text{ n radical signs }}=x$

Rewrite the given equation as $\underbrace{ \sqrt{ x+2\sqrt{x+2\sqrt{x+\ldots +2\sqrt{x+2\sqrt{x+2x}}}}}}_{\large \text{ n radical signs }}=x \qquad (1)$

On replacing the last letter $x$ on the LHS of the equation (1) by the value of $x$ expressed by equation (1), we get $\underbrace{ \sqrt{ x+2\sqrt{x+2\sqrt{x+\ldots +2\sqrt{x+2\sqrt{x+2x}}}}}}_{\large \text{ 2n radical signs }}=x$

Further lets replace the last letter $x$ by the same expression; again and again yields $\underbrace{ \sqrt{ x+2\sqrt{x+2\sqrt{x+\ldots +2\sqrt{x+2\sqrt{x+2x}}}}}}_{\large \text{ 3n radical signs }}=x$

$\underbrace{ \sqrt{ x+2\sqrt{x+2\sqrt{x+\ldots +2\sqrt{x+2\sqrt{x+2x}}}}}}_{\large \text{ 4n radical signs }}=x \\ . \\ . \\ . \\ .$

Hence we can write $x=\sqrt{x+2\sqrt{x+2\sqrt{}x+\ldots}}$

Now it follows that, $x=\sqrt{x+2 \left( \sqrt{x+2\sqrt{x+2\sqrt{x+\ldots}}}\right) }=\sqrt{x+2x}$

Now squaring both sides, we get $x^2=x+2x \Rightarrow x^2-3x=0 \Rightarrow x=0,3$ .

Now by putting the values $x=0,3$ in the original equation, we get that both the values satisfy the original equation. Hence verified the obtained solutions!.

Therefore the average of all the real roots is $\dfrac{0+3}{2}=1.5$ . $\square$

enjoy!

I think there should not be a "3" in the last radical. then the roots will come as 0,3 which average to 1.5