Two Two Two Two, Repeat

Let $\sqrt{ 2 + \sqrt{2 + \sqrt{ 2 + \cdots}}} =x$

$\Rightarrow \sqrt{2 + x} = x$

Squaring both sides ,

$2 + x = x^2$

$x^2 - x - 2 =0$

$(x + 1)(x - 2) = 0$

$\Rightarrow x + 1 = 0 \Rightarrow \boxed {x = -1}$

$\Rightarrow x - 2 = 0 \Rightarrow \boxed {x = 2}$ , which is there in the options.

$\sqrt{2+\sqrt{2+\sqrt{2+...}}} = k \\\\ (\sqrt{2+\sqrt{2+\sqrt{2+...}}})^2 = k^2 \\\\ 2 + \sqrt{2+\sqrt{2+\sqrt{2+...}}} = k^2 \\\\ 2 + k = k^2 \\\\ k^2 - k - 2 = 0 \\\\ (k - 2)(k + 1) = 0 \\\\ \begin{cases}\boxed{\boxed{k = 2}} \\ k = - 1 \end{cases}$

Let the given expression = x

Squaring, we get

x^2 = 2 + x

Then x = 2

or x = -1 .... (refused)

Let n be 2 So the expression be rewritten as (n+n)^1/2 4^1/2 = 2

_/2 = 1.41 rest of the 2 's will be getting smaller and smaller. ..so it is equal to 2 ..

that's a good mind washer......