Infinity square root

Make $x = \sqrt{2+\sqrt{2+\sqrt{2+...}}}$ Square both sides. Then $x^{2} = 2 + \sqrt{2+\sqrt{2+...}} = 2+x$ This results in a quadratic equation $x^{2} - x -2 = 0$ There are two solutions only, $x=-1$ and $x=2$ But, as $x > 0$ the solution is $x=2$

Let

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

Now the given expression is equal to $x.$

Now,

$x=\sqrt{2+x}$ { Since the interior part is also equal to $x.$ }

Squaring on both sides,

$x^2=2+x.$

$x^2-x-2=0.$

$x^2-2x+x-2=0.$

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

$(x-2)(x+1).$

Hence the two roots are $x=2,-1$

Since sum of anything can never be negative , the only answer is $2.$

$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2...}}}} = x$ $\large\sqrt{2+\underbrace{\sqrt{2+\sqrt{2+\sqrt{2...}}}}_x} = x$ $\sqrt{2+x} = x$ $2+x = x^2$ $x^2-x-2=0$ $(x-2)(x+1)=0$ $x=2; x=-1$ $\textbf{The answer=}\boxed{\color{#3D99F6}{2}}$

√2+(√2+(√2+(√2...)))=x 2+(√2+(√2+(√2+...)))=x² 2+x=x² x²-x-2=0 (x+1)(x-2) x=(-1), x=2 Positive.. =2 smile emoticon

√2+√2+√2+...........=x √2+x=x 2+x=x x 0=x x-x-2 then x=2