Question

please solve this problem​

Answer 1

$\displaystyle\longrightarrow\sf {x=2+\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}\quad\quad\dots (1)}$

$\displaystyle\longrightarrow\sf {x-2=\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}}$

$\displaystyle\longrightarrow\sf {(x-2)^2=2+\sqrt{2+\sqrt{2+\dots}}}$

That is,

$\displaystyle\longrightarrow\sf {(x-2)^2=x}$

$\displaystyle\longrightarrow\sf {x^2-4x+4=x}$

$\displaystyle\longrightarrow\sf {x^2-5x+4=0}$

$\displaystyle\longrightarrow\sf {x^2-x-4x+4=0}$

$\displaystyle\longrightarrow\sf {x(x-1)-4(x-1)=0}$

$\displaystyle\longrightarrow\sf {(x-1)(x-4)=0}$

This implies,

$\displaystyle\longrightarrow\sf {x=1\quad or\quad x=4}$

But from (1) we see that,

$\displaystyle\longrightarrow\sf {x\ \textgreater\ 2}$

Therefore,

$\displaystyle\longrightarrow\sf {\underline {\underline{x=4}}}$