Question

Find the value of $\sqrt{6+\sqrt{6+\sqrt{6+......to\infty }}}$

Answer 1

$\sqrt{6+\sqrt{6+\sqrt{6+.....+\infty }}}$

Let, $x=\sqrt{6+\sqrt{6+\sqrt{6+.....+\infty }}}$

$\therefore x=\sqrt{6+x}$

$⟹x^2=6+x$

$⟹x^2-x-6=0$

$⟹x^2-3x+2x-6=0$

$⟹x(x-3)+2(x-3)=0$

$⟹(x+2)(x-3)=0$

So, $x=-2$, $x=3$

$\Rightarrow x=3$