Square Root Scramble

$\begin{aligned} \sqrt{x+ \sqrt{x+ \sqrt{x+\cdots }}} & = 6 & \small \color{#3D99F6} \text{Squaring both sides} \\ x + \color{#3D99F6}\sqrt{x+ \sqrt{x+ \sqrt{x+\cdots }}} & = 36 & \small \blue{\text{Note that }\sqrt{x+ \sqrt{x+ \sqrt{x+\cdots }}} = 6} \\ x + \color{#3D99F6}6 & = 36 \\ \implies x & = \boxed{30}\end{aligned}$

Given : $\sqrt{x + \sqrt{x + \sqrt{x + \dotsb}}} = 6$

Now if we take one from infinity the result is still infinity .

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

$\implies x + 6 = 36$

$\implies x = 30$

Another way to solve it would be to take the equation and notice that the equation contains itself.So we get sqrtx+6=6,then x+6=36,and finally x=30.

Same as Cheongs.

Let $\[\sqrt{x + \sqrt{x + \sqrt{x + \dotsb}}}.$ \) *be called the "mess" which is equal to 6. If we square the equation the we get x + the"mess" = 36 because the square roots are infinite so x = 36 - "mess" = 36 - 6 = 30 *