Hi, in this type of questions we always follow these steps:
Step 1: let the a variable be equal to this expression.
For instance,
\(x=\sqrt{12+\sqrt{12+...}}\)
Step 2: Square both sides:
\(x^2=12+\sqrt{12+...}\)
But, notice, x was defined to be these "squareroots" right?
So: \(x^2=12+x\)
Step 3: We solve this equation.
\(x^2=12+x \\ \iff x^2-x-12=0 \\ \iff (x-4)(x+3)=0\\ \iff x=4 \\ \text{Note: We rejected x=-3 as we know x>0 (as the squareroot of any number is non-negative)}\)
\(\mbox{draw[blue, very thick] (0,0) rectangle (3,2)}\)Thanks, now I get it.
Well the next promblem is
\(\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}}\)