#3**+1 **

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.

\(x^2=12+x \\ \iff x^2-x-12=0 \\ \iff (x-4)(x+3)=0\\ \iff x=4 \\ \text{Note: We rejected x=-3 as we know x>0 (as the squareroot of any number is non-negative)}\)

Guest Jun 18, 2023

#4**0 **

\(\mbox{draw[blue, very thick] (0,0) rectangle (3,2)}\)Thanks, now I get it.

Well the next promblem is

\(\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}}\)

xud33 Jun 18, 2023