+0  
 
0
23
10
avatar+106 

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

 Jun 18, 2023
 #1
avatar+106 
0

I have no Idea of what to do.

 

 

 

I need help fast!!!

 Jun 18, 2023
edited by xud33  Jun 18, 2023
 #2
avatar
-1

The answer works out to 3.

 Jun 18, 2023
 #9
avatar+570 
0

 

3 is immediately seen obviously incorrect because  

the sqrt(12) by itself is already more than 3, before  

you start adding all those other things to it.    

Bosco  Jun 19, 2023
 #3
avatar
+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.

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

 Jun 18, 2023
 #4
avatar+106 
0

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

 

 

Well the next promblem is

 

 

 

 

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

 Jun 18, 2023
edited by xud33  Jun 18, 2023
 #5
avatar
0

Using the same method, that works out to 2.

Guest Jun 18, 2023
 #7
avatar
0

When the problem is "Minus Sqrt(12)", the answer is 3 and not 2.

Guest Jun 18, 2023
 #6
avatar+106 
0

It's the opposite if this promblem

 

 

The promblem is \(\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}}\)

 Jun 18, 2023
 #8
avatar
0

Well, the idea is the same:

\(x=\sqrt{12-\sqrt{12-\sqrt{12-...}}}\\ \iff x^2=12-x \\ \iff x^2+x-12=0 \\ \iff (x+4)(x-3)=0 \\ \iff x=3\)

Guest Jun 19, 2023
 #10
avatar+106 
0

Thanks, I think I might be able to solve the next promblem.

 Jun 19, 2023

0 Online Users