+0  
 
+10
104
10
avatar+201 

12+12+12+12+

 Jun 18, 2023
 #1
avatar+201 
+5

I have no Idea of what to do.

 

 

 

I need help fast!!!

 Jun 18, 2023
edited by xud33  Jun 18, 2023
 #2
avatar
-6

The answer works out to 3.

 Jun 18, 2023
 #9
avatar+1376 
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
+2

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=12+12+...

Step 2: Square both sides:

x2=12+12+...

But, notice, x was defined to be these "squareroots" right?

So: x2=12+x

Step 3: We solve this equation.

x2=12+xx2x12=0(x4)(x+3)=0x=4Note: We rejected x=-3 as we know x>0 (as the squareroot of any number is non-negative)

 Jun 18, 2023
 #4
avatar+201 
+5

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

 

 

Well the next promblem is

 

 

 

 

  121212

 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+201 
+5

It's the opposite if this promblem

 

 

The promblem is 121212

 Jun 18, 2023
 #8
avatar
+1

Well, the idea is the same:

x=121212...x2=12xx2+x12=0(x+4)(x3)=0x=3

Guest Jun 19, 2023
 #10
avatar+201 
+4

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

 Jun 19, 2023

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