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# ​ Evaluate!!!

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$$\sqrt{12 +\!\sqrt{12 + \!\sqrt{12 + \!\sqrt{12 + \cdots}}}}$$

Jun 18, 2023

#1
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I have no Idea of what to do.

I need help fast!!!

Jun 18, 2023
edited by xud33  Jun 18, 2023
#2
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The answer works out to 3.

Jun 18, 2023
#9
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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
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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
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$$\mbox{draw[blue, very thick] (0,0) rectangle (3,2)}$$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
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Using the same method, that works out to 2.

Guest Jun 18, 2023
#7
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When the problem is "Minus Sqrt(12)", the answer is 3 and not 2.

Guest Jun 18, 2023
#6
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It's the opposite if this promblem

The promblem is $$\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}}$$

Jun 18, 2023
#8
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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
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Thanks, I think I might be able to solve the next promblem.

Jun 19, 2023