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# $\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}}$.

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Nov 30, 2020

#1
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Could you retype your question so it is clearer? What is it that you are looking for?

Nov 30, 2020
#2
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infnate amouts of 12's square roots.

Nov 30, 2020
#3
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Okay. Thank you for the clarification.

So to solve the equation $$\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}}$$ we will use the following equation:

n(n-1)=12; with the answer to the infinite nested radical function (x) being equal to n-1.

So,

n(n-1)=12

n = 4

n-1=?

4-1=3

Thus, the solution is 3.

Also, just a tip, don't place your questions in the header. It makes it a bit difficult to know the question that you are asking. You are supposed to place your question where you said "help please." Just switch the places, and you'll be perfectly fine!

Spring  Nov 30, 2020
edited by Spring  Nov 30, 2020
#4
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n(n - 1) = 12, what is this going to give you? It gives you:

n = - 3  and   n = 4

Guest Nov 30, 2020
#5
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Here is a computer code that uses "successive approximations" to arrive at the answer:

c=0; a=listfor(n, 1, 15, listforeach(b,reverse (d=12), c=2#(b - c);printa

Answer =(3.464101615, 2.921625983, 3.013034022, 2.997826876, 3.000362165, 2.999939638, 3.00001006, 2.999998323, 3.000000279, 2.999999953, 3.000000008, 2.999999999, 3, 3, 3) == 3

Nov 30, 2020
#6
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I do not understand the algebraic answers above. I mean i do not see the logic that Spring has used

$$\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}} \qquad=\;\;?\\ \text{First, since the square root is in the question, the answer must be positive.}\\ let\\ y=\sqrt{12-\!\sqrt{12-\!\sqrt{12-\cdots}}} \\ then\\ y=\sqrt{12-y}\\ y^2=12-y\\ y^2+y-12=0\\ (y-3)(y+4)=0\\ y=3\;\;or \;\;-4\\ \text{y cannot be negative}\\ y=3$$

Nov 30, 2020