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# Square roots- I don't know what to do.

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If $$\sqrt{x+\!\sqrt{x+\!\sqrt{x+\!\sqrt{x+\cdots}}}}=9$$, find x.

I know that if this equation were switched around, where the x is 9 and 9 is x instead, we'd solve it like this:

$$\sqrt{9+\!\sqrt{9+\!\sqrt{9+\!\sqrt{9+\cdots}}}}=x\\ x^2=9+\sqrt{9+\!\sqrt{9+\!\sqrt{9+\cdots}}}\\ x^2=9+x\\ x^2-x-9=0\\ x=\frac{1+\sqrt{37}}{2}$$

I think I should be applying the same logic, but it's kind of difficult because of the x and it's making me confused. Could anyone help?

Jul 2, 2020

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Call that whose expression on the left side of the equal sign "expression" so you have

[expression]  =  9

Square both sides  --  you'll get:

x + [expression]  =  81

Since  [expression]  =  9  --  by substituting  '9'  for  [expression]  --  you'll now have:

x + 9  =  81

Therefore,  x  =  72

Jul 2, 2020
#2
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ah ok, I get it now. Thanks.

Jul 2, 2020