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

 Apr 27, 2020
 #1
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if you square it,

 x+ \(\sqrt{x+\sqrt{x+...}}\)=81

 

so x=72

 Apr 27, 2020
 #2
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wow, just wow, a very thoughtful explanation!

Guest Apr 27, 2020
 #3
avatar+633 
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Since the square root goes on for infinity:

\(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}}=\sqrt{x+\sqrt{x+\sqrt{x...}}}\)

 

Let us set \(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}}=y\)

 

This means: \(y=9\)

 

This also means: \(\sqrt{x+y}=9\)

Substitute \(y\)\(\sqrt{x+9}=9\)

Square both sides: \(x+9=81\)

 

\(\boxed{x=72}\)

.
 Apr 27, 2020

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