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What is the value of  for which \(\sqrt{x + \sqrt{x + \sqrt{x + \ldots}}} = 5?\)

 Jan 12, 2021

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 #1
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Let $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}=5.$

If you square $y$, you get $25=x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}=x+y$.

But since the value of y is 5, then you get $25=x+5$.

Therefore, $x=\boxed{20}$.

 Jan 12, 2021
 #1
avatar
+2
Best Answer

Let $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}=5.$

If you square $y$, you get $25=x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}=x+y$.

But since the value of y is 5, then you get $25=x+5$.

Therefore, $x=\boxed{20}$.

Guest Jan 12, 2021

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