Root problem

**Aditya3003** · December 21st 2012, 10:52 PM

Find x:
x = sq.root[1 + sq.root( 1 + sq.root(1 + sq.root(1 +....
"The dots mean that the series go till infinity.
By attempt the answer should be (1 +-sq.root(5))/2 but that is not the answer.

**jakncoke** · December 21st 2012, 11:03 PM

/the answer is $\frac{1 \frac{+}{-} \sqrt(5)}{2}$

The infinite nested radicals, $\sqrt(1 + \sqrt(1 + \sqrt(1 + ....$ = golden ratio.

Since golden ratio says $\frac{a+b}{a} = \frac{a}{b} = x$ then $1 + \frac{b}{a} = 1 + \frac{1}{x} = x$ so solving $x^2 - x - 1 = 0$

**ibdutt** · December 21st 2012, 11:08 PM

Could you please write as to what is the answer given. The answer you have got appears to be correct.
$Root problem-series.png$

**jakncoke** · December 21st 2012, 11:10 PM

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**Soroban** · December 22nd 2012, 05:06 AM

Hello, Aditya3003!

$\text{Evaluate: }\:x \:=\: \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}$

By attempt the answer should be: $\frac{1\pm\sqrt{5}}{2}$ , but that is not the answer.

Since $x$ is obviously positive,
. . you must reject the negative root: $\tfrac{1-\sqrt{5}}{2}$

While solving, you squared the equation, right?
. . This can introduce an extraneous root.

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