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I want to confirm if my answer in this problem is correct:

$$\sqrt{(x + \sqrt{(x + ...))}} = (1 + \sqrt{53}) / 2 $$

Solution:

$$x + \sqrt{(x + \sqrt{(x + ...))}} = (1 + \sqrt{53})^2 / 4 $$

$$x + (1 + \sqrt{53}) / 2 = (1 + 2\sqrt{53} + 53) / 4 $$

$$ x + (1 + \sqrt{53}) / 2 = (54 + 2\sqrt{53}) / 4 $$

$$x + (1 + \sqrt{53}) / 2 = (27 + \sqrt{53}) / 2 $$

$$2x + 1 + \sqrt{53} = 27 + \sqrt{53}$$

$2x = 26$

$x = 13$

I also tried solving something like $$\sqrt{x + \sqrt{x + \sqrt{x + \sqrt(x)}}} = (1 + \sqrt{53}) / 2 $$ and the answer is $x = 13.0006$

Please confirm if my answer is correct or not. Thanks!

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Yes your answer is correct. Let's write it more readable:

Let $$y=\sqrt{x+\sqrt{x+\cdots}}$$ hence the equation is $$y=\frac{1+\sqrt{53}}{2}$$ so squaring the terms of the last equality gives $$y^2=x+y=\frac{(1+\sqrt{53})^2}{4} $$ hence we have $$x=y^2-y=\frac{(1+\sqrt{53})^2}{4}-\frac{1+\sqrt{53}}{2}=\frac{1+2\sqrt{53}+53-2-2\sqrt{53}}{4}=\frac{52}{4}=13$$

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  • $\begingroup$ Thanks! I'm so sorry if I have troubled your reading. I don't know how to use what you've used in writing equations. $\endgroup$ – user133255 Mar 5 '14 at 11:15
  • $\begingroup$ @user133255: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – 5xum Mar 5 '14 at 11:19
  • $\begingroup$ @user133255 Not at all you didn't trouble me and you're welcome! $\endgroup$ – user63181 Mar 5 '14 at 11:19

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