I have a question which reads:

If $$\sqrt{12 + \sqrt{12 + \sqrt{12 + \cdots\cdots}}} = x$$ Then the value of $x$ is _.

I think that we can write $$x^2 - 12 = \sqrt{12 + \sqrt{12 + \sqrt{12 + \cdots\cdots}}}$$

But the square roots never end!

Can anyone please give me tips and hints for this.

Thanks a lot.



So we can write $$x^2-12=x$$ assuming the convergence of the series

Observe that $x>0$



On solving $x=4$ or $x=-3$

  • 1
    $\begingroup$ but clearly $x>0$ so? $\endgroup$ – user63181 Jan 4 '14 at 16:02
  • $\begingroup$ $x$ can be negative $\endgroup$ – Mathew George Jan 4 '14 at 16:05
  • $\begingroup$ If you take the nonstandard view and take the negative number as the square root, then $x=-3$. However, taking the positive square roots is too steeped in tradition and is the standard square root. So the answer is $4$. $\endgroup$ – user44197 Jan 4 '14 at 16:48

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