# Problem with indefinite square roots

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.

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HINT:

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

Observe that $x>0$

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Oh! I see! So easy question it was! Figured it out. thanks a lot :) –  Gaurang Tandon Jan 4 '14 at 16:03
@GaurangTandon, my pleasure. Related : math.stackexchange.com/questions/588414/… and math.stackexchange.com/questions/589288/… –  lab bhattacharjee Jan 4 '14 at 16:07
Oh! nice :) Always learning something new here! –  Gaurang Tandon Jan 4 '14 at 16:15

$$x^2-12=x$$

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

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but clearly $x>0$ so? –  Sami Ben Romdhane Jan 4 '14 at 16:02
@mathew Oh! I see! So easy question it was! Figured it out. thanks a lot :) –  Gaurang Tandon Jan 4 '14 at 16:04
$x$ can be negative –  Matt Jan 4 '14 at 16:05
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 $4$. –  user44197 Jan 4 '14 at 16:48