Problem with indefinite square roots

Question

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.

score 2 · Answer 1 · answered Jan 04 '14 at 16:01

2

HINT:

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

Observe that $x>0$

answered Jan 04 '14 at 16:01

lab bhattacharjee

@GaurangTandon, my pleasure. Related : http://math.stackexchange.com/questions/588414/textlet-y-sqrt5-sqrt5-sqrt5-sqrt5-what-is-the-nearest-val and http://math.stackexchange.com/questions/589288/sqrt7-sqrt7-sqrt7-sqrt7-sqrt7-cdots-approximation – lab bhattacharjee Jan 04 '14 at 16:07

score 2 · Answer 2 · answered Jan 04 '14 at 16:01

2

$$x^2-12=x$$

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

answered Jan 04 '14 at 16:01

Mathgrad

1

but clearly $x>0$ so? – Jan 04 '14 at 16:02
$x$ can be negative – Mathgrad Jan 04 '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 04 '14 at 16:48

2 Answers2