Continued addition and under rooting of 12

Question

$\sqrt{(12 + \sqrt{12......})}$ and so on....
How do I find its answer? This is a question in our class VII mats book.

P.S. - Answer is 4

Answer 1

Let $\sqrt{(12 + \sqrt{12......})}=x$

$$x^2=12+\sqrt{(12 + \sqrt{12......})}=12+x$$

$$x^2-x-12=0$$

$$(x-4)(x+3)=0$$

$$x=4 \text{ or } x=-3 \text{ (rejected as $x>0$)}$$

Answer 2

The answer appears to be $4$. Let $x=\sqrt{12+\sqrt{12+\ldots}}$. Then $x$ satifies the equation $x^2-12=x$ or $x^2-x-12=0$. The only solutions are $4$ and $-3$. Since $x$ is positive it must be $4$.