Question 5

1. Let $x = \sqrt{12 + \sqrt{12 + \sqrt{12 + \dots}}}$.

2. Squaring both sides yields:

   $$x^2 = 12 + x.$$

3. Rearrange into a standard quadratic equation:

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

4. Factor the quadratic:

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

5. Since the principal square root is non-negative, $x > 0$. Therefore:

   $$x = 4.000.$$

This matches option C.