A function problem

Question

Let the solution set of the equation:

$\sqrt{\left[x+[\frac{x}{2}]\right]}+\left[[\sqrt{x}]+[\frac{x}{3}]\right]=3 $ be $[a,b)$.

find $a\cdot b$, Where $[.]$ Is greatest integer function and $\{.\}$ Is fractional part function.

Answer 1

If you notice carefully in the equation, $$\sqrt{\left[x+[\frac{x}{2}]\right]}+\left[\sqrt{\{x\}}+[\frac{x}{3}]\right] = 3$$, this $\sqrt{\{x\}}$ is useless. So question becomes $$\sqrt{\left[x+[\frac{x}{2}]\right]} + \left[\frac{x}{3}\right] = 3$$.

It's quite clear that it won't have any solution in $[0, 3)$, it clearly satisfies for $[3, 4)$. I don't know how to explain it more. So, answer should be $3\cdot 4 = 12$.

Answer 2

Adding to Zenix' solution (read it first):

The way to start is to note that the second term on the left is an integer, so the first must be as well. Both are nonnegative, so they must be in $\{0,1,2,3\}$. Neither can be $3$ while the other is $0$, so one must be $1$ and the other must be $2$. If the second term is $2$, the first will be at least $2$, so $x \in [3,6)$ with $$\sqrt{\left[x+[\frac{x}{2}]\right]}=2\\ \left[\frac{x}{3}\right] = 1\\ \left[x+[\frac{x}{2}]\right]=4$$ which finally gives $x \in [3,4)$