Find the equation solution $\left[ x\right] +\left[ \dfrac{9}{x}\right]=6$ solve this following equation
$$\left[ x\right] +\left[ \dfrac{9}{x}\right]=6$$
 the equation obvious have a root $x=3$. But how to solve other roots?

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*for $x\le\lfloor{\frac{9}{6}}\rfloor$ there is no answer since $\lfloor{\frac{9}{x}}\rfloor$ is at least $6$ and when $\lfloor{x}\rfloor$ becomes zero, the other term is $7$.

*for $\frac{9}{6}<x \le \frac{9}{5}$, we have $\lfloor{\frac{9}{x}}\rfloor=5$ and $\lfloor{x}\rfloor=1$, hence all this range is a valid answer.

*for $\frac{9}{5}<x<2$, we have $\lfloor{\frac{9}{x}}\rfloor=4$ and $\lfloor{x}\rfloor=1$, hence no answer in this range.

*for $2\le x \le \frac{9}{4}$, we have $\lfloor{\frac{9}{x}}\rfloor=4$ and $\lfloor{x}\rfloor=2$, hence all this range is a valid answer.

*for $\frac{9}{4}<x<3$, we have $\lfloor{\frac{9}{x}}\rfloor=3$ and $\lfloor{x}\rfloor=2$, hence no answer in this range.

*for $x=3$, we have $\lfloor{\frac{9}{3}}\rfloor=3$ and it is an answer.

*for $3<x<4$, we have $\lfloor{\frac{9}{x}}\rfloor=2$ and $\lfloor{x}\rfloor=3$, hence no answer in this range.

*for $4\le x \le \frac{9}{2}$, we have $\lfloor{\frac{9}{x}}\rfloor=2$ and $\lfloor{x}\rfloor=4$, hence all this range is a valid answer.

*for $\frac{9}{2}<x<5$, we have $\lfloor{\frac{9}{x}}\rfloor=1$ and $\lfloor{x}\rfloor=4$, hence no answer in this range.

*for $5\le x <6$, we have $\lfloor{\frac{9}{x}}\rfloor=1$ and $\lfloor{x}\rfloor=5$, hence all this range is a valid answer.

*for $x\ge 6$ also there is no answer.

