Find the equation solution $\left[ x\right] +\left[ \dfrac{9}{x}\right]=6$

Question

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?

Answer 1

• 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.

Answer 2

Obviously $x>0$, suppose that $\lfloor x \rfloor =k$ (clearly $k$ must be between $0$ and $5$), then the solutions are $\lfloor \frac{9}{x} \rfloor = 6-k\iff 6-k\leq \frac{9}{x} < 6-k+1\iff \frac{9}{6-k+1}< x \leq \frac{9}{6-k}$.

So the solution set is $\bigcup\limits_{k=0}^{5}( (\frac{9}{6-k+1},\frac{9}{6-k}]\cap [k,k+1))$.

This is just a simple and fast calculation now.

Answer 3

First of all, it's clear that $x>0$, because for negative $x$, both terms on the left are negative. Therefore, both $\lfloor x\rfloor$ and $\left\lfloor\frac9x\right\rfloor$ are non-negative integers, so the pair $\left(\lfloor x\rfloor,\left\lfloor\frac9x\right\rfloor\right)$ can only be $(0,6)$, $(1,5)$, $(2,4)$, $(3,3)$, $(4,2)$, $(5,1)$ or $(6,0)$. Each pair can be dealt with as a pair of inequalities. For example:

$\left(\lfloor x\rfloor,\left\lfloor\frac9x\right\rfloor\right)=(1,5)$

is equivalent to:

$1\leq x < 2$ and $5\leq \frac9x < 6$.

The second inequality turns into $\frac16 < \frac{x}{9} \leq \frac15$, or

$\frac96 < x \leq \frac95$. The intersection of this interval with the interval $[1,2)$ is the interval $(1.5, 1.8]$, so that's part of your solution set. If you repeat this process for each of the seven cases, you'll get all solutions.