Solve (a-x)(x-1/a)>0

$$( a - x ) ( x - \frac { 1 } { a } ) > 0$$

Solve for x

$$\left\{\begin{matrix}x\in \left(a,\frac{1}{a}\right)\text{, }&a<-1\text{ or }\left(a>0\text{ and }a<1\right)\\x\in \left(\frac{1}{a},a\right)\text{, }&a>1\text{ or }\left(a>-1\text{ and }a<0\right)\end{matrix}\right.$$

Solution Steps

For the product to be positive, a-x and x-\frac{1}{a} have to be both negative or both positive. Consider the case when a-x and x-\frac{1}{a} are both negative.

$$a-x<0 x-\frac{1}{a}<0$$

The solution satisfying both inequalities is x\in \left(a,\frac{1}{a}\right)\text{, }a<-1\text{ or }\left(a>0\text{ and }a<1\right).

$$x\in \left(a,\frac{1}{a}\right)\text{, }a<-1\text{ or }\left(a>0\text{ and }a<1\right)$$

Consider the case when a-x and x-\frac{1}{a} are both positive.

$$x-\frac{1}{a}>0 a-x>0$$

The solution satisfying both inequalities is x\in \left(\frac{1}{a},a\right)\text{, }a>1\text{ or }\left(a>-1\text{ and }a<0\right).

$$x\in \left(\frac{1}{a},a\right)\text{, }a>1\text{ or }\left(a>-1\text{ and }a<0\right)$$

The final solution is the union of the obtained solutions.

$$\left\{\begin{matrix}x\in \left(a,\frac{1}{a}\right)\text{, }&a<-1\text{ or }\left(a>0\text{ and }a<1\right)\\x\in \left(\frac{1}{a},a\right)\text{, }&a>1\text{ or }\left(a>-1\text{ and }a<0\right)\end{matrix}\right.$$

Solve for a

$$\left\{\begin{matrix}a>x\text{, }&x\geq 1\\a>\frac{1}{x}\text{, }&x>0\text{ and }x<1\\a\in \left(\frac{1}{x},0\right)\text{, }&x<-1\\a\in \left(x,0\right)\text{, }&x\geq -1\text{ and }x<0\\a\in \left(0,\frac{1}{x}\right)\text{, }&x>1\\a\in \left(0,x\right)\text{, }&x>0\text{ and }x\leq 1\\a<x\text{, }&x\leq -1\\a<\frac{1}{x}\text{, }&x>-1\text{ and }x<0\end{matrix}\right.$$