Solve (x-3)(x-1)geq0

$$( x - 3 ) ( x - 1 ) \geq 0$$

Solve for x

$$x\in (-\infty,1]\cup [3,\infty)$$

Solution Steps

For the product to be ≥0, x-1 and x-3 have to be both ≤0 or both ≥0. Consider the case when x-1 and x-3 are both ≤0.

$$x-1\leq 0 x-3\leq 0$$

The solution satisfying both inequalities is x\leq 1.

$$x\leq 1$$

Consider the case when x-1 and x-3 are both ≥0.

$$x-3\geq 0 x-1\geq 0$$

The solution satisfying both inequalities is x\geq 3.

$$x\geq 3$$

The final solution is the union of the obtained solutions.

$$x\leq 1\text{; }x\geq 3$$