Solve 2x^2-7xgeq0

$$2 x ^ { 2 } - 7 x \geq 0$$

Solve for x

$$x\in (-\infty,0]\cup [\frac{7}{2},\infty)$$

Solution Steps

Factor out x.

$$x\left(2x-7\right)\geq 0$$

For the product to be ≥0, x and x-\frac{7}{2} have to be both ≤0 or both ≥0. Consider the case when x and x-\frac{7}{2} are both ≤0.

$$x\leq 0 x-\frac{7}{2}\leq 0$$

The solution satisfying both inequalities is x\leq 0.

$$x\leq 0$$

Consider the case when x and x-\frac{7}{2} are both ≥0.

$$x-\frac{7}{2}\geq 0 x\geq 0$$

The solution satisfying both inequalities is x\geq \frac{7}{2}.

$$x\geq \frac{7}{2}$$

The final solution is the union of the obtained solutions.

$$x\leq 0\text{; }x\geq \frac{7}{2}$$