Solve x^2geqx

$$x ^ { 2 } \geq x$$

Solve for x

$$x\in (-\infty,0]\cup [1,\infty)$$

Solution Steps

Subtract x from both sides.

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

Factor out x.

$$x\left(x-1\right)\geq 0$$

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

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

The solution satisfying both inequalities is x\leq 0.

$$x\leq 0$$

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

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

The solution satisfying both inequalities is x\geq 1.

$$x\geq 1$$

The final solution is the union of the obtained solutions.

$$x\leq 0\text{; }x\geq 1$$