Solve x^2leqx

$$x ^ { 2 } \leq x$$

Solve for x

$$x\in \begin{bmatrix}0,1\end{bmatrix}$$

Solution Steps

Subtract x from both sides.

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

Factor out x.

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

For the product to be ≤0, one of the values x and x-1 has to be ≥0 and the other has to be ≤0. Consider the case when x\geq 0 and x-1\leq 0.

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

The solution satisfying both inequalities is x\in \left[0,1\right].

$$x\in \begin{bmatrix}0,1\end{bmatrix}$$

Consider the case when x\leq 0 and x-1\geq 0.

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

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}0,1\end{bmatrix}$$