Solve -3x^2+27xgeq0

$$-3 { x }^{ 2 } +27x \geq 0$$

Solve for x

$$x\in \begin{bmatrix}0,9\end{bmatrix}$$

Solution Steps

Multiply the inequality by -1 to make the coefficient of the highest power in -3x^{2}+27x positive. Since -1 is negative, the inequality direction is changed.

$$3x^{2}-27x\leq 0$$

Factor out x.

$$3x\left(x-9\right)\leq 0$$

For the product to be ≤0, one of the values x and x-9 has to be ≥0 and the other has to be ≤0. Consider the case when x\geq 0 and x-9\leq 0.

$$x\geq 0 x-9\leq 0$$

The solution satisfying both inequalities is x\in \left[0,9\right].

$$x\in \begin{bmatrix}0,9\end{bmatrix}$$

Consider the case when x\leq 0 and x-9\geq 0.

$$x-9\geq 0 x\leq 0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}0,9\end{bmatrix}$$