Solve {l}{3x^2-12x}{+9leq0}

$$\left. \begin{array} { l } { 3 x ^ { 2 } - 12 x } \\ { + 9 \leq 0 } \end{array} \right.$$

Solve for x

$$x\in \begin{bmatrix}1,3\end{bmatrix}$$

Steps Using the Quadratic Formula

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$3x^{2}-12x+9=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 3 for a, -12 for b, and 9 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3\times 9}}{2\times 3}$$

Do the calculations.

$$x=\frac{12±6}{6}$$

Solve the equation x=\frac{12±6}{6} when ± is plus and when ± is minus.

$$x=3 x=1$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-3\right)\left(x-1\right)\leq 0$$

For the product to be ≤0, one of the values x-3 and x-1 has to be ≥0 and the other has to be ≤0. Consider the case when x-3\geq 0 and x-1\leq 0.

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

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-3\leq 0 and x-1\geq 0.

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

The solution satisfying both inequalities is x\in \left[1,3\right].

$$x\in \begin{bmatrix}1,3\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}1,3\end{bmatrix}$$