Solve x^2-8x+12leq0

$${ x }^{ 2 } -8x+12 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}2,6\end{bmatrix}$$

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.

$$x^{2}-8x+12=0$$

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 1 for a, -8 for b, and 12 for c in the quadratic formula.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\times 12}}{2}$$

Do the calculations.

$$x=\frac{8±4}{2}$$

Solve the equation x=\frac{8±4}{2} when ± is plus and when ± is minus.

$$x=6 x=2$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-6\right)\left(x-2\right)\leq 0$$

For the product to be ≤0, one of the values x-6 and x-2 has to be ≥0 and the other has to be ≤0. Consider the case when x-6\geq 0 and x-2\leq 0.

$$x-6\geq 0 x-2\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-6\leq 0 and x-2\geq 0.

$$x-2\geq 0 x-6\leq 0$$

The solution satisfying both inequalities is x\in \left[2,6\right].

$$x\in \begin{bmatrix}2,6\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}2,6\end{bmatrix}$$