Solve m^2-16m+48leq0

$$m ^ { 2 } - 16 m + 48 \leq 0$$

Solve for m

$$m\in \begin{bmatrix}4,12\end{bmatrix}$$

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.

$$m^{2}-16m+48=0$$

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, -16 for b, and 48 for c in the quadratic formula.

$$m=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 1\times 48}}{2}$$

Do the calculations.

$$m=\frac{16±8}{2}$$

Solve the equation m=\frac{16±8}{2} when ± is plus and when ± is minus.

$$m=12 m=4$$

Rewrite the inequality by using the obtained solutions.

$$\left(m-12\right)\left(m-4\right)\leq 0$$

For the product to be ≤0, one of the values m-12 and m-4 has to be ≥0 and the other has to be ≤0. Consider the case when m-12\geq 0 and m-4\leq 0.

$$m-12\geq 0 m-4\leq 0$$

This is false for any m.

$$m\in \emptyset$$

Consider the case when m-12\leq 0 and m-4\geq 0.

$$m-4\geq 0 m-12\leq 0$$

The solution satisfying both inequalities is m\in \left[4,12\right].

$$m\in \begin{bmatrix}4,12\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$m\in \begin{bmatrix}4,12\end{bmatrix}$$