Question: Consider the quadratic inequality $x^2 - 14x + 48 \leq 0$. What is the solution set?
1. **State the problem:** We need to solve the quadratic inequality $$x^2 - 14x + 48 \leq 0$$.
2. **Find the roots of the quadratic equation:** First, solve the equation $$x^2 - 14x + 48 = 0$$ to find critical points.
3. **Use the quadratic formula:**
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-14$, and $c=48$.
4. Calculate the discriminant:
$$\Delta = (-14)^2 - 4 \times 1 \times 48 = 196 - 192 = 4$$
5. Calculate the roots:
$$x = \frac{14 \pm \sqrt{4}}{2} = \frac{14 \pm 2}{2}$$
6. So the roots are:
$$x_1 = \frac{14 - 2}{2} = \frac{12}{2} = 6$$
$$x_2 = \frac{14 + 2}{2} = \frac{16}{2} = 8$$
7. **Analyze the inequality:** Since the parabola opens upwards ($a=1 > 0$), the quadratic expression is less than or equal to zero between the roots.
8. **Solution set:**
$$6 \leq x \leq 8$$
9. **Answer:** The solution set is $6 \leq x \leq 8$.