Solve the Quadratic Inequality: x² - 6x + 8 < 0

To solve this inequality $x^2 - 6x + 8 < 0$, we first identify the roots of the equation $x^2 - 6x + 8 = 0$.

Using the quadratic formula, where $a = 1$, $b = -6$, and $c = 8$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$

$x = \frac{6 \pm \sqrt{36 - 32}}{2}$

$x = \frac{6 \pm \sqrt{4}}{2}$

$x = \frac{6 \pm 2}{2}$

The solutions are:

$x = \frac{6 + 2}{2} = 4 \quad \text{and} \quad x = \frac{6 - 2}{2} = 2$

The roots are $x = 2$ and $x = 4$. These divide the number line into three intervals: $(-\infty, 2)$, $(2, 4)$, and $(4, \infty)$.

We test each interval to determine where the inequality is satisfied:

• In the interval $(-\infty, 2)$, select $x = 0$. Then:

$0^2 - 6 \times 0 + 8 = 8$, which is greater than 0. Inequality not satisfied.

• In the interval $(2, 4)$, select $x = 3$. Then:

$3^2 - 6 \times 3 + 8 = 9 - 18 + 8 = -1$, which is less than 0. Inequality satisfied.

• In the interval $(4, \infty)$, select $x = 5$. Then:

$5^2 - 6 \times 5 + 8 = 25 - 30 + 8 = 3$, which is greater than 0. Inequality not satisfied.

Therefore, the solution to the inequality $x^2 - 6x + 8 < 0$ is the interval $(2, 4)$.

Thus, the correct answer is $2 < x < 4$.