Solve the Quadratic Inequality: x²-8x+12>0 Step by Step

Let's proceed to solve the inequality $x^2 - 8x + 12 > 0$.

• Start by factoring the quadratic: $x^2 - 8x + 12$.
• Identify the factors of 12 that sum to 8: $6$ and $2$. This results in: $(x - 6)(x - 2) = 0$.

The factorization gives us the critical points $x = 6$ and $x = 2$. These points divide the number line into three intervals: $x < 2$, $2 < x < 6$, and $x > 6$.

Now, we evaluate the sign of the product $(x - 6)(x - 2)$ in each interval:

• For $x < 2$: Both $(x - 6)$ and $(x - 2)$ are negative, so their product is positive.
• For $2 < x < 6$: $(x - 2)$ is positive, $(x - 6)$ is negative, so their product is negative.
• For $x > 6$: Both $(x - 6)$ and $(x - 2)$ are positive, so their product is positive.

The inequality $(x - 6)(x - 2) > 0$ holds for $x < 2$ and $x > 6$.

Thus, the solution to the inequality $x^2 - 8x + 12 > 0$ is $x < 2$ or $x > 6$.

Therefore, the correct answer is $\boxed{x < 2, 6 < x}$.