Solve the Quadratic Inequality: x²-2x-8 > 0

Solve the following equation:

$x^2-2x-8>0$

To solve the inequality $x^2 - 2x - 8 > 0$, we first need to find the roots of the related equation $x^2 - 2x - 8 = 0$.

Step 1: Factor the quadratic
The quadratic $x^2 - 2x - 8$ can be factored as $(x - 4)(x + 2)$ because:

• The product is $-8$ and the sum is $-2$.
• Expanding $(x - 4)(x + 2)$, we get:
• $(x - 4)(x + 2) = x^2 + 2x - 4x - 8 = x^2 - 2x - 8$.

Step 2: Identify the roots
Set each factor to zero to find the roots:

• $x - 4 = 0 \Rightarrow x = 4$
• $x + 2 = 0 \Rightarrow x = -2$

Step 3: Determine the intervals
The critical points divide the number line into three intervals: $x < -2$, $-2 < x < 4$, and $x > 4$.

Step 4: Test each interval
Choose test points from each interval to check where $(x - 4)(x + 2) > 0$:

• For $x < -2$, take $x = -3$:
  $(-3 - 4)(-3 + 2) = (-7)(-1) = 7 > 0$
• For $-2 < x < 4$, take $x = 0$:
  $(0 - 4)(0 + 2) = (-4)(2) = -8 < 0$
• For $x > 4$, take $x = 5$:
  $(5 - 4)(5 + 2) = (1)(7) = 7 > 0$

Conclusion:
The solution to the inequality $x^2 - 2x - 8 > 0$ is on the intervals $x < -2$ and $x > 4$.

Final Answer:
The correct answer is: Answers (a) and (c)