Questions: x^2 - x - 2 > 0

Transcript text: $x^{2}-x-2>0$

Solution

Solution Steps

Step 1: Factor the Polynomial

We start with the inequality $ x^2 - x - 2 > 0 $. To solve this, we first factor the polynomial. The factorization yields: \[ x^2 - x - 2 = (x - 2)(x + 1) \]

Step 2: Identify Critical Points

Next, we identify the critical points where the expression changes sign. Setting each factor to zero gives us the critical points: \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \] Thus, the critical points are $ x = -1 $ and $ x = 2 $.

Step 3: Test Intervals

We will test the sign of the product $ (x - 2)(x + 1) $ in the intervals defined by the critical points: $ (-\infty, -1) $, $ (-1, 2) $, and $ (2, \infty) $.

Interval $ (-\infty, -1) $:
- Test Point: $ -2 $
- Value: $ (-2 - 2)(-2 + 1) = (-4)(-1) = 4 $
- Inequality holds: True
Interval $ (-1, 2) $:
- Test Point: $ 0.5 $
- Value: $ (0.5 - 2)(0.5 + 1) = (-1.5)(1.5) = -2.25 $
- Inequality holds: False
Interval $ (2, \infty) $:
- Test Point: $ 3 $
- Value: $ (3 - 2)(3 + 1) = (1)(4) = 4 $
- Inequality holds: True

Step 4: Conclusion

The inequality $ (x - 2)(x + 1) > 0 $ holds in the intervals $ (-\infty, -1) $ and $ (2, \infty) $. Thus, the solution to the inequality $ x^2 - x - 2 > 0 $ is: \[ x \in (-\infty, -1) \cup (2, \infty) \]

Final Answer

$\boxed{(-\infty, -1) \cup (2, \infty)}$

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