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Question 1209650: Determine the set of all real x satisfying
x^3 - 2x^2 - 3x < -25x^2 + 17x.
Enter your answer in interval notation.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 1. **Rewrite the inequality:**
Move all terms to one side:
x³ - 2x² - 3x + 25x² - 17x < 0
x³ + 23x² - 20x < 0
2. **Factor out x:**
x(x² + 23x - 20) < 0
3. **Solve the quadratic equation:**
Find the roots of x² + 23x - 20 = 0 using the quadratic formula:
x = (-b ± √(b² - 4ac)) / 2a
x = (-23 ± √(23² - 4(1)(-20))) / 2(1)
x = (-23 ± √(529 + 80)) / 2
x = (-23 ± √609) / 2
The roots are approximately x ≈ 0.839 and x ≈ -23.839
4. **Determine the intervals:**
The roots of the cubic equation are x = 0, x ≈ 0.839, and x ≈ -23.839. These roots divide the number line into four intervals:
* x < -23.839
* -23.839 < x < 0
* 0 < x < 0.839
* x > 0.839
5. **Test points in each interval:**
* x = -24: (-24)(-24² + 23(-24) - 20) < 0 => -24(576 - 552 - 20) < 0 => -24(4) < 0. True
* x = -1: (-1)(1 - 23 - 20) < 0 => (-1)(-42) < 0. False
* x = 0.5: (0.5)(0.25 + 11.5 - 20) < 0 => 0.5(-8.25) < 0. True
* x = 1: (1)(1 + 23 - 20) < 0 => 4 < 0. False
6. **Write the solution in interval notation:**
The inequality is satisfied when x < -23.839 or 0 < x < 0.839.
Final Answer: The final answer is $\boxed{(-\infty, \frac{-23 - \sqrt{609}}{2}) \cup (0, \frac{-23 + \sqrt{609}}{2})}$
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