Rational Inequality E485Ff

1. **State the problem:** Solve the inequality $$\frac{x^2 - 4}{x + 3} > 0$$ and express the solution as a union of intervals and on a number line. 2. **Rewrite the inequality:** Note that $$x^2 - 4$$ can be factored as a difference of squares: $$x^2 - 4 = (x - 2)(x + 2)$$ So the inequality becomes: $$\frac{(x - 2)(x + 2)}{x + 3} > 0$$ 3. **Identify critical points:** The numerator is zero at $$x = 2$$ and $$x = -2$$. The denominator is zero at $$x = -3$$ (which is a vertical asymptote and excluded from the domain). 4. **Determine intervals:** The critical points divide the real line into four intervals: $$(-\infty, -3), (-3, -2), (-2, 2), (2, \infty)$$ 5. **Test each interval:** Choose test points in each interval to determine the sign of the expression. - For $$x < -3$$, e.g., $$x = -4$$: $$\frac{(-4 - 2)(-4 + 2)}{-4 + 3} = \frac{(-6)(-2)}{-1} = \frac{12}{-1} = -12 < 0$$ - For $$-3 < x < -2$$, e.g., $$x = -2.5$$: $$\frac{(-2.5 - 2)(-2.5 + 2)}{-2.5 + 3} = \frac{(-4.5)(-0.5)}{0.5} = \frac{2.25}{0.5} = 4.5 > 0$$ - For $$-2 < x < 2$$, e.g., $$x = 0$$: $$\frac{(0 - 2)(0 + 2)}{0 + 3} = \frac{(-2)(2)}{3} = \frac{-4}{3} < 0$$ - For $$x > 2$$, e.g., $$x = 3$$: $$\frac{(3 - 2)(3 + 2)}{3 + 3} = \frac{(1)(5)}{6} = \frac{5}{6} > 0$$ 6. **Determine where the expression is positive:** From the tests, the expression is positive on intervals: $$(-3, -2) \cup (2, \infty)$$ 7. **Check endpoints:** - At $$x = -3$$, denominator zero, undefined, exclude. - At $$x = -2$$, numerator zero, expression equals zero, not greater than zero, exclude. - At $$x = 2$$, numerator zero, expression equals zero, exclude. 8. **Final solution:** $$\boxed{(-3, -2) \cup (2, \infty)}$$ 9. **Number line representation:** - Open circles at $$-3$$, $$-2$$, and $$2$$. - Shade the intervals $$(-3, -2)$$ and $$(2, \infty)$$. This completes the solution.