Polynomial Inequality D47210

1. **State the problem:** Solve the inequality $$x^2 - 12x + 27 > 0$$ and express the solution set in interval notation. 2. **Find the boundary points:** Solve the equation $$x^2 - 12x + 27 = 0$$ to find the critical points. 3. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-12$, and $c=27$. 4. Calculate the discriminant: $$\Delta = (-12)^2 - 4 \times 1 \times 27 = 144 - 108 = 36$$ 5. Find the roots: $$x = \frac{12 \pm \sqrt{36}}{2} = \frac{12 \pm 6}{2}$$ 6. So, $$x_1 = \frac{12 - 6}{2} = \frac{6}{2} = 3$$ $$x_2 = \frac{12 + 6}{2} = \frac{18}{2} = 9$$ 7. **Determine intervals:** The critical points divide the real line into three intervals: $$(-\infty, 3), \quad (3, 9), \quad (9, \infty)$$ 8. **Test each interval:** Choose test points in each interval to check the sign of $$f(x) = x^2 - 12x + 27$$. - For $$x=0$$ in $$(-\infty, 3)$$: $$f(0) = 0^2 - 12 \times 0 + 27 = 27 > 0$$ - For $$x=5$$ in $$(3, 9)$$: $$f(5) = 25 - 60 + 27 = -8 < 0$$ - For $$x=10$$ in $$(9, \infty)$$: $$f(10) = 100 - 120 + 27 = 7 > 0$$ 9. **Solution set:** The inequality $$f(x) > 0$$ holds for intervals where the function is positive: $$(-\infty, 3) \cup (9, \infty)$$ 10. **Express in interval notation:** $$\boxed{(-\infty, 3) \cup (9, \infty)}$$ 11. **Summary:** The solution to $$x^2 - 12x + 27 > 0$$ is all real numbers less than 3 or greater than 9.