Quadratic Inequalities 79252D

1. Solve $x^2 - 9x + 20 > 0$. - Factor the quadratic: $x^2 - 9x + 20 = (x - 4)(x - 5)$. - The roots are $x=4$ and $x=5$. - Since the parabola opens upward (coefficient of $x^2$ is positive), the inequality $>0$ holds outside the roots. - Solution: $(-\infty, 4) \cup (5, \infty)$. 2. Solve $x^2 + 2x - 15 \leq 0$. - Factor: $x^2 + 2x - 15 = (x + 5)(x - 3)$. - Roots: $x=-5$, $x=3$. - Parabola opens upward, so $\leq 0$ between roots. - Solution: $[-5, 3]$. 3. Solve $2x^2 - 5x - 3 \geq 0$. - Find roots using quadratic formula: $$x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm 7}{4}$$ - Roots: $x = 3$ and $x = -\frac{1}{2}$. - Parabola opens upward. - Inequality $\geq 0$ holds outside roots. - Solution: $(-\infty, -\frac{1}{2}] \cup [3, \infty)$. 4. Solve $3x^2 + 7x - 6 < 0$. - Find roots: $$x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 3 \cdot (-6)}}{2 \cdot 3} = \frac{-7 \pm \sqrt{49 + 72}}{6} = \frac{-7 \pm \sqrt{121}}{6} = \frac{-7 \pm 11}{6}$$ - Roots: $x=\frac{4}{6} = \frac{2}{3}$ and $x=\frac{-18}{6} = -3$. - Parabola opens upward. - Inequality $<0$ holds between roots. - Solution: $(-3, \frac{2}{3})$. 5. Solve $\frac{1}{2} x^2 - 4x + 6 > 0$. - Multiply both sides by 2 (positive, so inequality direction unchanged): $$\cancel{2} \cdot \left(\frac{1}{\cancel{2}} x^2 - 4x + 6\right) > \cancel{2} \cdot 0 \Rightarrow x^2 - 8x + 12 > 0$$ - Factor or use quadratic formula: $$x = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2}$$ - Roots: $x=2$ and $x=6$. - Parabola opens upward. - Inequality $>0$ holds outside roots. - Solution: $(-\infty, 2) \cup (6, \infty)$. 6. Solve $-x^2 + 4x - 5 \leq 0$. - Multiply both sides by $-1$ (flip inequality): $$\cancel{-1} \cdot (-x^2 + 4x - 5) \geq \cancel{-1} \cdot 0 \Rightarrow x^2 - 4x + 5 \geq 0$$ - Discriminant: $(-4)^2 - 4 \cdot 1 \cdot 5 = 16 - 20 = -4 < 0$. - No real roots, parabola opens upward. - Since $x^2 - 4x + 5 > 0$ always, original inequality holds for all $x$. - Solution: $(-\infty, \infty)$. 7. Solve $-2x^2 + 3x + 1 > 0$. - Multiply both sides by $-1$ (flip inequality): $$\cancel{-1} \cdot (-2x^2 + 3x + 1) < \cancel{-1} \cdot 0 \Rightarrow 2x^2 - 3x - 1 < 0$$ - Find roots: $$x = \frac{3 \pm \sqrt{9 + 8}}{4} = \frac{3 \pm \sqrt{17}}{4}$$ - Parabola opens upward. - Inequality $<0$ holds between roots. - Solution: $\left(\frac{3 - \sqrt{17}}{4}, \frac{3 + \sqrt{17}}{4}\right)$. 8. Solve $-x^2 - 2x - 3 \geq 0$. - Multiply both sides by $-1$ (flip inequality): $$\cancel{-1} \cdot (-x^2 - 2x - 3) \leq \cancel{-1} \cdot 0 \Rightarrow x^2 + 2x + 3 \leq 0$$ - Discriminant: $2^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8 < 0$. - No real roots, parabola opens upward. - Since $x^2 + 2x + 3 > 0$ always, inequality $\leq 0$ never holds. - Solution: $\emptyset$ (no solution).