Question: Solve each quadratic inequality and illustrate its solution set on the number line. Write your final answer in interval notation and set notation.

1. Solve $x^2 - 9x + 20 > 0$. Step 1: Factor the quadratic. $$x^2 - 9x + 20 = (x - 4)(x - 5)$$ Step 2: Find the roots. $$x - 4 = 0 \Rightarrow x = 4$$ $$x - 5 = 0 \Rightarrow x = 5$$ Step 3: Determine intervals based on roots: $(-\infty,4)$, $(4,5)$, $(5,\infty)$. Step 4: Test points in each interval for inequality $>0$. - For $x=3$: $(3-4)(3-5) = (-1)(-2) = 2 > 0$ (True) - For $x=4.5$: $(4.5-4)(4.5-5) = (0.5)(-0.5) = -0.25 < 0$ (False) - For $x=6$: $(6-4)(6-5) = (2)(1) = 2 > 0$ (True) Step 5: Solution intervals where inequality holds: $$(-\infty,4) \cup (5,\infty)$$ Step 6: Write in set notation: $$\{x \mid x < 4 \text{ or } x > 5\}$$ --- 2. Solve $x^2 + 2x - 15 \leq 0$. Step 1: Factor the quadratic. $$x^2 + 2x - 15 = (x + 5)(x - 3)$$ Step 2: Find roots. $$x = -5, 3$$ Step 3: Intervals: $(-\infty,-5)$, $[-5,3]$, $(3,\infty)$. Step 4: Test points: - $x=-6$: $( -6 + 5)(-6 - 3) = (-1)(-9) = 9 > 0$ (False) - $x=0$: $(0 + 5)(0 - 3) = (5)(-3) = -15 < 0$ (True) - $x=4$: $(4 + 5)(4 - 3) = (9)(1) = 9 > 0$ (False) Step 5: Solution interval: $$[-5,3]$$ Step 6: Set notation: $$\{x \mid -5 \leq x \leq 3\}$$ --- 3. Solve $2x^2 - 5x - 3 \geq 0$. Step 1: Factor the quadratic. $$2x^2 - 5x - 3 = (2x + 1)(x - 3)$$ Step 2: Roots: $$2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$$ $$x - 3 = 0 \Rightarrow x = 3$$ Step 3: Intervals: $(-\infty,-\frac{1}{2})$, $[-\frac{1}{2},3]$, $(3,\infty)$. Step 4: Test points: - $x=-1$: $(2(-1)+1)(-1-3) = (-2+1)(-4) = (-1)(-4) = 4 > 0$ (True) - $x=0$: $(2(0)+1)(0-3) = (1)(-3) = -3 < 0$ (False) - $x=4$: $(2(4)+1)(4-3) = (8+1)(1) = 9 > 0$ (True) Step 5: Solution intervals: $$(-\infty,-\frac{1}{2}] \cup [3,\infty)$$ Step 6: Set notation: $$\{x \mid x \leq -\frac{1}{2} \text{ or } x \geq 3\}$$ --- 4. Solve $3x^2 + 7x - 6 < 0$. Step 1: Factor the quadratic. $$3x^2 + 7x - 6 = (3x - 2)(x + 3)$$ Step 2: Roots: $$3x - 2 = 0 \Rightarrow x = \frac{2}{3}$$ $$x + 3 = 0 \Rightarrow x = -3$$ Step 3: Intervals: $(-\infty,-3)$, $[-3,\frac{2}{3}]$, $(\frac{2}{3},\infty)$. Step 4: Test points: - $x=-4$: $(3(-4)-2)(-4+3) = (-12-2)(-1) = (-14)(-1) = 14 > 0$ (False) - $x=0$: $(3(0)-2)(0+3) = (-2)(3) = -6 < 0$ (True) - $x=1$: $(3(1)-2)(1+3) = (3-2)(4) = (1)(4) = 4 > 0$ (False) Step 5: Solution interval: $$(-3, \frac{2}{3})$$ Step 6: Set notation: $$\{x \mid -3 < x < \frac{2}{3}\}$$ --- 5. Solve $\frac{1}{2}x^2 - 4x + 6 > 0$. Step 1: Multiply entire inequality by 2 to clear fraction: $$x^2 - 8x + 12 > 0$$ Step 2: Factor quadratic. $$x^2 - 8x + 12 = (x - 2)(x - 6)$$ Step 3: Roots: $$x = 2, 6$$ Step 4: Intervals: $(-\infty,2)$, $(2,6)$, $(6,\infty)$. Step 5: Test points: - $x=0$: $(0-2)(0-6) = (-2)(-6) = 12 > 0$ (True) - $x=4$: $(4-2)(4-6) = (2)(-2) = -4 < 0$ (False) - $x=7$: $(7-2)(7-6) = (5)(1) = 5 > 0$ (True) Step 6: Solution intervals: $$(-\infty,2) \cup (6,\infty)$$ Step 7: Set notation: $$\{x \mid x < 2 \text{ or } x > 6\}$$ --- 6. Solve $-x^2 + 4x - 5 \leq 0$. Step 1: Multiply entire inequality by $-1$ (reverse inequality): $$x^2 - 4x + 5 \geq 0$$ Step 2: Calculate discriminant: $$\Delta = (-4)^2 - 4(1)(5) = 16 - 20 = -4 < 0$$ Step 3: Since discriminant is negative, quadratic has no real roots and is always positive (since leading coefficient is positive). Step 4: Therefore, $$x^2 - 4x + 5 > 0 \text{ for all } x$$ Step 5: So original inequality holds for all $x$: $$\mathbb{R} = (-\infty, \infty)$$ Step 6: Set notation: $$\{x \mid x \in \mathbb{R}\}$$ --- 7. Solve $-2x^2 + 3x + 1 > 0$. Step 1: Multiply entire inequality by $-1$ (reverse inequality): $$2x^2 - 3x - 1 < 0$$ Step 2: Factor quadratic. $$2x^2 - 3x - 1 = (2x + 1)(x - 1)$$ Step 3: Roots: $$2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$$ $$x - 1 = 0 \Rightarrow x = 1$$ Step 4: Intervals: $(-\infty,-\frac{1}{2})$, $[-\frac{1}{2},1]$, $(1,\infty)$. Step 5: Test points for $2x^2 - 3x - 1 < 0$: - $x=-1$: $(2(-1)+1)(-1-1) = (-2+1)(-2) = (-1)(-2) = 2 > 0$ (False) - $x=0$: $(2(0)+1)(0-1) = (1)(-1) = -1 < 0$ (True) - $x=2$: $(2(2)+1)(2-1) = (4+1)(1) = 5 > 0$ (False) Step 6: Solution interval for $2x^2 - 3x - 1 < 0$ is: $$\left(-\frac{1}{2}, 1\right)$$ Step 7: Since original inequality was $-2x^2 + 3x + 1 > 0$, solution is the same interval: $$\left(-\frac{1}{2}, 1\right)$$ Step 8: Set notation: $$\{x \mid -\frac{1}{2} < x < 1\}$$ --- 8. Solve $-x^2 - 2x - 3 \geq 0$. Step 1: Multiply entire inequality by $-1$ (reverse inequality): $$x^2 + 2x + 3 \leq 0$$ Step 2: Calculate discriminant: $$\Delta = 2^2 - 4(1)(3) = 4 - 12 = -8 < 0$$ Step 3: Since discriminant is negative and leading coefficient positive, quadratic is always positive. Step 4: So $x^2 + 2x + 3 > 0$ for all $x$, thus $x^2 + 2x + 3 \leq 0$ has no real solutions. Step 5: Therefore, original inequality $-x^2 - 2x - 3 \geq 0$ has no solution. Step 6: Solution set is empty: $$\emptyset$$