Quadratic Inequalities E0B6Bb

1. **Problem 1:** Solve the quadratic inequality $$x^2 - 5x + 6 < 0$$ and express the solution in interval notation and set-builder notation. 2. **Problem 2:** Solve the quadratic inequality $$x^2 + 4x - 12 \geq 0$$ and express the solution in interval notation and set-builder notation. 3. **Problem 3:** Solve the quadratic inequality $$-x^2 + 3x + 4 > 0$$ and express the solution in interval notation and set-builder notation. --- ### Step-by-step solutions: **1. Solve $$x^2 - 5x + 6 < 0$$** 1. Factor the quadratic: $$x^2 - 5x + 6 = (x - 2)(x - 3)$$. 2. The inequality becomes $$(x - 2)(x - 3) < 0$$. 3. The product of two factors is less than zero when one factor is positive and the other is negative. 4. Find critical points: $$x = 2$$ and $$x = 3$$. 5. Test intervals: - For $$x < 2$$, choose $$x=1$$: $$(1-2)(1-3) = (-1)(-2) = 2 > 0$$ (false). - For $$2 < x < 3$$, choose $$x=2.5$$: $$(2.5-2)(2.5-3) = (0.5)(-0.5) = -0.25 < 0$$ (true). - For $$x > 3$$, choose $$x=4$$: $$(4-2)(4-3) = (2)(1) = 2 > 0$$ (false). 6. Solution set is $$2 < x < 3$$. 7. Express in interval notation: $$(2, 3)$$. 8. Express in set-builder notation: $$\{x \mid 2 < x < 3\}$$. --- **2. Solve $$x^2 + 4x - 12 \geq 0$$** 1. Factor the quadratic: $$x^2 + 4x - 12 = (x + 6)(x - 2)$$. 2. The inequality becomes $$(x + 6)(x - 2) \geq 0$$. 3. Critical points: $$x = -6$$ and $$x = 2$$. 4. Test intervals: - For $$x < -6$$, choose $$x = -7$$: $$(-7+6)(-7-2) = (-1)(-9) = 9 \geq 0$$ (true). - For $$-6 < x < 2$$, choose $$x=0$$: $$(0+6)(0-2) = (6)(-2) = -12 < 0$$ (false). - For $$x > 2$$, choose $$x=3$$: $$(3+6)(3-2) = (9)(1) = 9 \geq 0$$ (true). 5. Include points where product equals zero: $$x = -6$$ and $$x = 2$$. 6. Solution set is $$(-\infty, -6] \cup [2, \infty)$$. 7. Set-builder notation: $$\{x \mid x \leq -6 \text{ or } x \geq 2\}$$. --- **3. Solve $$-x^2 + 3x + 4 > 0$$** 1. Rewrite inequality: $$-x^2 + 3x + 4 > 0$$. 2. Multiply both sides by $$-1$$ (remember to reverse inequality): $$x^2 - 3x - 4 < 0$$. 3. Factor: $$x^2 - 3x - 4 = (x - 4)(x + 1)$$. 4. Inequality becomes $$(x - 4)(x + 1) < 0$$. 5. Critical points: $$x = -1$$ and $$x = 4$$. 6. Test intervals: - For $$x < -1$$, choose $$x = -2$$: $$(-2-4)(-2+1) = (-6)(-1) = 6 > 0$$ (false). - For $$-1 < x < 4$$, choose $$x=0$$: $$(0-4)(0+1) = (-4)(1) = -4 < 0$$ (true). - For $$x > 4$$, choose $$x=5$$: $$(5-4)(5+1) = (1)(6) = 6 > 0$$ (false). 7. Solution set is $$-1 < x < 4$$. 8. Interval notation: $$(-1, 4)$$. 9. Set-builder notation: $$\{x \mid -1 < x < 4\}$$. --- ### Final answers: - Problem 1: $$(2, 3)$$ or $$\{x \mid 2 < x < 3\}$$. - Problem 2: $$(-\infty, -6] \cup [2, \infty)$$ or $$\{x \mid x \leq -6 \text{ or } x \geq 2\}$$. - Problem 3: $$(-1, 4)$$ or $$\{x \mid -1 < x < 4\}$$.