1. **State the problem:** We have three sets: - $\xi = \{x: x \text{ is an integer and } -4 \leq x \leq 2\}$ - $P = \{x: 3x - 1 < 2\}$ - $Q = \{x: 2x^2 + 7x + 3 = 0\}$ We need to find: (a) $P'$ (the complement of $P$ relative to $\xi$), (b) $P \cap Q$ (the intersection of $P$ and $Q$), (c) $P \cup Q$ (the union of $P$ and $Q$). 2. **Find the elements of $\xi$:** $$\xi = \{-4, -3, -2, -1, 0, 1, 2\}$$ 3. **Find the elements of $P$:** Solve the inequality: $$3x - 1 < 2$$ Add 1 to both sides: $$3x - 1 + 1 < 2 + 1$$ $$3x < 3$$ Divide both sides by 3: $$\cancel{3}x < \cancel{3}$$ $$x < 1$$ Since $x$ must be an integer in $\xi$, elements of $P$ are all integers less than 1 in $\xi$: $$P = \{-4, -3, -2, -1, 0\}$$ 4. **Find the elements of $Q$:** Solve the quadratic equation: $$2x^2 + 7x + 3 = 0$$ Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=7$, $c=3$. Calculate the discriminant: $$\Delta = 7^2 - 4 \times 2 \times 3 = 49 - 24 = 25$$ Calculate roots: $$x = \frac{-7 \pm \sqrt{25}}{2 \times 2} = \frac{-7 \pm 5}{4}$$ Two roots: $$x_1 = \frac{-7 + 5}{4} = \frac{-2}{4} = -\frac{1}{2}$$ $$x_2 = \frac{-7 - 5}{4} = \frac{-12}{4} = -3$$ Since $Q$ contains integers in $\xi$ satisfying the equation, only $x = -3$ is in $Q$ (because $-\frac{1}{2}$ is not an integer). So: $$Q = \{-3\}$$ 5. **Find (a) $P'$:** $P'$ is the complement of $P$ in $\xi$, i.e., elements in $\xi$ not in $P$: $$P' = \xi \setminus P = \{-4, -3, -2, -1, 0, 1, 2\} \setminus \{-4, -3, -2, -1, 0\} = \{1, 2\}$$ 6. **Find (b) $P \cap Q$:** Intersection of $P$ and $Q$: $$P \cap Q = \{-4, -3, -2, -1, 0\} \cap \{-3\} = \{-3\}$$ 7. **Find (c) $P \cup Q$:** Union of $P$ and $Q$: $$P \cup Q = \{-4, -3, -2, -1, 0\} \cup \{-3\} = \{-4, -3, -2, -1, 0\}$$ **Final answers:** - (a) $P' = \{1, 2\}$ - (b) $P \cap Q = \{-3\}$ - (c) $P \cup Q = \{-4, -3, -2, -1, 0\}$