1. Let's start with problem (a): Find the set $A_1 = \{x \in \mathbb{Z} : 2x^2 - 3 = x\}$. 2. Rearrange the equation: $$2x^2 - 3 = x \implies 2x^2 - x - 3 = 0$$ 3. Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=2$, $b=-1$, $c=-3$: $$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm \sqrt{25}}{4}$$ 4. Simplify the roots: $$x = \frac{1 \pm 5}{4}$$ 5. Calculate each root: - $x = \frac{1 + 5}{4} = \frac{6}{4} = \frac{3}{2}$ (not an integer) - $x = \frac{1 - 5}{4} = \frac{-4}{4} = -1$ (integer) 6. Since $x$ must be an integer, the only solution is $x = -1$. 7. Therefore, $$A_1 = \{-1\}$$ This is the interval notation for a single integer: $$[-1, -1]$$ --- For the other parts (b), (c), and (d), the total number of distinct problems is 4, but per instructions, only the first problem is solved here.