1. **Problem (a):** Given that $\frac{5}{2}$ and $\frac{3}{5}$ are solutions of the quadratic equation $ax^2 + bx + c = 0$ with $a,b,c \in \mathbb{Z}$, find $a$, $b$, and $c$. 2. Since the roots are $r_1 = \frac{5}{2}$ and $r_2 = \frac{3}{5}$, the quadratic can be written as: $$a(x - r_1)(x - r_2) = 0$$ 3. Expanding: $$a\left(x - \frac{5}{2}\right)\left(x - \frac{3}{5}\right) = 0$$ 4. Multiply the factors: $$a\left(x^2 - \left(\frac{5}{2} + \frac{3}{5}\right)x + \frac{5}{2} \times \frac{3}{5}\right) = 0$$ 5. Calculate the sum and product of roots: $$\frac{5}{2} + \frac{3}{5} = \frac{25}{10} + \frac{6}{10} = \frac{31}{10}$$ $$\frac{5}{2} \times \frac{3}{5} = \frac{15}{10} = \frac{3}{2}$$ 6. So the quadratic is: $$a\left(x^2 - \frac{31}{10}x + \frac{3}{2}\right) = 0$$ 7. To clear denominators and have integer coefficients, multiply inside by 10: $$a\left(10x^2 - 31x + 15\right) = 0$$ 8. Choose $a=1$ for simplicity (since $a$ can be any integer, but usually $a=1$ is standard): $$10x^2 - 31x + 15 = 0$$ 9. Therefore: $$a = 10, \quad b = -31, \quad c = 15$$ --- 10. **Problem (b)(i):** Write $g(x) = 2x^2 + 4x - 5$ in the form $g(x) = a(x + h)^2 + k$ where $a,h,k \in \mathbb{Z}$. 11. Factor out the coefficient of $x^2$ from the first two terms: $$g(x) = 2(x^2 + 2x) - 5$$ 12. Complete the square inside the parentheses: $$x^2 + 2x = (x + 1)^2 - 1^2 = (x + 1)^2 - 1$$ 13. Substitute back: $$g(x) = 2\left((x + 1)^2 - 1\right) - 5 = 2(x + 1)^2 - 2 - 5$$ 14. Simplify constants: $$g(x) = 2(x + 1)^2 - 7$$ 15. Thus: $$a = 2, \quad h = 1, \quad k = -7$$