1. **Problem:** Given the quadratic equation $$2x^2 - 5x + 2 = 0$$ with roots $$p$$ and $$q$$, show that $$2p + p^2 q + 2q - (pq)^2 = \frac{13}{2}$$. 2. **Formula and rules:** - Sum of roots: $$p + q = -\frac{b}{a} = \frac{5}{2}$$ (M1) - Product of roots: $$pq = \frac{c}{a} = 1$$ (M1) 3. **Calculate each term:** - $$2p + 2q = 2(p + q) = 2 \times \frac{5}{2} = 5$$ (A1) - $$p^2 q + q^2 p = pq(p + q) = 1 \times \frac{5}{2} = \frac{5}{2}$$ (M1) - Note that $$p^2 q + q^2 p = pq(p + q)$$ because $$p^2 q + q^2 p = pq(p + q)$$ (A1) - $$ (pq)^2 = 1^2 = 1$$ (A1) 4. **Sum all terms:** $$2p + p^2 q + 2q - (pq)^2 = (2p + 2q) + p^2 q - (pq)^2 = 5 + \frac{5}{2} - 1 = \frac{10}{2} + \frac{5}{2} - \frac{2}{2} = \frac{13}{2}$$ (A1) **Final answer:** $$2p + p^2 q + 2q - (pq)^2 = \frac{13}{2}$$ (A1) --- 1. **Problem:** Find an equation with integral coefficients whose roots are $$2p + q$$ and $$p + 2q$$. 2. **Use known sums and products:** - Sum of new roots: $$S = (2p + q) + (p + 2q) = 3(p + q) = 3 \times \frac{5}{2} = \frac{15}{2}$$ (M1) - Product of new roots: $$P = (2p + q)(p + 2q) = 2p^2 + 5pq + 2q^2$$ 3. **Express product in terms of $$p+q$$ and $$pq$$:** - Note $$p^2 + q^2 = (p + q)^2 - 2pq = \left(\frac{5}{2}\right)^2 - 2 \times 1 = \frac{25}{4} - 2 = \frac{17}{4}$$ - So, $$P = 2p^2 + 5pq + 2q^2 = 2(p^2 + q^2) + 5pq = 2 \times \frac{17}{4} + 5 \times 1 = \frac{34}{4} + 5 = \frac{34}{4} + \frac{20}{4} = \frac{54}{4} = \frac{27}{2}$$ (A1) 4. **Form quadratic equation:** $$x^2 - Sx + P = 0 \Rightarrow x^2 - \frac{15}{2}x + \frac{27}{2} = 0$$ 5. **Clear denominators:** Multiply through by 2: $$2x^2 - 15x + 27 = 0$$ (A1) **Final equation:** $$2x^2 - 15x + 27 = 0$$ (A1)