1. **Problem Statement:** Given the quadratic equation $$2x^2 - 3x - 1 = 0$$ with roots \(\alpha\) and \(\beta\), find new equations with integer coefficients whose roots are: (a) (i) $$\frac{1}{\alpha}$$ and $$\frac{1}{\beta}$$ (a) (ii) $$\frac{\alpha^2}{\beta}$$ and $$\frac{\beta^2}{\alpha}$$ (b) (i) $$\alpha^3$$ and $$\beta^3$$ (b) (ii) $$3\lambda \alpha$$ and $$3\lambda \beta$$ (where \(\lambda\) is a constant) (b) (iii) $$4\alpha + 3\beta$$ and $$4\beta + 3\alpha$$ 2. **Find sum and product of roots for the original equation:** For $$2x^2 - 3x - 1 = 0$$, sum of roots, $$\alpha + \beta = -\frac{b}{a} = \frac{3}{2}$$ product of roots, $$\alpha \beta = \frac{c}{a} = -\frac{1}{2}$$ --- **(a)(i) Roots: $$\frac{1}{\alpha}$$ and $$\frac{1}{\beta}$$** 3. Sum of new roots: $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{\frac{3}{2}}{-\frac{1}{2}} = -3$$ 4. Product of new roots: $$\frac{1}{\alpha} \times \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{-\frac{1}{2}} = -2$$ 5. Equation with these roots: $$x^2 - (\text{sum})x + (\text{product}) = 0$$ $$x^2 - (-3)x + (-2) = x^2 + 3x - 2 = 0$$ --- **(a)(ii) Roots: $$\frac{\alpha^2}{\beta}$$ and $$\frac{\beta^2}{\alpha}$$** 6. Express these roots as $$\alpha \times \frac{\alpha}{\beta}$$ and $$\beta \times \frac{\beta}{\alpha}$$. Sum: $$\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \alpha \cdot \frac{\alpha}{\beta} + \beta \cdot \frac{\beta}{\alpha} = \frac{\alpha^2 \beta + \beta^2 \alpha}{\alpha \beta} = \frac{\alpha \beta (\alpha + \beta)}{\alpha \beta} = \alpha + \beta$$ Check carefully: Actually, $$\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha \beta}$$ Use formula for sum of cubes: $$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3 \alpha \beta (\alpha + \beta)$$ Calculate it: $$= \left(\frac{3}{2}\right)^3 - 3 \times \left(-\frac{1}{2}\right) \times \frac{3}{2} = \frac{27}{8} + \frac{9}{4} = \frac{27}{8} + \frac{18}{8} = \frac{45}{8}$$ Sum of roots: $$\frac{45/8}{-1/2} = \frac{45}{8} \times \left(-2\right) = -\frac{90}{8} = -\frac{45}{4}$$ 7. Product: $$\frac{\alpha^2}{\beta} \times \frac{\beta^2}{\alpha} = \frac{\alpha^2 \beta^2}{\alpha \beta} = \alpha \beta = -\frac{1}{2}$$ 8. Equation: $$x^2 - (\text{sum}) x + (\text{product}) = 0$$ $$x^2 + \frac{45}{4} x - \frac{1}{2} = 0$$ Multiply through by 4 to clear denominators: $$4x^2 + 45x - 2 = 0$$ --- **(b)(i) Roots: $$\alpha^3$$ and $$\beta^3$$** 9. Sum: $$\alpha^3 + \beta^3 = \frac{45}{8}$$ (from above) 10. Product: $$\alpha^3 \beta^3 = (\alpha \beta)^3 = \left(-\frac{1}{2}\right)^3 = -\frac{1}{8}$$ 11. Equation: $$x^2 - \frac{45}{8}x - \frac{1}{8} = 0$$ Multiply through by 8: $$8x^2 - 45x -1 = 0$$ --- **(b)(ii) Roots: $$3\lambda \alpha$$ and $$3\lambda \beta$$** 12. Sum: $$3\lambda (\alpha + \beta) = 3\lambda \times \frac{3}{2} = \frac{9}{2} \lambda$$ 13. Product: $$ (3\lambda)^2 \alpha \beta = 9 \lambda^2 \times \left(-\frac{1}{2}\right) = -\frac{9}{2} \lambda^2 $$ 14. Equation: $$x^2 - \frac{9}{2} \lambda x - \frac{9}{2} \lambda^2 = 0$$ Multiply by 2: $$2x^2 - 9 \lambda x - 9 \lambda^2 = 0$$ --- **(b)(iii) Roots: $$4\alpha + 3\beta$$ and $$4\beta + 3\alpha$$** 15. Sum: $$ (4\alpha + 3\beta) + (4\beta + 3\alpha) = 7(\alpha + \beta) = 7 \times \frac{3}{2} = \frac{21}{2} $$ 16. Product: Calculate carefully: $$ (4\alpha + 3\beta)(4\beta + 3\alpha) = 16 \alpha \beta + 12 \alpha^2 + 12 \beta^2 + 9 \alpha \beta = 25 \alpha \beta + 12 (\alpha^2 + \beta^2) $$ Recall: $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta = \left(\frac{3}{2}\right)^2 - 2 \times \left(-\frac{1}{2}\right) = \frac{9}{4} + 1 = \frac{13}{4}$$ So, $$ \text{Product} = 25 \times \left(-\frac{1}{2}\right) + 12 \times \frac{13}{4} = -\frac{25}{2} + 39 = \frac{-25 + 78}{2} = \frac{53}{2} $$ 17. Equation: $$x^2 - \frac{21}{2} x + \frac{53}{2} = 0$$ Multiply by 2: $$2x^2 - 21x + 53 = 0$$ **Final answers:** (a)(i): $$x^2 + 3x - 2 = 0$$ (a)(ii): $$4x^2 + 45x - 2 = 0$$ (b)(i): $$8x^2 - 45x - 1 = 0$$ (b)(ii): $$2x^2 - 9\lambda x - 9\lambda^2 = 0$$ (b)(iii): $$2x^2 - 21x + 53 = 0$$