1. Given the quadratic equation $x^2 - 6x + 5 = 0$, let its roots be $\alpha$ and $\beta$. From the equation, the sum of roots $\alpha + \beta = 6$ and the product $\alpha \beta = 5$. (a) (i) To find $\frac{1}{\alpha} + \frac{1}{\beta}$, use the identity: $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{6}{5}$$ (ii) To find $\frac{1}{\alpha \beta}$: $$\frac{1}{\alpha \beta} = \frac{1}{5}$$ (b) (i) To find $\alpha^2 + \beta^2$, use the identity: $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = 6^2 - 2 \times 5 = 36 - 10 = 26$$ (ii) To find $\alpha^2 \beta^2$: $$\alpha^2 \beta^2 = (\alpha \beta)^2 = 5^2 = 25$$ 2. For the equation $2x^2 - 4x + 1 = 0$, roots $\alpha, \beta$ satisfy: $$\alpha + \beta = \frac{4}{2} = 2, \quad \alpha \beta = \frac{1}{2}$$ (a) (i) To find $\alpha^3 + \beta^3$, use: $$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta) = 2^3 - 3 \times \frac{1}{2} \times 2 = 8 - 3 = 5$$ (ii) To find $\alpha^3 \beta^3$: $$\alpha^3 \beta^3 = (\alpha \beta)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$$ (b) (i) To find $(\alpha - \beta)^2$: $$ (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta = 2^2 - 4 \times \frac{1}{2} = 4 - 2 = 2$$ (ii) To find $\alpha - \beta$ where $\alpha > \beta$: $$\alpha - \beta = \sqrt{(\alpha - \beta)^2} = \sqrt{2}$$ 3. For $2x^2 - 8x + 1 = 0$, roots satisfy: $$\alpha + \beta = \frac{8}{2} = 4, \quad \alpha \beta = \frac{1}{2}$$ (a) (i) Find $(\alpha + 1) + (\beta + 1)$: $$ (\alpha + 1) + (\beta + 1) = (\alpha + \beta) + 2 = 4 + 2 = 6$$ (ii) Find $(\alpha + 1)(\beta + 1)$: $$ (\alpha + 1)(\beta + 1) = \alpha \beta + (\alpha + \beta) + 1 = \frac{1}{2} + 4 + 1 = 5.5$$ (b) (i) Find $\frac{\beta}{\alpha} + \frac{\alpha}{\beta}$: $$ \frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\alpha^2 + \beta^2}{\alpha \beta}$$ Calculate $\alpha^2 + \beta^2$: $$ (\alpha + \beta)^2 - 2\alpha \beta = 4^2 - 2 \times \frac{1}{2} = 16 - 1 = 15$$ So, $$ \frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{15}{\frac{1}{2}} = 30$$ (ii) Find $(\frac{\beta}{\alpha})(\frac{\alpha}{\beta})$: $$ = 1$$ 4. For $2x^2 - 4x - 3 = 0$, roots satisfy: $$\alpha + \beta = \frac{4}{2} = 2, \quad \alpha \beta = \frac{-3}{2} = -1.5$$ (a) (i) Find $(2\alpha + 1) + (2\beta + 1)$: $$ 2(\alpha + \beta) + 2 = 2 \times 2 + 2 = 6$$ (ii) Find $(2\alpha + 1)(2\beta + 1)$: $$ 4\alpha \beta + 2(\alpha + \beta) + 1 = 4 \times (-1.5) + 2 \times 2 + 1 = -6 + 4 + 1 = -1$$ (b) (i) Find $(\alpha - 2) + (\beta - 2)$: $$ (\alpha + \beta) - 4 = 2 - 4 = -2$$ (ii) Find $(\alpha - 2)(\beta - 2)$: $$ \alpha \beta - 2(\alpha + \beta) + 4 = -1.5 - 4 + 4 = -1.5$$ 5. For $3x^2 - 6x - 4 = 0$, roots satisfy: $$\alpha + \beta = \frac{6}{3} = 2, \quad \alpha \beta = \frac{-4}{3} = -\frac{4}{3}$$ (a) (i) Find $\alpha^2 \beta + \beta^2 \alpha$: $$ \alpha \beta (\alpha + \beta) = -\frac{4}{3} \times 2 = -\frac{8}{3}$$ (ii) Find $(\alpha^2 \beta)(\beta^2 \alpha) = (\alpha \beta)^3 = \left(-\frac{4}{3}\right)^3 = -\frac{64}{27}$$ (b) (i) Find $(\alpha + \frac{1}{\beta}) + (\beta + \frac{1}{\alpha})$: $$ (\alpha + \beta) + \left(\frac{1}{\beta} + \frac{1}{\alpha}\right) = 2 + \frac{\alpha + \beta}{\alpha \beta} = 2 + \frac{2}{-\frac{4}{3}} = 2 - \frac{3}{2} = \frac{1}{2}$$ (ii) This is the same as (i), so answer is also $\frac{1}{2}$. 6. Form quadratic equations with given roots: (a) Roots: 1, -2 Equation: $x^2 - (1 - 2)x + (1)(-2) = x^2 + x - 2 = 0$ (b) Roots: 2, 3 Equation: $x^2 - 5x + 6 = 0$ (c) Roots: -4, -3 Equation: $x^2 + 7x + 12 = 0$ (d) Roots: $-\frac{1}{3}, \frac{1}{2}$ Sum: $-\frac{1}{3} + \frac{1}{2} = \frac{-2 + 3}{6} = \frac{1}{6}$ Product: $-\frac{1}{3} \times \frac{1}{2} = -\frac{1}{6}$ Equation: $6x^2 - x - 1 = 0$ (e) Roots: $\frac{1}{3}, \frac{2}{5}$ Sum: $\frac{1}{3} + \frac{2}{5} = \frac{5 + 6}{15} = \frac{11}{15}$ Product: $\frac{1}{3} \times \frac{2}{5} = \frac{2}{15}$ Equation: $15x^2 - 11x + 2 = 0$ (f) Roots: $\frac{3}{4}, \frac{2}{3}$ Sum: $\frac{3}{4} + \frac{2}{3} = \frac{9 + 8}{12} = \frac{17}{12}$ Product: $\frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$ Equation: $12x^2 - 17x + 6 = 0$ 7. Form quadratic equations with given sum and product: (a) Sum = 4, Product = 3 Equation: $x^2 - 4x + 3 = 0$ (b) Sum = 2, Product = -8 Equation: $x^2 - 2x - 8 = 0$ (c) Sum = -7, Product = 10 Equation: $x^2 + 7x + 10 = 0$ (d) Sum = $-\frac{5}{2}$, Product = $-\frac{3}{2}$ Multiply by 2 to clear denominators: Equation: $2x^2 + 5x - 3 = 0$ (e) Sum = $-\frac{7}{6}$, Product = $\frac{1}{3}$ Multiply by 6: Equation: $6x^2 + 7x + 2 = 0$ (f) Sum = $\frac{11}{12}$, Product = $\frac{1}{6}$ Multiply by 12: Equation: $12x^2 - 11x + 2 = 0$ 8. For $5x^2 - 6x - 2 = 0$, roots $\alpha, \beta$ satisfy: $$\alpha + \beta = \frac{6}{5}, \quad \alpha \beta = \frac{-2}{5}$$ (a) Roots of new equation: $\alpha + 3, \beta + 3$ Sum: $$ (\alpha + 3) + (\beta + 3) = (\alpha + \beta) + 6 = \frac{6}{5} + 6 = \frac{36}{5}$$ Product: $$ (\alpha + 3)(\beta + 3) = \alpha \beta + 3(\alpha + \beta) + 9 = \frac{-2}{5} + 3 \times \frac{6}{5} + 9 = \frac{-2}{5} + \frac{18}{5} + 9 = \frac{16}{5} + 9 = \frac{61}{5}$$ Equation: $$ x^2 - \frac{36}{5}x + \frac{61}{5} = 0$$ Multiply by 5: $$ 5x^2 - 36x + 61 = 0$$ (b) Roots: $\alpha^2, \beta^2$ Sum: $$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = \left(\frac{6}{5}\right)^2 - 2 \times \left(-\frac{2}{5}\right) = \frac{36}{25} + \frac{4}{5} = \frac{36}{25} + \frac{20}{25} = \frac{56}{25}$$ Product: $$ (\alpha \beta)^2 = \left(-\frac{2}{5}\right)^2 = \frac{4}{25}$$ Equation: $$ x^2 - \frac{56}{25}x + \frac{4}{25} = 0$$ Multiply by 25: $$ 25x^2 - 56x + 4 = 0$$ 9. For $-3x^2 + x + 1 = 0$, roots satisfy: $$\alpha + \beta = -\frac{1}{3}, \quad \alpha \beta = -\frac{1}{3}$$ (a) Roots: $\frac{1}{\alpha}, \frac{1}{\beta}$ Sum: $$ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-\frac{1}{3}}{-\frac{1}{3}} = 1$$ Product: $$ \frac{1}{\alpha} \times \frac{1}{\beta} = \frac{1}{\alpha \beta} = -3$$ Equation: $$ x^2 - x - 3 = 0$$ (b) Roots: $\frac{1 + \alpha}{\beta}, \frac{1 + \beta}{\alpha}$ Sum: $$ \frac{1 + \alpha}{\beta} + \frac{1 + \beta}{\alpha} = \frac{\alpha + 1}{\beta} + \frac{\beta + 1}{\alpha} = \frac{\alpha + 1}{\beta} + \frac{\beta + 1}{\alpha}$$ Rewrite as: $$ \frac{\alpha + 1}{\beta} + \frac{\beta + 1}{\alpha} = \frac{\alpha^2 + \alpha}{\alpha \beta} + \frac{\beta^2 + \beta}{\alpha \beta} = \frac{\alpha^2 + \beta^2 + \alpha + \beta}{\alpha \beta}$$ Calculate numerator: $$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = \left(-\frac{1}{3}\right)^2 - 2 \times \left(-\frac{1}{3}\right) = \frac{1}{9} + \frac{2}{3} = \frac{1}{9} + \frac{6}{9} = \frac{7}{9}$$ Sum $\alpha + \beta = -\frac{1}{3} = -\frac{3}{9}$ So numerator: $$ \frac{7}{9} - \frac{3}{9} = \frac{4}{9}$$ Denominator: $$ \alpha \beta = -\frac{1}{3}$$ Sum: $$ \frac{4/9}{-1/3} = \frac{4}{9} \times \left(-3\right) = -\frac{12}{9} = -\frac{4}{3}$$ Product: $$ \left(\frac{1 + \alpha}{\beta}\right) \times \left(\frac{1 + \beta}{\alpha}\right) = \frac{(1 + \alpha)(1 + \beta)}{\alpha \beta}$$ Calculate numerator: $$ (1 + \alpha)(1 + \beta) = 1 + (\alpha + \beta) + \alpha \beta = 1 - \frac{1}{3} - \frac{1}{3} = 1 - \frac{2}{3} = \frac{1}{3}$$ Denominator: $$ \alpha \beta = -\frac{1}{3}$$ Product: $$ \frac{1/3}{-1/3} = -1$$ Equation with roots sum $-\frac{4}{3}$ and product $-1$: $$ x^2 + \frac{4}{3}x - 1 = 0$$ Multiply by 3: $$ 3x^2 + 4x - 3 = 0$$ 10. For $-2x^2 + 3x + 1 = 0$, roots $\alpha, \beta$ satisfy: $$\alpha + \beta = -\frac{3}{-2} = \frac{3}{2}, \quad \alpha \beta = \frac{1}{-2} = -\frac{1}{2}$$ (a) Roots: $\alpha^3 \beta, \alpha \beta^3$ Sum: $$ \alpha^3 \beta + \alpha \beta^3 = \alpha \beta (\alpha^2 + \beta^2)$$ Calculate $\alpha^2 + \beta^2$: $$ (\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}$$ Sum: $$ -\frac{1}{2} \times \frac{13}{4} = -\frac{13}{8}$$ Product: $$ (\alpha^3 \beta)(\alpha \beta^3) = (\alpha \beta)^4 = \left(-\frac{1}{2}\right)^4 = \frac{1}{16}$$ Equation: $$ x^2 + \frac{13}{8}x + \frac{1}{16} = 0$$ Multiply by 16: $$ 16x^2 + 26x + 1 = 0$$ (b) Roots: $\alpha^2 + \frac{1}{\beta}, \beta^2 + \frac{1}{\alpha}$ Sum: $$ (\alpha^2 + \beta^2) + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)$$ Calculate each term: $$ \alpha^2 + \beta^2 = \frac{13}{4}$$ $$ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{\frac{3}{2}}{-\frac{1}{2}} = -3$$ Sum: $$ \frac{13}{4} - 3 = \frac{13}{4} - \frac{12}{4} = \frac{1}{4}$$ Product: $$ \left(\alpha^2 + \frac{1}{\beta}\right) \left(\beta^2 + \frac{1}{\alpha}\right) = \alpha^2 \beta^2 + \alpha^2 \frac{1}{\alpha} + \beta^2 \frac{1}{\beta} + \frac{1}{\alpha \beta}$$ Simplify terms: $$ \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(-\frac{1}{2}\right)^2 = \frac{1}{4}$$ $$ \alpha^2 \frac{1}{\alpha} = \alpha$$ $$ \beta^2 \frac{1}{\beta} = \beta$$ $$ \frac{1}{\alpha \beta} = -2$$ Sum: $$ \frac{1}{4} + (\alpha + \beta) - 2 = \frac{1}{4} + \frac{3}{2} - 2 = \frac{1}{4} + \frac{3}{2} - 2 = \frac{1}{4} + \frac{3}{2} - \frac{4}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$$ Equation: $$ x^2 - \frac{1}{4}x - \frac{1}{4} = 0$$ Multiply by 4: $$ 4x^2 - x - 1 = 0$$