Right mary, let's go. Here are the evaluations for question 2a: Given the quadratic equation $2x^2 - 4x + 7 = 0$. For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product of the roots is $\alpha\beta = \frac{c}{a}$. In this case, $a=2$, $b=-4$, $c=7$. Sum of roots: $$ \alpha + \beta = -\frac{-4}{2} = \frac{4}{2} = 2 $$ Product of roots: $$ \alpha\beta = \frac{7}{2} $$ i) Evaluate $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$: Step 1: Combine the fractions. $$ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} $$ Step 2: Express $\alpha^2 + \beta^2$ in terms of $\alpha + \beta$ and $\alpha\beta$. $$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta $$ Step 3: Substitute the expressions and values. $$ \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{(2)^2 - 2\left(\frac{7}{2}\right)}{\frac{7}{2}} $$ Step 4: Simplify the expression. $$ \frac{4 - 7}{\frac{7}{2}} = \frac{-3}{\frac{7}{2}} = -3 \times \frac{2}{7} = \boxed{-\frac{6}{7}} $$ ii) Evaluate $\frac{1}{\alpha} + \frac{1}{\beta}$: Step 1: Combine the fractions. $$ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} $$ Step 2: Substitute the values. $$ \frac{2}{\frac{7}{2}} = 2 \times \frac{2}{7} = \boxed{\frac{4}{7}} $$ iii) Evaluate $\alpha^2 + \beta^2$: Step 1: Use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. Step 2: Substitute the values. $$ (2)^2 - 2\left(\frac{7}{2}\right) = 4 - 7 = \boxed{-3} $$ iv) Evaluate $\alpha^3 + \beta^3$: Step 1: Use the identity $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)$. Step 2: Substitute $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. $$ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 2\alpha\beta - \alpha\beta) = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) $$ Step 3: Substitute the values. $$ (2)\left((2)^2 - 3\left(\frac{7}{2}\right)\right) = 2\left(4 - \frac{21}{2}\right) $$ Step 4: Simplify the expression. $$ 2\left(\frac{8}{2} - \frac{21}{2}\right) = 2\left(-\frac{13}{2}\right) = \boxed{-13} $$ v) Evaluate $\frac{1}{\alpha^2} + \frac{1}{\beta^2}$: Step 1: Combine the fractions. $$ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{(\alpha\beta)^2} $$ Step 2: Substitute the value of $\alpha^2 + \beta^2 = -3$ (from part iii) and $\alpha\beta = \frac{7}{2}$. $$ \frac{-3}{\left(\frac{7}{2}\right)^2} = \frac{-3}{\frac{49}{4}} $$ Step 3: Simplify the expression. $$ -3 \times \frac{4}{49} = \boxed{-\frac{12}{49}} $$ Send me the next one 📸