Question

If alpha and beta are zeroes of polynomial 3x² + 2x - 6 find the value of alpha-beta, alpha²+ beta², alpha³+ beta³, and 1/alpha + 1/beta.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{polynomial}\mathrm{is}: \\ 3x^2+2x-6 \\ \mathrm{Comparing}\mathrm{this}\mathrm{polynomial}\mathrm{with}\mathrm{standard}\mathrm{polynomial}a{x}^2+bx+c,\mathrm{we}\mathrm{get}, \\ a=3,b=2andc=-6 \\ \mathrm{Since},\alpha \&\beta \mathrm{are}\mathrm{the}\mathrm{zeroes}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}3x^2+2x-6,\mathrm{then}, \\ \alpha +\beta =-\frac{b}{a}=-\frac{2}{3} \\ \mathrm{And}\alpha \beta =\frac{c}{a}=-\frac{6}{3}=-2 \\ \left(\mathbf{1}\right) \\ \mathrm{We}\mathrm{know}\mathrm{tha}t{\left(\alpha -\beta \right)}^2={\left(\alpha +\beta \right)}^2-4\alpha \beta \\ \Rightarrow {\left(\alpha -\beta \right)}^2={\left(-\frac{2}{3}\right)}^2-4\times \left(-2\right) \\ \Rightarrow {\left(\alpha -\beta \right)}^2=\frac{4}{9}+8 \\ \Rightarrow {\left(\alpha -\beta \right)}^2=\frac{4+72}{9}=\frac{76}{9} \\ \mathrm{Therefore},\alpha -\beta =\frac{\sqrt{76}}{3}=\frac{\sqrt{4\times 19}}{3}=\frac{2\sqrt{19}}{3} \end{gathered}$$

$$\begin{gathered} \left(\mathbf{2}\right) \\ \mathrm{We}\mathrm{know}\mathrm{that} \\ {\left(\alpha +\beta \right)}^2={\alpha }^2+{\beta }^2+2\alpha \beta \\ \Rightarrow {\left(-\frac{2}{3}\right)}^2={\alpha }^2+{\beta }^2+2\left(-2\right) \\ \Rightarrow \frac{4}{9}={\alpha }^2+{\beta }^2-4 \\ \Rightarrow \frac{4}{9}+4={\alpha }^2+{\beta }^2 \\ \Rightarrow \frac{4+36}{9}={\alpha }^2+{\beta }^2 \\ \Rightarrow {\alpha }^2+{\beta }^2=\frac{40}{9} \end{gathered}$$

$$\begin{gathered} \left(\mathbf{3}\right) \\ \mathrm{We}\mathrm{know}\mathrm{that} \\ {\left(\alpha +\beta \right)}^3={\alpha }^3+{\beta }^3+3\alpha \beta \left(\alpha +\beta \right) \\ \Rightarrow {\left(-\frac{2}{3}\right)}^3={\alpha }^3+{\beta }^3+3\left(-2\right)\left(-\frac{2}{3}\right) \\ \Rightarrow -\frac{8}{27}={\alpha }^3+{\beta }^3+4 \\ \Rightarrow -\frac{8}{27}-4={\alpha }^3+{\beta }^3 \\ \Rightarrow \frac{-8-108}{27}={\alpha }^3+{\beta }^3 \\ \Rightarrow {\alpha }^3+{\beta }^3=-\frac{116}{27} \end{gathered}$$

$$\begin{gathered} \left(\mathbf{4}\right) \\ \frac{1}{\alpha }+\frac{1}{\beta } \\ =\frac{\beta +\alpha }{\alpha \beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-{\displaystyle \frac{2}{3}}}{-{\displaystyle \frac{2}{1}}}=-\frac{2}{3}\times \left(-\frac{1}{2}\right)=\frac{1}{3} \end{gathered}$$