Question

α and β are zeroes of the quadratic polynomial 3x2

4x 5, find the values of :(i) α-1+ β-1(ii) α2+ β2(iii) α3 + β3(iv) α2β + αβ2

Answer 1

Taking the quadratic polynomial as $3x^2+4x+5$;
Since $\alpha \mathrm{and}\beta$ are the zeroes of the given quadratic polynomial $3x^2+4x+5$, so we have;
Sum of the roots = $\alpha +\beta =\frac{-4}{3}...\left(\mathrm{i}\right)$
And product of the roots = $\alpha \beta =\frac{5}{3}...\left(\mathrm{ii}\right)$
$$\begin{gathered} 1)\alpha -1+\beta -1 \\ =\alpha +\beta -2 \\ =\frac{-4}{3}-2\{\mathrm{using}\left(\mathrm{i}\right)\} \\ =\frac{-10}{3} \\ 2){\alpha }^2+{\beta }^2 \\ ={\left(\alpha +\beta \right)}^2-2\alpha \beta \\ ={\left(\frac{-4}{3}\right)}^2-2\times \frac{5}{3}\{\mathrm{using}(\mathrm{i})\mathrm{and}(\mathrm{ii}\left)\right\} \\ =\frac{16}{9}-\frac{10}{3} \\ =\frac{16-30}{9} \\ =\frac{-14}{9} \\ 3){\alpha }^3+{\beta }^3 \\ ={\left(\alpha +\beta \right)}^3-3\alpha \beta \left(\alpha +\beta \right) \\ ={\left(\frac{-4}{3}\right)}^3-3\times \frac{5}{3}\times \left(\frac{-4}{3}\right)\{\mathrm{using}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\} \\ =\frac{-64}{27}+\frac{20}{3} \\ =\frac{-64+180}{27} \\ =\frac{116}{27} \\ 4){\alpha }^2\beta +\alpha {\beta }^2 \\ =\alpha \beta \left(\alpha +\beta \right) \\ =\frac{5}{3}\times \left(\frac{-4}{3}\right)\{\mathrm{using}(\mathrm{i})\mathrm{and}(\mathrm{ii}\left)\right\} \\ =\frac{-20}{9} \end{gathered}$$