Question

if alpha beta are the zeros of quadrilateral p(x) =2x²-5x+7

find polynomial whose zeros are 2 alpha beta and 3 alpha +2 beta

Answer 1

Dear student
$$\begin{gathered} \mathrm{Since}\mathrm{\alpha }\mathrm{and}\mathrm{\beta }\mathrm{are}\mathrm{the}\mathrm{zeros}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{polynomial}. \\ \mathrm{Therefore},\mathrm{\alpha }+\mathrm{\beta }=\frac{5}{2}\mathrm{and}\mathrm{\alpha \beta }=\frac{7}{2} \\ \mathrm{Now},\mathrm{Let}\mathrm{S}\mathrm{and}\mathrm{P}\mathrm{be}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{zeros}\mathrm{and}\mathrm{product}\mathrm{of}\mathrm{zeros}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{polynomial}. \\ \mathrm{So},\mathrm{S}=\left(2\mathrm{\alpha }+3\mathrm{\beta }\right)+\left(3\mathrm{\alpha }+2\mathrm{\beta }\right) \\ =5\mathrm{\alpha }+5\mathrm{\beta } \\ =5(\mathrm{\alpha }+\mathrm{\beta }) \\ =5\times \frac{5}{2} \\ =\frac{25}{2} \\ \mathrm{and}\mathrm{P}=\left(2\mathrm{\alpha }+3\mathrm{\beta }\right)\left(3\mathrm{\alpha }+2\mathrm{\beta }\right) \\ =6{\mathrm{\alpha }}^2+4\mathrm{\alpha \beta }+9\mathrm{\alpha \beta }+6{\mathrm{\beta }}^2 \\ =6\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2\right)+13\mathrm{\alpha \beta } \\ =6\left[{\left(\mathrm{\alpha }+\mathrm{\beta }\right)}^2-2\mathrm{\alpha \beta }\right]+13\mathrm{\alpha \beta } \\ =6\left[{\left(\frac{5}{2}\right)}^2-2\left(\frac{7}{2}\right)\right]+13\left(\frac{7}{2}\right) \\ =6\left[\frac{25}{4}-7\right]+\frac{91}{2} \\ =6\left(\frac{25-28}{4}\right)+\frac{91}{2} \\ =6\left(\frac{-3}{4}\right)+\frac{91}{2} \\ =41 \\ \mathrm{So},\mathrm{the}\mathrm{required}\mathrm{polynomial}\mathrm{is}\mathrm{k}\left\{{\mathrm{x}}^2-\mathrm{Sx}+\mathrm{P}\right\}=\mathrm{k}\left[{\mathrm{x}}^2-\frac{25}{2}\mathrm{x}+41\right]\mathrm{where}\mathrm{k}\mathrm{is}\mathrm{non}-\mathrm{negative}\mathrm{teg}. \end{gathered}$$
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