Question

If α and β are the zeroes of the polynomial kx² + 3x + 2 such that ${\alpha }^2+{\beta }^2+\alpha \beta =-\frac{1}{25}$, find the value of k.
[Hint: ${\alpha }^2+{\beta }^2+\alpha \beta ={\left(\alpha +\beta \right)}^2-\alpha \beta$]

Answer 1

$$\begin{gathered} Hi, \\ k{x}^2+3x+2=0 \\ sosumofroots\alpha +\beta =\frac{-coefficientofx}{coefficientof{x}^2}=\frac{-3}{k} \\ productofroots\alpha \beta =\frac{\cos n\tan tterm}{coefficientof{x}^2}=\frac{2}{k} \\ sohere \\ {\alpha }^2+{\beta }^2+\alpha \beta =\frac{-1}{25} \\ \left({\alpha }^2+{\beta }^2+2\alpha \beta \right)-\alpha \beta =\frac{-1}{25} \\ {\left(\alpha +\beta \right)}^2-\alpha \beta =\frac{-1}{25} \\ puttingvalues \\ {\left(\frac{-3}{k}\right)}^2-\frac{2}{k}=\frac{-1}{25} \\ \frac{9}{k^2}-\frac{2}{k}=\frac{-1}{25} \\ \frac{9-2k}{k^2}=-\frac{1}{25} \\ 225-50k=-k^2 \\ {k}^2-50k+225=0 \\ {k}^2-5k-45k+225=0 \\ k\left(k-5\right)-45\left(k-5\right)=0 \\ \left(k-5\right)\left(k-45\right)=0 \\ k=5,45 \end{gathered}$$