Question

If alpha and beta are zeroes of quadratic polynomial 2x^2+5x+k, find value of k such that (alpha+beta)^2-alphaxbeta=24.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{polynomial}\mathrm{is}, \\ p\left(x\right)=2x^2+5x+k \\ \mathrm{Since}\alpha \&\beta \mathrm{are}\mathrm{zeroes}\mathrm{of}p\left(x\right),\mathrm{then}, \\ \alpha +\beta =\frac{-\mathrm{coefficient}\mathrm{of}x}{\mathrm{coefficient}\mathrm{of}{x}^2}=-\frac{5}{2} \\ \alpha \times \beta =\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{x}^2}=\frac{k}{2} \\ \mathrm{Now}, \\ {\left(\alpha +\beta \right)}^2-2\alpha \beta =24 \\ \Rightarrow {\left(-\frac{5}{2}\right)}^2-2\times \frac{k}{2}=24 \\ \Rightarrow \frac{25}{4}-k=24 \\ \Rightarrow \frac{25}{4}-\frac{24}{1}=k \\ \Rightarrow k=\frac{25-96}{4}=-\frac{71}{4} \\ \Rightarrow k=-\frac{71}{4} \end{gathered}$$