Question

If α, β be the zero of the polynomial 2x² + 5x + k such that α² + β² + αβ = $\frac{21}{4}$, then k = ?
(a) 3
(b) −3
(c) −2
(d) 2

Answer 1

$$\begin{gathered} \left(\mathrm{d}\right)2 \\ {\text{Since α and β are the zeroes of 2x}}^2+5x+k,\mathrm{we}\mathrm{have}: \\ \mathrm{\alpha }+\mathrm{\beta }=\frac{-5}{2}\mathrm{and}\mathrm{\alpha \beta }=\frac{k}{2} \\ \mathrm{Also,}\mathrm{it}\mathrm{is}\mathrm{given\; that}{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+\mathrm{\alpha \beta }=\frac{21}{4}. \\ =>{(\mathrm{\alpha }+\mathrm{\beta })}^2-\mathrm{\alpha \beta }=\frac{21}{4} \\ =>{\left(\frac{-5}{2}\right)}^2-\frac{k}{2}=\frac{21}{4} \\ =>\frac{25}{4}-\frac{k}{2}=\frac{21}{4} \\ =>\frac{k}{2}=\frac{25}{4}-\frac{21}{4}=\frac{4}{4}=1 \\ =>k=2 \end{gathered}$$