Question

If α, β are the zeros of the polynomial x² + 6x + 2, then $\left(\frac{1}{\mathrm{\alpha }}+\frac{1}{\mathrm{\beta }}\right)=?$
(a) 3
(b) −3
(c) 12
(d) −12

Answer 1

$$\begin{gathered} \left(\text{b}\right)\text{-3} \\ \text{Since}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{are the zeroes of}{x}^2+6x+2\text{,we have:} \\ \mathrm{\alpha }+\mathrm{\beta }=-6\text{and}\mathrm{\alpha \beta }=2 \\ \therefore (\frac{1}{\alpha }+\frac{1}{\beta })=\left(\frac{\alpha +\beta }{\alpha \beta }\right)=\frac{-6}{2}=-3 \end{gathered}$$