Question

A quadratic polynomial whose zeros are $\frac{3}{5}$ and $\frac{-1}{2}$, is
(a) 10x² + x + 3
(b) 10x² + x − 3
(c) 10x² − x + 3
(d) 10x² – x – 3

Answer 1

$$\begin{gathered} \left(\text{d}\right)10x^2–x–3 \\ \text{Here, the zeroes are}\frac{3}{5}\text{and}\frac{-1}{2}\text{.} \\ \mathrm{Let}\alpha =\frac{3}{5}and\beta =\frac{-1}{2} \\ \text{So, sum of the zeroes, α+β=}\frac{3}{5}+\left(\frac{-1}{2}\right)=\frac{1}{10} \\ \begin{array}{l}\text{Also, product of the zeroes, αβ=}\frac{3}{5}\times \left(\frac{-1}{2}\right)=\frac{-3}{10}\\ \mathrm{The}\mathrm{polynomial}\mathrm{will}\mathrm{be}{x}^2-(\alpha +\beta )x+\alpha \beta .\end{array} \\ \therefore \mathrm{The}\mathrm{r}\text{equired polynomial is}{x}^2-\frac{1}{10}x-\frac{3}{10}. \end{gathered}$$
Multiply by 10, we get
$10x^2-x-3$