Question

Find the quadratic polynomial whose zeros are $\frac{2}{3}$ and $\frac{-1}{4}$. Verify the relation between the coefficients and the zeros of the polynomial.

Answer 1

$\begin{array}{l}\text{Let}\alpha \text{=}\frac{2}{3}\text{and}\beta \text{=}\frac{-1}{4}.\\ \text{Sum of the zeroes = (}\alpha +\beta )=\frac{2}{3}+\left(\frac{-1}{4}\right)=\frac{8-3}{12}=\frac{5}{12}\\ \text{Product of the zeroes =}\alpha \beta =\frac{2}{3}\times \left(\frac{-1}{4}\right)=\frac{-{\sout{2}}^1}{\sout{1}{\sout{2}}^6}=\frac{-1}{6}\\ \\ \text{∴ Required polynomial =}{x}^2-\text{(}\alpha +\beta )x+\alpha \beta =x^2-\frac{5}{12}x+\left(\frac{-1}{6}\right)\\ \text{=}{x}^2-\frac{5}{12}x-\frac{1}{6}\\ \begin{array}{l}\text{Sum of the zeroes =}\frac{5}{12}=\frac{-\left(\text{coefficient of}x\right)}{\left(\text{coefficient of}{x}^2\right)}\\ \text{Product of the zeroes =}\frac{-1}{6}=\frac{\text{constant term}}{\text{coefficient of}{x}^2}\end{array}\\ \\ \end{array}$