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
Let $\alpha$ and $\beta$ are the zeros then
$\alpha +\beta=\frac{2}{3}+\Big(-\frac{1}{4}\Big)$
$=\frac{8-3}{12}=\frac{5}{12}$
$\alpha\beta=\frac{2}{3}\times\Big(-\frac{1}{4}\Big)$
$=-\frac{2}{12}=-\frac{1}{6}$
$\therefore$ quadratic polynomial whose zeros are $\alpha,\beta$ is:
$\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$=\text{x}^2-\Big(\frac{5}{12}\Big)\text{x}+\Big(-\frac{1}{6}\Big)$
$=\frac{1}{12}(\text{12x}^2-\text{5x}-2)$
Sum of zeros $=-\frac{\text{Coefficient of x}}{\text{Coefficients of }\text{x}^2}$
$=-\frac{-5}{12}=\frac{5}{12}$
Also sum of zeros $=\frac{2}{3}+\Big(-\frac{1}{4}\Big)$
$=\frac{5}{12}$
Product of zeros $=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
$=\frac{-2}{12}=\frac{-1}{6}$
Also product of zeros $=\frac{2}{3}\times\Big(-\frac{1}{4}\Big)$
$=\frac{-2}{12}=\frac{-1}{6}$
$\alpha +\beta=\frac{2}{3}+\Big(-\frac{1}{4}\Big)$
$=\frac{8-3}{12}=\frac{5}{12}$
$\alpha\beta=\frac{2}{3}\times\Big(-\frac{1}{4}\Big)$
$=-\frac{2}{12}=-\frac{1}{6}$
$\therefore$ quadratic polynomial whose zeros are $\alpha,\beta$ is:
$\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta$
$=\text{x}^2-\Big(\frac{5}{12}\Big)\text{x}+\Big(-\frac{1}{6}\Big)$
$=\frac{1}{12}(\text{12x}^2-\text{5x}-2)$
Sum of zeros $=-\frac{\text{Coefficient of x}}{\text{Coefficients of }\text{x}^2}$
$=-\frac{-5}{12}=\frac{5}{12}$
Also sum of zeros $=\frac{2}{3}+\Big(-\frac{1}{4}\Big)$
$=\frac{5}{12}$
Product of zeros $=\frac{\text{Constant term}}{\text{Coefficient of }\text{x}^2}$
$=\frac{-2}{12}=\frac{-1}{6}$
Also product of zeros $=\frac{2}{3}\times\Big(-\frac{1}{4}\Big)$
$=\frac{-2}{12}=\frac{-1}{6}$
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