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Find the quadratic polynomial whose zeroes are $\frac{2}{3}$ and $\frac{-1}{4}$. Verify the relation between the coefficients and the zeroes of the polynomial.
Find the quadratic polynomial whose zeroes are $\frac{2}{3}$ and $\frac{-1}{4}$. Verify the relation between the coefficients and the zeroes of the polynomial.
CBSE | Class 10 | Excercise 2A | Maths | Polynomials | RS Aggarwal
Find the quadratic polynomial whose zeroes are $\frac{2}{3}$ and $\frac{-1}{4}$. Verify the relation between the coefficients and the zeroes of the polynomial.

Let $\alpha=\frac{2}{3}$ and $\beta=\frac{-1}{4}$.

Sum of the zeroes $=(\alpha+\beta)=\frac{2}{3}+\left(\frac{-1}{4}\right)=\frac{8-3}{12}=\frac{5}{12}$

Product of the zeroes, $\alpha \beta=\frac{2}{3} \times\left(\frac{-1}{4}\right)=\frac{-2}{12}=\frac{-1}{6}$

$\therefore$ Required polynomial $=\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta=\mathrm{x}^{2}-\frac{5}{12} \mathrm{x}+\left(\frac{-1}{6}\right)$ $=x^{2}-\frac{5}{12} x-\frac{1}{6}$

Sum of the zeroes $=\frac{5}{12}=\frac{-(\text { coefficient of } x)}{\left(\text { coef ficient of } x^{2}\right)}$

Product of zeroes $=\frac{-1}{6}=\frac{\text { constant term }}{\left(\text { coef ficient of } x^{2}\right)}$

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