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A quadratic polynomial whose zeroes are $\frac{3}{5}$ and $\frac{-1}{2}$, is (a) $10 x^{2}+x+3$ (b) $10 x^{2}+x-3$ (c) $10 x^{2}-x+3$ (d) $x^{2}-\frac{1}{10} x-\frac{3}{10}$
A quadratic polynomial whose zeroes are $\frac{3}{5}$ and $\frac{-1}{2}$, is (a) $10 x^{2}+x+3$ (b) $10 x^{2}+x-3$ (c) $10 x^{2}-x+3$ (d) $x^{2}-\frac{1}{10} x-\frac{3}{10}$
CBSE | Class 10 | Excercise 2D | Maths | Polynomials | RS Aggarwal
A quadratic polynomial whose zeroes are $\frac{3}{5}$ and $\frac{-1}{2}$, is (a) $10 x^{2}+x+3$ (b) $10 x^{2}+x-3$ (c) $10 x^{2}-x+3$ (d) $x^{2}-\frac{1}{10} x-\frac{3}{10}$

The correct option is option (d) $x^{2}-\frac{1}{10} x-\frac{3}{10}$

Since, the zeroes are $\frac{3}{5}$ and $\frac{-1}{2}$

$\alpha=\frac{3}{5}$ and $\beta=\frac{-1}{2}$

sum of the zeroes, $\alpha+\beta=\frac{3}{5}+\left(\frac{-1}{2}\right)=\frac{1}{10}$

product of the zeroes, $\alpha \beta=\frac{3}{5} \times\left(\frac{-1}{2}\right)=\frac{-3}{10}$

The polynomial will be $\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta$.

$\therefore$ The required polynomial is $x^{2}-\frac{1}{10} x-\frac{3}{10}$.

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