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Without actually calculating the zeroes, form a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial 

5x^{2}+2x-3

 

 
 
 
 
 

Answers (2)

\\ $ Let $ \alpha $ and $ \beta $ be zeroes of the given quadratic $ \\ $Sum of the roots = $ \frac{-b}{a} \\ \alpha+\beta=\frac{-2}{5} \\ $Product of the roots = $ \frac{c}{a} \\ \alpha \beta=\frac{-3}{5} \\ $ Now $ \\\\ \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta} \\ =\frac{\frac{-2}{5}}{\frac{-3}{5}}\\\\=\frac{2}{3} \\ \frac{1}{\alpha \beta}=\frac{-5}{3} \\ $ Required Polynomial is $ \\ x^{2}-($Sum of the roots $)x+ ($product of the roots $) = 0 \\ x^{2}-\frac{2}{3}x+ (\frac{-5}{3}) = 0 \\ 3x^2 - 2x - 5 = 0

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Safeer PP

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Yogesh Kumar

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