A) \[{{x}^{2}}-x+10=0\]
B) \[{{x}^{2}}-x-10=0\]
C) \[{{x}^{2}}+x+10=0\]
D) \[{{x}^{2}}+x-10=0\]
Correct Answer: D
Solution :
Since \[\alpha ,\,\,\beta \] are roots of the equation\[5{{x}^{2}}-x-2=0\]. \[\therefore \] \[\alpha +\beta =\frac{1}{5}\]and\[\alpha \beta =\frac{-2}{5}\] Now, sum of roots of required equation \[=\frac{2}{\alpha }+\frac{2}{\beta }=\frac{2(\alpha +\beta )}{\alpha \beta }\] \[=\frac{2\left( \frac{1}{5} \right)}{\left( \frac{-2}{5} \right)}=-1\] and product of roots of required equation \[=\frac{2}{\alpha }\cdot \frac{2}{\beta }=\frac{4}{\alpha \beta }=\frac{4}{\frac{-2}{5}}=-10\] Hence, required equation is \[{{x}^{2}}-(-1)x+(-10)=0\] or \[{{x}^{2}}+x-10=0\]
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