A) \[b{{x}^{2}}+cx+a=0\]
B) \[b{{x}^{2}}+ax+c=0\]
C) \[c{{x}^{2}}+ax+b=0\]
D) \[c{{x}^{2}}+bx+a=0\]
Correct Answer: D
Solution :
Let roots of the equation\[a{{x}^{2}}+bx+c=0\]are \[\alpha \]and\[\beta \]. \[\therefore \] \[\alpha +\beta =-\frac{b}{a}\]and \[\alpha \beta =\frac{c}{a}\] Now, \[\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }\] \[=\frac{-b/a}{c/a}=\frac{-b}{c}\]and \[\frac{1}{\alpha }\times \frac{1}{\beta }=\frac{1}{c/a}=\frac{a}{c}\] \[\therefore \]Required equation is \[{{x}^{2}}-\left( -\frac{b}{c} \right)x+\frac{a}{c}=0\Rightarrow c{{x}^{2}}+bx+a=0\] Alternative method: To find the equation of reciprocal roots, interchange the coefficients of x2 and constant term in the given equation, then required equation is \[c{{x}^{2}}+bx+a=0\]
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