Question

If $ \alpha $ and $ \beta $ are the roots of the equation $ a{{x}^{2}}+bx+c=0, $ $ (c\ne 0), $ then the equation whose roots are $ \frac{1}{a\alpha +b} $ and $ \frac{1}{a\beta +b} $ is

• A. $ ac{{x}^{2}}-bx+1=0 $
• B. $ {{x}^{2}}-acx+bc+1=0 $
• C. $ ac{{x}^{2}}+bx-1=0 $
• D. $ {{x}^{2}}+acx-bc+11=0 $
• E. $ ac{{x}^{2}}-bx-11=0 $

Answer 1

Answer: (A)

$ \because $ $ \alpha ,\beta $ are roots of equation, $ a{{x}^{2}}+bx+c=0 $ $ (c\ne 0) $
$ \therefore $ $ \alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a} $
$ \therefore $ Required equation is $ {{x}^{2}}-\left( \frac{1}{a\alpha +b}+\frac{1}{\alpha \beta +b} \right)x+\left( \frac{1}{a\alpha +b}.\frac{1}{a\beta +b} \right) $
$ \Rightarrow $ $ {{x}^{2}}-\left( \frac{a(\alpha +\beta )+2b}{{{a}^{2}}\alpha \beta +ab(\alpha +\beta )+{{b}^{2}}} \right)x $ $ +\left( \frac{1}{{{a}^{2}}\alpha \beta +ab(\alpha +\beta )+{{b}^{2}}} \right)=0 $
$ \Rightarrow $ $ {{x}^{2}}-\left( \frac{a\left( -\frac{b}{a} \right)+2b}{{{a}^{2}}.\frac{c}{a}+ab\left( -\frac{b}{a} \right)+{{b}^{2}}} \right)x $ $ +\frac{1}{{{a}^{2}}.\frac{c}{a}+ab\left( -\frac{b}{a} \right)+{{b}^{2}}}=0 $
$ \Rightarrow $ $ {{x}^{2}}-\frac{b}{ac}x+\frac{1}{ac}=0 $
$ \Rightarrow $ $ ac{{x}^{2}}-bx+1=0 $