A) \[{{\alpha }^{-1}},{{\beta }^{-1}}\]
B) \[{{\alpha }^{2}},{{\beta }^{2}}\]
C) \[\alpha {{\beta }^{-1}},{{\alpha }^{-1}}\beta \]
D) \[{{\alpha }^{-2}},{{\beta }^{-2}}\]
Correct Answer: D
Solution :
\[a{{x}^{2}}-bx+c=0\] \[\alpha +\beta =\frac{b}{a},\alpha \beta =\frac{c}{a}\] Also, \[{{(a+cy)}^{2}}={{b}^{2}}y\] \[\Rightarrow \] \[{{c}^{2}}{{y}^{2}}-({{b}^{2}}-2ac)y+{{a}^{2}}=0\] \[\Rightarrow \]\[{{\left( \frac{c}{a} \right)}^{2}}{{y}^{2}}-\left\{ {{\left( \frac{b}{a} \right)}^{2}}-\left( \frac{c}{a} \right) \right\}y+1=0\] \[\Rightarrow \]\[{{(\alpha \beta )}^{2}}{{y}^{2}}-({{\alpha }^{2}}+{{\beta }^{2}})y+1=0\] \[\Rightarrow \]\[{{y}^{2}}-({{\alpha }^{-2}}+{{\beta }^{-2}})y+{{\alpha }^{-2}}{{\beta }^{-2}}=0\] \[\Rightarrow \]\[(y-{{\alpha }^{-2}})(y-{{\beta }^{-2}})=0\] So, the roots are \[{{\alpha }^{-2}}\]and\[{{\beta }^{-2}}.\]
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