A) \[{{l}^{4}}{{x}^{2}}-nl({{m}^{2}}-2nl)x+{{n}^{4}}=0\]
B) \[{{l}^{4}}{{x}^{2}}-nl({{m}^{2}}-2nl)x+{{n}^{4}}=0\]
C) \[{{l}^{4}}{{x}^{2}}+nl({{m}^{2}}-2nl)x-{{n}^{4}}=0\]
D) \[{{l}^{4}}{{x}^{2}}-nl({{m}^{2}}+2nl)x+{{n}^{4}}=0\]
Correct Answer: A
Solution :
Since, \[\alpha ,\,\,\beta \] are the roots of the equation \[l{{x}^{2}}+mx+n=0\] \[\therefore \] \[\alpha +\beta =-\frac{m}{l},\,\,\,\,\alpha \beta =\frac{n}{l}\] Now, \[{{\alpha }^{3}}\beta +\alpha {{\beta }^{3}}=\alpha \beta ({{\alpha }^{2}}+{{\beta }^{2}})\] \[=\alpha \beta [{{(\alpha +\beta )}^{2}}-2\alpha \beta ]\] \[=\frac{n}{l}\left[ {{\left( \frac{-m}{l} \right)}^{2}}-\frac{2n}{l} \right]\] \[=\frac{n}{l}\left[ \frac{{{m}^{2}}}{{{l}^{2}}}-\frac{2n}{l} \right]\] and \[{{\alpha }^{3}}\beta \cdot \alpha {{\beta }^{3}}={{(\alpha \beta )}^{4}}\] \[=\frac{{{n}^{4}}}{{{l}^{4}}}\] \[\therefore \]Required quadratic equation is \[{{x}^{2}}-({{\alpha }^{3}}\beta +\alpha {{\beta }^{3}})x+{{\alpha }^{3}}\beta \cdot \alpha {{\beta }^{3}}=0\] \[\Rightarrow \] \[{{x}^{2}}-\frac{n}{l}\left[ \frac{{{m}^{2}}}{{{l}^{2}}}-\frac{2n}{l} \right]x+\frac{{{n}^{4}}}{{{l}^{4}}}=0\] \[\Rightarrow \] \[{{l}^{4}}{{x}^{2}}-nl({{m}^{2}}-2nl)x+{{n}^{4}}=0\]
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