A) \[pq{{x}^{2}}-pqx+{{p}^{2}}+{{q}^{2}}+4pq=0\]
B) \[{{p}^{2}}{{q}^{2}}{{x}^{2}}-{{p}^{2}}{{q}^{2}}x+{{p}^{3}}+{{q}^{3}}-4pq=0\]
C) \[{{p}^{3}}{{q}^{3}}{{x}^{2}}-{{p}^{3}}{{q}^{3}}x+{{p}^{4}}+{{q}^{4}}-4{{p}^{2}}{{q}^{2}}=0\]
D) \[(p+q){{x}^{2}}-(p+q)x+{{p}^{2}}+{{q}^{2}}+pq=0\]
Correct Answer: B
Solution :
Here, \[\alpha +\beta =p, \,\alpha \beta =q\] \[{{\alpha }_{1}}+{{\beta }_{1}} =\,\,p,\,\,{{\alpha }_{1}}{{\beta }_{1}}=p\] Sum of given roots \[=\,\,\,\left( \frac{1}{{{\alpha }_{1}}\beta }+\frac{1}{\alpha {{\beta }_{1}}} \right)+\left( \frac{1}{\alpha {{\alpha }_{1}}}+\frac{1}{\beta {{\beta }_{1}}} \right)\] \[=\,\,\,\frac{\alpha {{\beta }_{1}}+{{\alpha }_{1}}\beta +\beta {{\beta }_{1}}+\alpha {{\alpha }_{1}}}{\alpha \beta {{\alpha }_{1}}{{\beta }_{1}}}\] Product of given roots \[=\,\,\left( \frac{1}{{{\alpha }_{1}}\beta }+\frac{1}{\alpha \beta 1} \right)\left( \frac{1}{\alpha {{\alpha }_{1}}}+\frac{1}{\beta {{\beta }_{1}}} \right)\] \[=\,\,-\frac{(\alpha {{\beta }_{1}}+{{\alpha }_{1}}\beta )(\alpha {{\alpha }_{1}}+\beta {{\beta }_{1}})}{{{\alpha }^{2}}{{\beta }^{2}}\alpha _{1}^{2}\beta _{1}^{2}}\] \[=\,\,\,\frac{\alpha \beta (\alpha _{1}^{2}+\beta _{1}^{2})+{{\alpha }_{1}}{{\beta }_{1}}({{\alpha }^{2}}+{{\beta }^{2}})}{{{\alpha }^{2}}{{\beta }^{2}}\alpha _{1}^{2}\beta _{1}^{2}}\] \[=\,\,\,\frac{\alpha \beta \left[ {{({{\alpha }_{1}}+{{\beta }_{1}})}^{2}}-2{{\alpha }_{1}}{{\beta }_{1}} \right]+{{\alpha }_{1}}{{\beta }_{1}}\left[ {{(\alpha +\beta )}^{2}}-2\alpha \beta \right]}{{{(\alpha \beta )}^{2}}{{({{\alpha }_{1}}{{\beta }_{1}})}^{2}}}\] \[=\,\,\,\frac{q({{q}^{2}}-2p)+p({{p}^{2}}-2q)}{{{q}^{2}}{{p}^{2}}}=\frac{{{p}^{3}}+{{q}^{3}}-4pq}{{{p}^{2}}{{q}^{2}}}\] Hence, the required equation is \[\left( {{p}^{2}}{{x}^{2}} \right){{x}^{2}}-\,\left( {{p}^{2}}{{q}^{2}} \right)x+{{p}^{3}} +{{q}^{3}}-4pq = 0\]
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