Complex Numbers And Quadratic Equations question 695
Question: If $ \alpha ,\beta $ be the roots of the equation $ x^{2}-px+q=0 $ and $ {\alpha_1},{\beta_1} $ the roots of the equation $ x^{2}-qx+p=0, $ then the equation whose roots are $ \frac{1}{{\alpha_1}\beta }+\frac{1}{\alpha {\beta_1}} $ and $ \frac{1}{\alpha {\alpha_1}}+\frac{1}{\beta {\beta_1}} $ is
Options:
A) $ pqx^{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}-4p^{2}q^{2}=0 $
D) $ (p+q)x^{2}-(p+q)x+p^{2}+q^{2}+pq=0 $
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Answer:
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 $ =( \frac{1}{{\alpha_1}\beta }+\frac{1}{\alpha {\beta_1}} )+( \frac{1}{\alpha {\alpha_1}}+\frac{1}{\beta {\beta_1}} ) $ $ =\frac{\alpha {\beta_1}+{\alpha_1}\beta +\beta {\beta_1}+\alpha {\alpha_1}}{\alpha \beta {\alpha_1}{\beta_1}} $ Product of given roots $ =( \frac{1}{{\alpha_1}\beta }+\frac{1}{\alpha \beta 1} )( \frac{1}{\alpha {\alpha_1}}+\frac{1}{\beta {\beta_1}} ) $ $ =-\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 [ {{({\alpha_1}+{\beta_1})}^{2}}-2{\alpha_1}{\beta_1} ]+{\alpha_1}{\beta_1}[ {{(\alpha +\beta )}^{2}}-2\alpha \beta ]}{{{(\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 $ ( p^{2}x^{2} )x^{2}-( p^{2}q^{2} )x+p^{3} +q^{3}-4pq = 0 $
BETA
