Discussion Forum

1)

If the roots  of the equation

$\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{5}{2}$ are p and q (p>q)  and the roots of the equation

$(p+q)x^{4}-pqx^{2}+\frac{p}{q}=0$   are $\alpha,\beta , \gamma ,\delta$ then   

$(\sum \alpha)^{2}-\sum \alpha\beta+\alpha\beta\gamma\delta=0$

Answer:

Option B

Explanation:

 Let   $\sqrt{\frac{x}{1-x}}=y$

  So,    $y+\frac{1}{y}=\frac{5}{2}$

  $\Rightarrow$      $2y^{2}-5y+2=0$

$\Rightarrow$       $y=2, \frac{1}{2}$

 So,          $x= \frac{4}{5}, \frac{1}{5}$

 $\because$   p>q

 $\therefore$       $p=\frac{4}{5}, q=\frac{1}{5}$

 Now,  for the equation    $(p+q)x^{4}-pqx^{2}+\frac{p}{q}=0$

$\sum \alpha=0,\sum\alpha\beta=-\frac{pq}{p+q} $   and    $ \alpha\beta\gamma\delta=\frac{p}{q(p+q)}$

 So,  $(\sum \alpha)^{2}-\sum \alpha\beta+\alpha\beta\gamma\delta=0+\frac{pq}{p+q}+\frac{p}{q(p+q)}$

    =  $\frac{p}{(p+q)}\left(q+\frac{1}{q}\right)$

    =  $\frac{\frac{4}{5}}{\frac{4}{5}+\frac{1}{5}}\left(\frac{1}{5}+5\right)=\frac{4}{5}\times\frac{26}{5}=\frac{104}{25}$