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1)

If the roots  of the equation

x1x+1xx=52 are p and q (p>q)  and the roots of the equation

(p+q)x4pqx2+pq=0   are α,β,γ,δ then   

(α)2αβ+αβγδ=0


A) 0

B) 10425

C) 254

D) 164

Answer:

Option B

Explanation:

 Let   x1x=y

  So,    y+1y=52

        2y25y+2=0

       y=2,12

 So,          x=45,15

    p>q

        p=45,q=15

 Now,  for the equation    (p+q)x4pqx2+pq=0

α=0,αβ=pqp+q   and    αβγδ=pq(p+q)

 So,  (α)2αβ+αβγδ=0+pqp+q+pq(p+q)

    =  p(p+q)(q+1q)

    =  4545+15(15+5)=45×265=10425