Question

Let a, b,c,d be distinct real numbers such that a, b are roots of x2 – 5cx – 6d = 0, and c, d are
roots of x2 - 5ax - 6b = 0. Then b + d is
(A) 180
(B) 162
(C) 144
(D) 126​

Answer 1

Here, a and b are roots of the equation x2−5cx−6d=0

∴  a+b=1−(−5)c=5c               ----- ( 1 )

And

⇒  a.b=1−6d=−6d               ----- ( 2 )

Now, c and d are roots of the equation x2−5ax−6d=0

∴  c+d=1−(−5a)=5a            ----- ( 3 )

⇒  cd=1−6b=−6b         ----- ( 4 )

Adding ( 1 ) and ( 2 ),

⇒  a+b+c+d=5c+5a

⇒  b+d=4c+4a

⇒  b+d=4(c+a)                   ---- ( 5 )

Now dividing ( 1 ) by ( 2 ) we get,

⇒  c+da+b=5a5c

⇒  5a2+5ab=5c2+5cd

⇒  5a2+5(−6d)=5c2+5(−6b)                   [ From ( 2 ) and ( 4 ) ]

⇒  5a2−30d=5c2−30b

⇒  5(a2−c2)=−30b+30d

⇒  (a+c)(a−c)=−6b+6d

⇒  (a+c)(a−c)=−6(5c−a)+6(5a−c)                 [ From ( 1 ) and ( 2 ) ]

⇒  (a+c)(a−c)=−30c+6a+30a−6c

⇒  (a+c)(a−c)=−36c+36a

⇒  (a+c)(a−c)=36(a−c)

⇒  a+c=36                

Substituting value of a+c in equation ( 5 ) we get,

⇒  b+d=4(36)

∴  b+d=144