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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
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B
162
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C
144
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D
126
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Solution

The correct option is C 144
x25cx6d=0......(1)x25ax6b=0......(2)

Given that a,b are the roots of the equation (1) and c,d are the roots of the equation (2)

a+b=5c, ab=6d& c+d=5a, cd=6b

Solving these equation
b+d=4(a+c).....(3)db=6(ac).....(4)

Now a and c are the roots of equation (1) and (2) respectively.
a25ac6d=0......(5)c25ac6b=0.......(6)

(5)(6)a2c2=6(db)(a+c)(ac)=6(db)(b+d4)(db6)=6(db) (eq. (3) & (4))b+d=144

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