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Question

# Let a,b,c,d be distinct real numbers such that a,b are roots of x2âˆ’5cxâˆ’6d=0, and c,d are roots of x2âˆ’5axâˆ’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 144x2−5cx−6d=0......(1)x2−5ax−6b=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)d−b=6(a−c).....(4) Now a and c are the roots of equation (1) and (2) respectively. a2−5ac−6d=0......(5)c2−5ac−6b=0.......(6) (5)−(6)⇒a2−c2=6(d−b)⇒(a+c)(a−c)=6(d−b)⇒(b+d4)(d−b6)=6(d−b) (∵eq. (3) & (4))⇒b+d=144

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