let  a, b be the roots of the equation  x^2-10cx-11d=0 and c,d be the roots of x^2 -10ax-11b=0, Find the value of a+b+c+d.

110 Points
13 years ago
As a + b = 10c and c + d = 10a
ab = -11d , cd = -11b
ac = 121 and (b + d) = 9(a + c)
a2 - 10ac - 11d = 0
c2 - 10ac - 11b = 0
a2 + c2 - 20ac - 11(b + d) = 0
(a + c)2 - 22(121) - 11 × 9(a + c) = 0
(a + c) = 121 or -22 (rejected)
a + b + c + d = 1210.

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Kushagra Mishra
35 Points
8 years ago
how is ac=121?
I dont get it xD :/
Premanshu Rao
25 Points
8 years ago
Why did we reject -22?
Akshay
185 Points
8 years ago
you will not reject -22, a=-11,b=-99,c=-11,d=-99 is also a possible solution. Just substitute and check
Rishabh kumar
13 Points
5 years ago
a b c d must be distinct numbers so that (-22) is rejected in the real question of iit jee it was written that a ,b,c,d must be distinct