KVPY 2019 SA Exam Previous Papers

Question : 61 of 80

Marks: +1, -0

Let a, b, c, d be distinct real numbers such that a, b are roots of

x^2 - 5cx - 6d = 0

, and c, d are roots of

x^2-5ax -6b = 0.

Then b + d is :

Solution:

a + b = 5c and ab = –6d
c + d = 5a and cd = –6b
⇒ ac = 36
Now a satisfies first equation and c second one, so

a^2 - 5ac -6d = 0

c^2- 5ac - 6b = 0

adding these 2 are get

(a + c)^2 - 12ac -24(a + c) = 0

⇒

(a + c)^2 -24(a + c) -12(36) = 0

⇒

a + c = 36 \or a + c

= -12(a = c = -6)

hence a + c = 36
so b + d = 144

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