341(a-b)+(b-c)+(c-a) = 0 and hence at least one of them is negative.
If an equation has a non-real root, the roots are conjugates and their product is positive.
So all three roots cannot be non-real.
Given reals numbers a, b, c.
Prove that at least one of three euqations
x2+(a-b)x+(b-c)=0, x2+(b-c)x+(c-a)=0, x2+(c-a)x+(a-b)=0
has a real root.
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3 Answers
66Lets assume the contrary, i.e. none of the given equations have real roots. Then the discriminant of each of the equations must be negative. Accordingly we get the following set of inequalities:
(a-b)2 - 4(b-c) < 0
(b-c)2 - 4(c-a) < 0
(c-a)2 - 4(a-b) < 0
But adding these, we get
(a-b)2 + (b-c)2 + (c-a)2 < 0
which is a contradiction.
As such, at least one of the given equations must have a real root.
341
62Good question Bhargav..
Not because it was solved by Anant sir adn Prophet sir but because it is much better in terms of JEE preparations.. :)
