Q) Find X in terms of a b, and c: , X ≠ a, b, c

Ans:

Given equation is:

∴

∴ [a (X – b) + b (X – a)] (X – c) = 2 c (X – a) (X – b)

∴ (a X – a b + b X – a b )(X – c) = 2 c (X – a) (X – b)

∴ (a X + b X – 2 a b) (X – c) = 2 c (X² – a X – b X + a b)

∴ a X² + b X² – 2 a b X – a c X – b c X + 2 a b c = 2 c X² – 2 a c X – 2 b c X + 2 a b c

∴ a X² + b X² – 2 c X² – 2 a b X + a c X + b c X = 0

∴ X (a X + b X – 2 c X – 2 a b  + a c  + b c)  = 0

∴ X [ X (a  + b  – 2 c ) – (2 a b  – a c  – b c )] = 0

∴ X = 0 and X (a + b  – 2 c ) – (2 a b  – a c  – b c ) = 0

∴ X = 0 and X =  

Therefore, X = 0 and X =   are the two roots of the given equation.

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