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If the roots of the equation (a-b)x² + (b-c)x + (c - a ) = 0 are equal, prove that b+c = 2a.

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Agam answered 1 year(s) ago

If the roots of the equation (a-b)x2 + (b-c)x + (c-a) = 0 are equal

If the roots of the equation (a-b)x2 + (b-c)x + (c-a) = 0 are equal , the prove 2a=b+c.

Class-X Maths

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Asked by Rajesh kumar pandey

Aug 8

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Syeda , SubjectMatterExpert

Member since Jan 25 2017

Answer.

let r & s be the roots, then:
r + s = -(b - c)/(a - b)
but r = s:
2r = -(b - c)/(a - b)
r = -(b - c)/[2(a - b)]
also:
r * s = r^2 = (c - a)/(a - b)
(b - c)^2/[4(a - b)^2] = (c - a)/(a - b)
(b - c)^2/[4(a - b)] = (c - a)
b^2 - 2bc + c^2 = 4ac - 4a^2 - 4bc + 4ab
4a^2 - 4ab + b^2 - 4ac + 2bc + c^2 = 0
(2a - b)^2 - 2c(2a - b) + c^2 = 0
[(2a - b) - c]^2 = 0
2a - b - c = 0
2a = b + c

Hope it's help you
Please mark me in brain list answer

Answer 2

Heya !!




The given equation is (a-b)X² + ( B - C ) X + ( C - A ) = 0.





Here,


A = ( a - b ) , B = ( b - c ) and C = ( c - a ).





Discriminant ( D ) = ( B² - 4AC )



=> ( b-c)² - 4 × ( a - b ) ( c - a ).


For real and equal roots , we must have : D = 0




Now,




D = 0



( b - c )² - 4 ( a - b ) ( c - a ) = 0


=> 4a² + b² + c² - 4ab + 2bc - 4ca = 0



=> (-2a)² + b² + c² + 2 (-2b) b + 2bc + 2c (-2a ) = 0




=> (-2a + b + c )² = 0



=> -2a + b + c = 0


=> b + c = 2a......PROVED.....