if the roots of equation (a-b)x^2 + (b-c)x + (c-a) = 0 are equal , prove that 2a = b+c
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Given, (a−b)x
2
+(b−c)x+(c−a)=0 are equal.
Then the discriminant =0.
Then
(b−c)
2
−4(c−a)(a−b)=0
or, (b
2
−2bc+c
2
)−4(ac−bc−a
2
+ab)=0
or, (b
2
+2bc+c
2
)−2a(b+c)+4a
2
=0
or, (b+c)
2
−2a(b+c)+4a
2
=0
or, (b+c−2a)
2
=0
or, 2a=b+c.
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