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Question

If roots is the equation (a–b)x2+(b–c)x+(c–a)=0 are areare equal then prove that 2a=b+c.

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Solution

Using Discriminant,

D = B^2-4AC as compared with the general quadratic equation Ax^2+Bx+C=0

so, A = a-b

B = b-c

C = c-a

For roots to be equal, D=0

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

b^2+c^2-2bc -4(ac-a^2-bc+ab) =0

b^2+c^2-2bc -4ac+4a^2+4bc-4ab=0

4a^2+b^2+c^2+2bc-4ab-4ac=0

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

i.e. 2a-b-c =0

2a= b+c


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