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If the roots of the equation (a-b)x^2 +(b-c)x+(c-a) =0 then prove that b+c=2a

If the roots of the equation (a-b)x^2 +(b-c)x+(c-a) =0 then prove that b+c=2a

We know that,

If the quadratic equation ax²+bx+c=0

Whose roots are equal then it’s deteminant is equal to zero

We have, (a-b)x²+(b-c)x+(c-a)=0

Deteminant = D= b² -4ac = 0

(b-c)² -4(a-b)(c-a)=0

b²+c²-2bc-4ac+4a²+4bc-4ab=0

b²+c²+4a²+4bc-4ac-4ab=0

b²+c²+(-2a)²+2bc+2c(-2a)+2(-2a)b=0

(b+c-2a)²=0

b+c-2a=0

b+c=2a

Hence proved.

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