When roots are equal

Question

If the roots of the equation $(b-c)x^2$$+ (c-a)x +(a-b)=0$ are equal then we have to prove $2b=a+c$ .

I tried it by doing D=0 , but not able to solve it . Please help

Answer 1

You have $$A x^2+Bx+C=0\qquad \qquad(A=b-c\qquad B=c-a\qquad C=a-b)$$ Then $$\Delta=B^2-4AC=(c-a)^2-4 (a-b) (b-c)=a^2-4 a b+2 a c+4 b^2-4 b c+c^2$$ $$A=(a-2 b+c)^2$$

Answer 2

You can easily note by inspection that one of the roots of the given quadratic is equal to 1. You can infer this by simply adding up all the three coefficients, and if they add up to zero, then one of the roots of the given quadratic eqn. is equal to 1.

Now since we already know a root we can find the other root as well by using the relation between the roots and the coefficients. In this case the other root will be also be equal to 1, as mentioned in the question.

Hence we have:

(a-b)/(b-c) = 1

Therefore:

a-b=b-c

2b=a+c (Hence Proved)...

Answer 3

Hint:  rewrite it as $b(x^2-1)-c (x^2-x) - a(x-1)=0$.

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[ EDIT ]   Then: $\big(x-1\big)\big(b(x+1) - c x - a\big) = 0$