If the roots of the equation $(b-c)x^2+(c-a)x+(a-b)=0$ are equal, then prove that $2b=a+c$.
Given:
Given quadratic equation is $(b-c)x^2+(c-a)x+(a-b)=0$. The roots of the given quadratic equation are equal.
To do:
We have to prove that $2b=a+c$.
Solution:
$(b-c)x^2+(c-a)x+(a-b)=0$
Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,
$a=(b-c), b=(c-a)$ and $c=(a-b)$.
The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.
$D=(c-a)^2-4(b-c)(a-b)$
$D=c^2+a^2-2ac-4ab+4b^2+4ca-4bc$
$D=a^2+4b^2+c^2+2ac-4ab-4bc$
$D=(a-2b+c)^2$ ($(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$)
The given quadratic equation has equal roots if $D=0$.
This implies,
$(a-2b+c)^2=0$
$a-2b+c=0$
$a+c=2b$
Hence proved.
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