Question

If roots of the equation $(a-b)x^2+(b-c)x+(c-a)=0$ are equal, prove that $2a=b+c$.

Answer 1

Given quadratic equation is $(a-b)x^2+(b-c)x+(c-a)=0$

Since the root are equal, discriminant of the quadratic equation $=0$

Comparing above equation with the standard form $A{x}^2+Bx+C=0$

We get, $A=(a-b),B=(b-c),C=(c-a)$

Hence, discriminant

$D=B^2-4AC=0\Rightarrow B^2=4AC\Rightarrow (b-c{)}^2=4(a-b\left)\right(c-a)$
$\Rightarrow b^2+c^2-2bc=4(ac-a^2-bc+ab)\Rightarrow b^2+c^2-2bc=4ac-4a^2-4bc+4ab\Rightarrow b^2+c^2+4a^2+2bc-4ac-4ab=0\Rightarrow b^2+c^2+(-2a{)}^2+2(b\left)\right(c)+2(-2a)c+2(-2a)b=0\Rightarrow (b+c-2a{)}^2=0\Rightarrow b+c-2a=0\therefore 2a=b+c$