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

Given: $a(b-c)x^2+b(c-a)x+c(a-b)=0$

To do: To prove $b(a+c) =2ac$.

Solution:

$a(b-c)x^2+b(c-a)x+c(a-b)=0$, also zeroes of given equation are equal.

Therefore, Discriminant$=0$

$\Rightarrow B^2-4AC=0$

$\Rightarrow [b(c-a)]^2-4[a(b-c)c(a-b)]=0$

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

$\Rightarrow b^2c^2+a^2b^2-2ab^2c-4(a^2bc-ab^2c-a^2c^2+abc^2)=0$

$\Rightarrow b^2c^2+a^2b^2-2ab^2c-4a^2bc+4ab^2c+4a^2c^2-4abc^2=0$

$\Rightarrow b^2c^2+a^2b^2+2ab^2c-4a^2bc +4a^2c^2-4abc^2=0$

$\Rightarrow b^2c^2+a^2b^2+4a^2c^2+2ab^2c -4a^2bc-4abc^2=0$

We know that,

$\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=(a+b+c)^2$

By above identity we get,

$\Rightarrow (bc+ab-2ac)^2=0$

$\Rightarrow bc+ab-2ac=0$

$\Rightarrow b(a+c)=2ac$

$\Rightarrow b=\frac{2ac}{(a+c)}$

Hence if zeroes of given quadratic equation are equal then,

$b=\frac{2ac}{(a+c)}$