Question

suppose that a,b and c are distinct numbers such that (b-a)^2 -4(b-c) (c-a)=0,find (b-c)/(c-a)

Answer 1

$\begin{aligned}&(b-a)^2-4(b-c)(c-a)&=&\ \ 0\\ \\ \Longrightarrow\ \ &(b^2-2ab+a^2)-4(bc-ab-c^2+ac)&=&\ \ 0\\ \\ \Longrightarrow\ \ &b^2-2ab+a^2-4bc+4ab+4c^2-4ac&=&\ \ 0\\ \\ \Longrightarrow\ \ &4c^2+a^2+b^2-4ac+2ab-4bc&=&\ \ 0\\ \\ \Longrightarrow\ \ &(2c-a-b)^2&=&\ \ 0\\ \\ \Longrightarrow\ \ &2c-a-b&=&\ \ 0\\ \\ \Longrightarrow\ \ &2c&=&\ \ a+b\\ \\ \Longrightarrow\ \ &c+c&=&\ \ a+b\\ \\ \Longrightarrow\ \ &c-a&=&\ \ b-c\end{aligned}$

Here it seems that  $b-c\ =\ c-a$.

So,

$\large \boxed{\dfrac{b-c}{c-a}\ =\ \Large \textbf{1}}$