Question

If the roots of the equation (a-b)x2 +(b-c)x+(c-a) =0 then prove that 2a=b-c

Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

When you done a ton of questions on this topic than you find that when a quadratic equation is written with some order (same as your question) . Than almost in all of such cases the root is 1.

$(a - b) {x}^{2} + (b - c)x + (c - a) = 0 \\ \\ product \: of \: root = \frac{constant \: term}{coefficient \: of \: {x}^{2} } \\ 1 = \frac{c - a}{a - b} \\ a - b = c - a \\ \\ 2a = b + c$