Question

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

Answer 1

D = 0

$\Rightarrow$ $(c-a{)}^2$ - 4 (b - c) (a - b) = 0

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

$\Rightarrow$ $(c+a-2ab{)}^2=\Rightarrow c+a-2b=0$

ALITER

Clearly, x = 1 satisfies the given equation. Since it has equal roots.

So, both roots are equal to 1.

$\therefore$ Product of the roots = 1. $\Rightarrow \frac{a-b}{b-c}=1\Rightarrow a-b=b-c\Rightarrow 2b=a+c$