Question

If the roots of the equation $(b-c)x^2+(c-a)x+(a-b)=0$ are equal, then $n\times b=a+c$. Find $n$.

Answer 1

Given roots of

$(b-c)x^2+(c-a)x+(a-b)=0$ are equal

$\therefore △=$discriminant $=0$

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

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

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

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

$\therefore a-2b+c=0$

$\therefore 2b=a+c$

Given that $nb=a+c$

$\therefore n=2(\because 2b=a+c)$