Question

If the roots $QE(a-b)x^2+(b-c)x+(c-a)=0$ are equal then $P.T$ $2a=b+c$

Answer 1

For equal roots

$D=0$

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

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

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

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

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

$2a=b+c$