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

LET THE ROOTS OF EQUATION BE α
on comparing given equation with Ax²+Bx+C=o
A=(b-c), B=(c-a), C=(a-b)
now sum of roots=-B/A
                   α+α=-(c-a)/(b-c)
                      2α=a-c/(b-c)......1
product of roots=C/A
                    α²=(a-b)/(b-c)
putting the value of α from 1
  $\frac{ (a-c)^{2} }{4 (b-c)^{2} } = \frac{a-b}{b-c}$
on solving we get 2b=a+c