# if the equation [b-c]x2+[c-a]x+a-c=0,,,,has equal roots then prove 2b=a+c

Vinay Goyal IITB
37 Points
11 years ago

I think there is some problem in question. As you have told equation has equal roots then its descriminant will be 0

so                        B2=4AC

[c-a]2=4[b-c][a-c]     on solving this you get 4b=3c+a

jagdish singh singh
173 Points
11 years ago

$Here equation is (b-c)^2+(c-a)x+(a-b)=0\\\\ So Clearly one Root is x=1\\\\ Let This Root be \alpha=1 and other Root be \beta\\\\ so \alpha.\beta=\frac{a-b}{b-c}\Leftrightarrow \beta=\frac{a-b}{b-c}\\\\ so Here It is Given \alpha=\beta\Leftrightarrow 1=\frac{a-b}{b-c}\\\\ so a-b=b-c\Leftrightarrow \boxed{\boxed{2b=a+c}}$