Question

If the roots of a quadratic equation (m-n)x²+(n-l)x+(l-m)=0 are equal, show that l,m,n are in A.P

Answer 1

       

$Proving by the concept given below.$

$\sf{If} \ \ x,\ y,\ z\ \ \sf{are three consecutive terms in an AP, then,} \ \\ \\ \boxed{2y=x+z} \\ \\ \\ \sf{because, \\ \\ let the three consecutive terms of the AP be} \ \ a,\ a+d,\ a+2d. \\ \\ \\ \sf{such that,} \ \\ \\ \\ (a)+(a+2d) \\ \\ \Rightarrow a+a+2d \\ \\ \Rightarrow\ 2a+2d \\ \\ \Rightarrow\ 2[(a+d)]$

$Okay, proving...!!!$

$p(x)=(m-n)x^2+(n-l)x+(l-m)=0 \\ \\ \\ a=m-n \\ \\ b=n-l \\ \\ c=l-m$

$As the roots are equal, the discriminant is 0.$

$b^2-4ac=0 \\ \\ \Rightarrow\ (n-l)^2-(4(m-n)(l-m))=0 \\ \\ \Rightarrow\ n^2-2nl+l^2-(4(lm-m^2-ln+mn))=0 \\ \\ \Rightarrow\ n^2-2nl+l^2-(4lm-4m^2-4ln+4mn)=0 \\ \\ \Rightarrow\ n^2-2nl+l^2-4lm+4m^2+4ln-4mn=0 \\ \\ \Rightarrow\ 4m^2+l^2+n^2-4lm-4mn+2nl=0 \\ \\ \Rightarrow\ (2m-l-n)^2=0 \\ \\ \Rightarrow\ 2m-l-n=0 \\ \\ \Rightarrow\ 2m=l+n$

$As \ \ 2m=l+n\ \ seems like \ \ 2y=x+z \ \ which is mentioned earlier , \\ \\ l,\ m,\ n\ \ are three consecutive terms in an AP. \\ \\ \\ Hence proved!!!$

$Thank you. :-))$

$\mathfrak{\#adithyasajeevan}$