Question

If root of the equation ${(q-r)x}^2+(r-p)x+(p-q)=0$ are equal, then $p,q,r$ are in

• A. $AP$
• B. $GP$
• C. $HP$
• D. $Noneofthese$

Answer 1

The correct option is A $AP$

$(q-r)x^2+(r-p)x+(p-q)=0$

If roots are equal,

$D=0$

$\therefore (r-p{)}^2=4(p-q\left)\right(q-r)$

$\therefore r^2+p^2-2rp=4(p-q)(q-r)$

$\therefore r^2+p^2-2rp=4(pq-pr-q^2+2r)$

$\therefore r^2+p^2-2rp=4p2-4pr-4q^2+4qr$

$\therefore (r+p{)}^2=4pq-4q^2+4qr$

$\therefore (r+p{)}^2=4(pq-q^2+qr)$

$\therefore 2q=p+r$

Hence $p,q,r$ are in $A.P$.