Question

if the roots of equation : p(q-r)x×x + q(r-p)x + r(p-q) =0 are equal ,show that :1/p+ 1/r = 2/q

Answer 1

here something mistake
I think equation is ---
p (q-r) x^2 +q (r-p) x +r (p-q)=0

question ask ,
roots are equal
so, D =b^2 -4ac =0
b^2 =4ac
hence ,
{q (r - p) }^2 = 4pr (q -r )(p-q )
q^2 (r-p)^2 =4pr (q-r)(p-q)

q^2r^2+q^2p^2-2q^2pr=4pr (pq-q^2-pr+rq)

(qr)^2+(qp)^2-2q (pqr)=4p (pqr)-4q (pqr)-4 (pr)^2+4r (pqr)

(qr)^2+(qp)^2+2q (pqr)=4p (pqr)-4 (pr)^2+4r (pqr)

(qr)^2+(qp)^2+4 (pr)^2 +2q (pqr)-4p (pqr)-4r (pqr)=0

(qr)^2+(qp)^2+(2pr)^2+2.(qr)(qp)+2(qp)(-2pr)+2 (-2pr)(qr)=0

use formula ,
(a+b-c)^2 =a^2+b^2+c^2+2ab-2bc-2ca

{qr+qp-2pr}^2 =0

qr + qp -2pr =0

qr + qp = 2pr

divided pqr both side

1/p + 1/r = 2 /q
hence proved