Question

if p(q-r)x 2 + q(r-p)x + r(p-q) = 0 has equal roots prove that 2/q = 1/p + 1/r (cyclic order) i.e. p,q,r are in h.p

Answer 1

Nice question .

U have to use some formula.

U can see that on putting x=1,we get the equation=0

I.e one root is 1,other root =c/a

=>Other root=r(p-q)/p(q-r)

Also,if both roots are equal then,other root is also equals to 1

I.e =[r(p-q)/p(q-r)]=1

=>pr-qr=pq-pr

=>2pr=pq+qr

=>Dividing both sides by pqr,we get

=>2/q=1/r+1/p

I.e p,q,r are in ho

Hope it helps