if the equation p(p-r)x²+q(r-p)x+r(p-q)=0 has equal roots find 2/q?

Use D=0 and you will get the value of p in terms of Q.

In this fashion you get the value of 2/q

I think the question should be to show that 1/p+1/r=2/q. So the answer will be:The given equation is, p(q-r) x2 +q(r-p)x +r(p-q) = 0 and it has both the roots equal, Now put x = 1 we get, p(q-r) +q(r-p) +r(p-q) = 0So x = 1 is a roots of the equation p(q-r) x2 +q(r-p)x +r(p-q) = 0​Since both the roots are equal so second root will also be =1 Now, product of roots = 1*1 = 1 So, r(p−q)p(q−r)= 1r(p−q) = p(q−r) 2pr = pq + qrNow dividing by pqr on both the side we get, 2pr/pqr =pq/pqr+qr/pqr Or 2/q=1/r+1/pHence proved

I think the question should be to show that 1/p+1/r=2/q. So the answer will be:The given equation is, p(q-r) x2 +q(r-p)x +r(p-q) = 0 and it has both the roots equal, Now put x = 1 we get, p(q-r) +q(r-p) +r(p-q) = 0So x = 1 is a roots of the equation p(q-r) x2 +q(r-p)x +r(p-q) = 0​Since both the roots are equal so second root will also be =1 Now, product of roots = 1*1 = 1 So, r(p−q)p(q−r)= 1r(p−q) = p(q−r) 2pr = pq + qrNow dividing by pqr on both the side we get, 2pr/pqr =pq/pqr+qr/pqr Or 2/q=1/r+1/pHence proved.