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

Sourabh Singh IIT Patna
6 years ago
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
pink pheme
19 Points
5 years ago
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
pink pheme
19 Points
5 years ago
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.
22 Points
one year ago
Let A=p(q-r), B=q(r-p) & C=r(p-q). A+B+C=0 => B=-(A+C). Since the equation has equal roots, B²-4AC=0
=> (A+C)²-4AC=0 => (A-C)²=0 => A=C
=> p(q-r)=r(p-q)
=> (q-r)/qr=(p-q)/pq
=> 1/r-1/q=1/q-1/p
=> 1/p+1/r=2/q
Hence QED

p,q,r are in Harmonic Progression.
Ram Kushwah
110 Points
one year ago
The given equation is:
p(r-p)x² + q(r-p)x+r(p-q)=0..................(1)
so here we have
a=p(r-p)
b=q(r-p)
c=r(p-q)
We observe
a+b+c=p(q-r)+q(r-p)+r(p-q)
=pq-pr+qr-pq+pr-qr=0
Thus we get
a+b+c=0
Or b=-(a+c)..................(2)
The roots are equal if  b²=4ac
Putting value of b from(2)
(a+c)²=4ac
a²+c²+2ac=4ac
a²+c²-2ac=0
(a-c)²=0
a-c=0
Or a=c
Now plugging values of a and c
p(q-r)=r(p-q)
pq-rp=rp-qr
2rp=pq+qr
dividing both sides by pqr
2rp/pqr=pq/pqr+qr/pqr
2/q=1/r+1/p
Hence  2/q=1/p+1/r