Asked by Praveen Kumar | 2nd May, 2014, 08:26: PM

Expert Answer:

1/s+p +1/s+p =1/r
 
r(s+p+s+q)=(s+p)(s+q)
 
r(s+p+s+q)=s2+sp+sq+pq
 
s2+s(p+q-2r)+pq-pr-qr=0
 
acoording to the given condition that both roots are of equal in maginitute and opposite in sign, then
 
sum of roots = 0
 
p+q-2r=0
 
p+q=2r  ... (i)
 
Now,
 
the product of roots=pq-pr-qr
 
                            =pq-r(p+q)
 
                            = pq-2r2    [From (i)]

Answered by Avinash Soni | 4th May, 2014, 07:42: AM

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