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if the roots of the equation `1/ (x+p) + 1/ (x+q) = 1/r` are equal in magnitude but opposite in sign, show that p+q = 2r & that the product of roots is equal to `(-1/2)(p^2+q^2)`.

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We have `1/(x+p)+1/(x+q)=1/r`
`implies((x+q)+(x+p))/(x^(2)+(p+q)x+pq)=1/r`
`impliesx^(2)+(p+q-2r)x+pq-(p+q)r=0`
Now since the roots are equal in magnitudes but opposite in sign. Therefore
Sum of the roots `=0`
`impliesp+q-2r=0`
`impliesp+q=2r`....i
and product fo the roots `=pq-(p+q)r`
`=pq-(p+q)((p+q)/2)` [from Eq (i) ]
`=(2pq-p^(2)-q^(2)-2pq)/2`
`=-(p^(2)+q^(2))/2`
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