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```
The equation A 2 /(x-a)+B 2 /(x-b)+C 2 /(x-c)+....+H 2 /(x-h)=k,has (a)no real root (b)at the most one real root (c) no complex root (d) at the most 2 complex roots

```
11 years ago

```							The option is C , it has all real roots i.e., no complex roots
T o prove this
lets take (p + iq) as a root of above eqn, the for x=(p + iq)
we have
A2/((p + iq)-a) + B2/((p + iq)-b) + ...............+ H2/((p + iq)-h)   - k  = 0
we can write  this eqn in simplified form as
A2.[((p -a)- iq)) / ((p-a)2+q2)] + B2.[((p -b)- iq)) / ((p-b)2+q2)]  + ...............+ H2.[((p -h)- iq)) / ((p-h)2+q2)]   - k  = 0
equating imaginery partson both sides to 0 , we have

q* {( A2 / ((p-a)2+q2)+B2/ ((p-b)2+q2)+ .................+H2 / ((p-h)2+q2)}  = 0
so q = 0           [since A2 , B2 ,............etc are all positive ]
hence all roots are REAL

```
11 years ago
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