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# The equation A2/(x-a)+B2/(x-b)+C2/(x-c)+....+H2/(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

Ramesh V
70 Points
12 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