If `A_(1),A_(2),A_(3),...,A_(n),a_(1),a_(2),a_(3),...a_(n),a,b,c in R` show that the roots of the equation
`(A_(1)^(2))/(x-a_(1))+(A_(2)^(2))/(x-a_(2))+(A_(3)^(2))/(x-a_(3))+…+(A_(n)^(2))/(x-a_(n))`
`=ab^(2)+c^(2) x+ac` are real.

Explanation

Assume `alpha + i beta` is a complex root of the given equation, then conjugate of this root i.e. `alpha-ibeta` is also root of this equation.
On putting `x=alpha+ibeta` and `x=alpha-ibeta` in the given equation we get
`(A_(1)^(2))/(alpha+ibeta-a_(1))+(A_(2)^(2))/(alpah+i beta-a_(2))+(A_(3)^(2))/(alpha+i beta-a_(3))+....+(A_(n)^(2))/(alpha+ibeta-a_(n))`
`=ab^(2)+c^(2)(alpha+ibeta)+ac`...i
and `(A_(1)^(2))/(alpha -ibeta-a_(1))+(A_(2)^(2))/(alpha-ibeta-a_(2))+(A_(3)^(2))/(alpha-ibeta-a_(3))+..........+(A_(n)^(2))/(alpha-i beta-a_(n))`
`=ab^(2)+c^(2)(alpha-i beta)+ac` ...........ii
On subtracting Eq. i from Eq ii we get
`2i beta[(A_(1)^(2))/((alpha-a_(1)^(2)+beta^(2))+(A_(2)^(2))/((alpha-a_(2))^(2)+beta^(2))+(A_(3)^(2))/((alpha-a_(3))^(2)+beta^(2))`
`+...........+(A_(n)^(2))/((alpha-a_(n))^(2)+beta^(2))+c^(2)]=0`
The expression in bracket `!=0`
`:.2ibeta=0impliesbeta=0`
Hence all roots of the given equation are real.