Let p,qandr be real numbers (p≠q,r≠0), such that the roots of the equation 1x+p+1x+q=1r are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :
A
p2+q22
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B
2(p2+q2)
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C
p2+q2
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D
p2+q2+r2
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Solution
The correct option is Cp2+q2 Converting the given equation in general form
we get [(x+q)+(x+p)]r=(x+p)(x+q)