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+q2
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B
p2+q22
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C
2(p2+q2)
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D
p2+q2+r2
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Solution
The correct option is Ap2+q2 Converting the given equation in general form we get (2x+p+q)r=(x+p)(x+q)x2+(p+q–2r)x+pq–pr–qr=0p+q=2r........(1) α=−β (α+β)2−2αβ=α2+β2 [∵(α+β)=0] α2+β2=−2αβ =−2(pq−pr−qr)=−2pq+2r(p+q)from(1)α2+β2=p2+q2