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If the roots of the equation 1x+p+1x+q=1rare equal in magnitude but opposite in sign, then the product of the roots will be

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a
p2+q22
b
p2+q22
c
p2q22
d
p2q22

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detailed solution

Correct option is B

Given equation 1x+p+1x+q=1rcan be rewritten as
x2+x(p+q2r)+pqprqr=0……………………….(i)
Let roots are α and - α, then the product of roots
α2=pqprqr=pqr(p+q)………………………(ii)
and            sum of roots, 0=(p+q2r)
                          r=p+q2……………………….(iii)
On solving Eqs. (ii) and (iii), we get
α2=pqp+q2(p+q)=12(p+q)22pq α2=p2+q22

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