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Question

If px2 + qx + r = 0 has no real roots and p, q, r are real such that p + r > 0 then


A

p - q + r < 0

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B

p - q + r > 0

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C

p + r = q

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D

All of these

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Solution

The correct option is B

p - q + r > 0


Let α+iβ,αiβ be the roots. Then α2+β2=rp>0. So p, r are of the same sign. Also p + r > 0. So p, r are both positive.
If q < 0, p - q + r > 0
If q > 0, (p+r)2(pr)2=4prq2( roots are nonreal)
(p+r)2q2+(pr)2
q2 p + r > q


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