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Question

If p,q,r and s are real numbers such that pr=2(q+s), then show that at least one of the equations x2+px+q=0 and x2+rx+s=0 has real roots.

Solution
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We have x2+px+q=0....(i)
x2+rx+s=0....(ii)

Let D1 and D2 be the discriminants of equations (i) and (ii). Then

D1=b24ac=p24q
similarly,
D2=r24s

D1+D2=p24q+r24s=(p2+r2)4(q+s)

D1+D2=p2+r24(pr2)[pr=2(q+s)q+s=pr2]

D1+D2=p2+r22pr=(pr)20

At least one of D1 and D2 is greater than or equal to zero
At least one of the two equations has real roots.

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