Math
posted by Majek
Given that the equation x(x2p)=q(xp) has real roots for all real values of p and q. If q=3, find a nonzero value for p so that the roots are rational.

Steve
x(x2p)=3(xp)
x^22px = 3x3p
x^2(2p+3) + 3p = 0
for rational roots, the discriminant must be a perfect square. That is,
(2p+3)^212p
= 4p^2+12p+912p
= 4p^2+9
must be a perfect square.
p=2 is one solution
check:
x(x4) = 3(x2)
x^27x+6 = 0
(x1)(x6) = 0
Not only rational, but integers!
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