x^2 + px + q = 0 has one negative solution and one positive solution. If p = 3, then what are all the possible values of q?
q must be negative ( if it were positive, we would have factors of ( x + m) ( x + n) = 0, and the roots m, n would both be negative)
So......any negative for q would work because the discriminant
3^2 - 4(q)(1) will be positive
q < 0