> > > > > > Find the largest values of the integer a for which ax^2-5x-3 is > > > > > > negative for all values of x. > > > > > set a = (5x + 3) / x^2 and find the minima of this function. This > > > > > will, with your constraints, give the answer: a = -3 > > > > 5/x + 3/x^2; -5/x^2 - 6/x^3 = 0; -5x + -6 = 0; x = -6/5 > > > > a = (-6 + 3)/(36/25) = -25/12 > > > > Why does the minimum work? > a < f(x) = (5x + 3) / x^2, x <> 0 What's f? > If a < f(x) for all x, then a has to be less than the global minimum of f(x). > The global minimum of f(x) = -25/12. The largest integer less than this is -3.