There are two postive integers c for which the equation \(5x^2+11x+c=0\)has rational solutions. What is the product of those two values of c?
This can be solved using the Quadratic Formula!
If this has rational solutions, then the discriminant must be a perfect square > 0
So we have
11^2 - 4 (5)c
121 - 20c
Note that when c =2
121 - 20(2) =
121 - 40 = 81 which is a perfect square
And when c = 6
121 - 20(6) =
121 - 120 = 1 which is also a perfect square
So....the product of these two values of c = 2 * 6 = 12
So, the only way for the solutions to be rational is if the discriminant (\(\sqrt{b^2-4ac}\)) is rational. Substituting a and c, we get \(\sqrt{121-20c}\). Since this has to be rational, \(121-20c\) has to be a perfect square. The only possible values of c wherein this is a perfect square are 0 (edit: 6) and 2, resulting in 1 and 81 respectively. Getting the product of 6 and 2, we get, well, 12.
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