+817 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!
+129852 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
+349 oh no did i do something wrong
EDIT: Yes I did. Zero is not positive xD
+349 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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