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!

mathtoo
Aug 11, 2018

#1**+2 **

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

CPhill
Aug 11, 2018

#3

#2**+2 **

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**.

CPhill's cognitive power is unmatched.

Mathhemathh
Aug 11, 2018