+247 I'm not quite sure about how to solve a problem like this, but I'll mess around with it and see what I get.
Lets try to factor it by grouping. To factor nicely the 11x should split up with one of the parts a multiple of 5.
5x^2+5x+6x+c
5x(x+1)+6(x+c/6)
If this is going to be factored, the second part needs to match the first part. For this to happen c/6=1 so c=6.
(5x+6)(x+1)
The solitutions of this are x=-1 and x=-6/5. These are both rational.
There is one more way 11x can be split up with a multiple of 5.
5x^2+10x+x+c
5x(x+2)+1(x+c/1)
For this one to factor, c/1=2 so c=2.
(5x+1)(x+2)
The solitutions to this one are x=-2 and x=-1/5 and they are also rational.
So when c=6 and c=2 the quadratic has rational solitutions and the answer you are looking for is \(2\cdot6=\boxed{12}\)
+26367 There are two positive integers c for which the equation
\(5x^2 + 11x+c=0\) has rational solutions.
What is the product of those two values of \(c\)?
\(\begin{array}{|rcll|} \hline \mathbf{ 5x^2 + 11x+c} &=& \mathbf{0} \\\\ x &=& \dfrac{-11\pm \sqrt{121-4\cdot 5c} }{2\cdot 5} \\ \hline 121-4\cdot 5c &>& 0 \\ 121-20c&>& 0 \\ 121&>&20c \\ \mathbf{\dfrac{121}{20}} &>& \mathbf{c} \\ \hline \end{array}\)
The possible values of positive integers \(c\) are \(\{1,2,3,4,5,6\}\)
\(\begin{array}{|c|l|c|} \hline c & \sqrt{121-20c} & \text{rational solutions} \\ \hline \mathbf{6} & \sqrt{121-20\cdot 6}=\sqrt{1}= 1 & \checkmark \\ \hline 5 & \sqrt{121-20\cdot 5}=\sqrt{21} & \\ \hline 4 & \sqrt{121-20\cdot 4}=\sqrt{41} & \\ \hline 3 & \sqrt{121-20\cdot 3}=\sqrt{61} & \\ \hline \mathbf{2} & \sqrt{121-20\cdot 2} =\sqrt{81} = 9 & \checkmark\\ \hline 1 & \sqrt{121-20\cdot 1}=\sqrt{101} & \\ \hline \end{array}\)
So \(c=6\) and \(c=2\) and \(2\cdot 6 = 12\)
