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deleted.

 Jun 9, 2019
edited by sinclairdragon428  Nov 20, 2019
 #1
avatar+195 
+1

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 \(2\cdot6=\boxed{12}\)

.
 Jun 10, 2019
 #2
avatar+23786 
+2

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 \(2\cdot 6 = 12\)

 

laugh

 Jun 10, 2019

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