Question 909043
in order for the solutions to be rational, (b^2 - 4ac) must be a perfect square.


the equation is 5x^2 + 11x + c = 0


a = 5
b = 11
c = c


the quadratic formula is {{{x = (-b + sqrt(b^2-4ac)) / (2a)}}} and x = {{{(-b - sqrt(b^2-4ac))/(2a)}}}


replace a with 5 and b with 11 and you get:


{{{x = (-11 + sqrt(121-20c)) / (10)}}} and x = {{{(-11 - sqrt(121-20c))/(10)}}}


(121-20c) must be a perfect square.


when c = 0, 121 is a perfect square.
but c has to be positive and 0 is not positive.


when c = 1, 121 - 20 = 101 is not a perfect square.
when c = 2, 121 - 40 = 81 is a perfect square **************
when c = 3, 121 - 60 = 61 is not a perfect square.
when c = 4, 121 - 80 = 41 is not a perfect square.
when c = 5, 121 - 100 = 21 is not a perfect square.
when c = 6, 121 - 120 = 1 is a perfect square **************


looks like your solution is c = 2 and c = 6


when c = 2:


{{{x = (-11 + sqrt(121-20c)) / (10)}}} and x = {{{(-11 - sqrt(121-20c))/(10)}}} becomes:


{{{x = (-11 + sqrt(121-40)) / (10)}}} and x = {{{(-11 - sqrt(121-40))/(10)}}} which becomes:


{{{x = (-11 + sqrt(81)) / (10)}}} and x = {{{(-11 - sqrt(81))/(10)}}} which becomes:


{{{x = (-11 + 9) / (10)}}} and x = {{{(-11 - 9)/(10)}}} which becomes:


{{{x = (-2) / (10)}}} and x = {{{(-20)/(10)}}} which becomes:


{{{x = -1/5}}} and {{{x = -2}}}


you can confirm the solution is good by replacing c with 2 in the equation and then solving for f(-2/10) and f(-2)


the equation you started with is:


5x^2 + 11x + 2 = 0


set f(x) = 5x^2 + 11x + 2


f(-2/10) = 0


f(-2) = 0


I confirmed with my calculator and I also graphed the equation to show the solution graphically.


the graph when c = 2 is shown below.
the zero points are at x = -2 and x = -1/5.
<img src = "http://theo.x10hosting.com/2014/100303.jpg" alt="$$$" </>


the graph when c = 6 is shown below.
the zero points are at x = -6/5 and x = -1.
<img src = "http://theo.x10hosting.com/2014/100304.jpg" alt="$$$" </>