SOLUTION: 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?

Question 909043: 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?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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 and x =

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

and x =

(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:

and x = becomes:

and x = which becomes:

and x = which becomes:

and x = which becomes:

and x = which becomes:

and

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.
$$$

the graph when c = 6 is shown below.
the zero points are at x = -6/5 and x = -1.
$$$

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