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?

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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) About Me  (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 x+=+%28-b+%2B+sqrt%28b%5E2-4ac%29%29+%2F+%282a%29 and x = %28-b+-+sqrt%28b%5E2-4ac%29%29%2F%282a%29

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

x+=+%28-11+%2B+sqrt%28121-20c%29%29+%2F+%2810%29 and x = %28-11+-+sqrt%28121-20c%29%29%2F%2810%29

(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+=+%28-11+%2B+sqrt%28121-20c%29%29+%2F+%2810%29 and x = %28-11+-+sqrt%28121-20c%29%29%2F%2810%29 becomes:

x+=+%28-11+%2B+sqrt%28121-40%29%29+%2F+%2810%29 and x = %28-11+-+sqrt%28121-40%29%29%2F%2810%29 which becomes:

x+=+%28-11+%2B+sqrt%2881%29%29+%2F+%2810%29 and x = %28-11+-+sqrt%2881%29%29%2F%2810%29 which becomes:

x+=+%28-11+%2B+9%29+%2F+%2810%29 and x = %28-11+-+9%29%2F%2810%29 which becomes:

x+=+%28-2%29+%2F+%2810%29 and x = %28-20%29%2F%2810%29 which becomes:

x+=+-1%2F5 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.
$$$

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