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# Positive Inters, of the day

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There are two postive integers c for which the equation $$5x^2+11x+c=0$$has rational solutions. What is the product of those two values of c?

This can be solved using the Quadratic Formula!

Aug 11, 2018

#1
+100483
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If this has rational solutions, then the discriminant must be a perfect square >  0

So we have

11^2  - 4 (5)c

121 - 20c

Note that when c  =2

121 - 20(2)  =

121 - 40 =  81  which is a perfect square

And when  c  = 6

121 - 20(6)  =

121  - 120 =  1  which is also a perfect square

So....the product of these  two values of  c   =  2 * 6    = 12

Aug 11, 2018
#3
+349
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oh no did i do something wrong

EDIT: Yes I did. Zero is not positive xD

Mathhemathh  Aug 11, 2018
edited by Mathhemathh  Aug 11, 2018
#4
+100483
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That's OK, Mathhemathh, I almost forgot that, too !!!

CPhill  Aug 11, 2018
#2
+349
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So, the only way for the solutions to be rational is if the discriminant ($$\sqrt{b^2-4ac}$$) is rational. Substituting a and c, we get $$\sqrt{121-20c}$$. Since this has to be rational, $$121-20c$$ has to be a perfect square. The only possible values of c wherein this is a perfect square are 0 (edit: 6) and 2, resulting in 1 and 81 respectively. Getting the product of 6 and 2, we get, well, 12.

CPhill's cognitive power is unmatched.

Aug 11, 2018
edited by Mathhemathh  Aug 11, 2018
#5
+100483
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HAHAHA!!!.... I don't know about that....most of the time....I just make lucky guesses....LOL!!!!

CPhill  Aug 11, 2018
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