+0  
 
+1
72
6
avatar+621 

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!

mathtoo  Aug 11, 2018
 #1
avatar+91027 
+2

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

 

 

cool cool cool

CPhill  Aug 11, 2018
 #3
avatar+349 
+2

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
avatar+91027 
+1

That's OK, Mathhemathh, I almost forgot that, too !!!

 

 

cool cool cool

CPhill  Aug 11, 2018
 #2
avatar+349 
+2

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.

 

wink

 

CPhill's cognitive power is unmatched.

Mathhemathh  Aug 11, 2018
edited by Mathhemathh  Aug 11, 2018
 #5
avatar+91027 
+1

HAHAHA!!!.... I don't know about that....most of the time....I just make lucky guesses....LOL!!!!

 

 

cool cool cool

CPhill  Aug 11, 2018
 #6
avatar+621 
+1

Thanks! Very helpful! 

mathtoo  Aug 11, 2018

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