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For how many integer values of \(a\) does the equation \(x^2+ax+5a=0\) have integer solutions for \(x\)?

 Dec 1, 2018
 #1
avatar+94545 
+1

Here's what I find.....

 

Let a = m....the discriminant is

 

m^2 -  4*5 * (m)  = m^2 - 20m  = m(m - 20)

This will be a perfect square when m = 20   ⇒ a = 20

And

x^2  + 20x +5(20) =

x^2 + 20x + 100       has  the root (-10, 0) 

 

And if a = 0.....this has the root   (0, 0)  

 

So

 

a = 0   or  a = 20

 

cool cool cool

 Dec 1, 2018
 #2
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0

what about a=36 and a=-16....those work too

Guest Dec 1, 2018

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