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For how many positive integer values of \(k\) does \(kx^2+10x+k\)\(=0\) have rational solutions?

Guest Nov 8, 2018
 #1
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\(x= {-b \pm \sqrt{b^2-4ac} \over 2a}={-10\pm \sqrt{10^2-4k^2} \over 2k}\) If  rational solutions means x real so must \(10^2-4k^2> 0 \)

So \(k=1,2,3,4\) 

Total 4 positive integer values of k

 

Hope this helps! 

Dimitristhym  Nov 8, 2018
edited by Dimitristhym  Nov 8, 2018
 #2
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If it is 5, then the disciminant is 0, but the problem says solutions, which implies that there is more than one. Am I just being dumb or do we actually have to consider this?

Guest Nov 8, 2018
 #3
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Yes right question because the exercise say solutionwe will have 4 positive integer values of k

Dimitristhym  Nov 8, 2018
 #4
avatar+316 
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This question its like "we will take the 0" But is not positive so no we will not take it!

 

Hope I help you to understand​!

Dimitristhym  Nov 8, 2018
edited by Dimitristhym  Nov 8, 2018
 #5
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Wait.. if k=1, the discriminat is 100-4*1^2=96, which is not a perfect square... so the solutions wouldn't be rational...right?

Guest Nov 8, 2018
 #6
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so wouldn't it be 2?

Guest Nov 10, 2018
 #7
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Hello?

Guest Nov 12, 2018

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