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# Discriminant help plz

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

Nov 8, 2018

### 7+0 Answers

#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!

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?

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
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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?

Nov 8, 2018