For how many positive integer values of \(k\) does \(kx^2+10x+k\)\(=0\) have rational solutions?
\(x= {-b \pm \sqrt{b^2-4ac} \over 2a}={-10\pm \sqrt{10^2-4k^2} \over 2k}\) 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!
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?
Yes right question because the exercise say solutions we will have 4 positive integer values of k
This question its like "we will take the 0" But 0 is not positive so no we will not take it!
Hope I help you to understand!