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

Guest Nov 8, 2018

#1**0 **

\(x= {-b \pm \sqrt{b^2-4ac} \over 2a}={-10\pm \sqrt{10^2-4k^2} \over 2k}\) 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

#2**+1 **

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**0 **

Yes right question because the exercise say solution**s **we will have 4 positive integer values of k

Dimitristhym
Nov 8, 2018

#4**0 **

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 ****understandâ€‹!**

Dimitristhym
Nov 8, 2018