We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
194
2
avatar+116 

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

 Mar 21, 2019
 #1
avatar+6042 
+1

\(\text{Rational solutions means the discriminant is a positive perfect square}\\ D = 10^2 - 4(k)(k) = 10^2 - 4k^2 = 100-4k^2\\ \text{The perfect squares less than 100 are}\\ 81, 64, 49, 36, 25, 16, 9, 4, 1\\ 100-4k^2 \text{ is going to be even so we can ignore the odd squares}\\ 64 = 100- 36 = 100-4(3)^2,~k=3 \text{ works}\\ 36=100-64^2 = 100-4(4)^2,~k=4 \text{ works}\\ 16 = 100-84 = 100 - 4(21), \text{ nope}\\ 4 = 100-96 = 100-4(24),~\text{ nope}\\ \text{only }k=3,4\)

.
 Mar 21, 2019
 #2
avatar+103972 
+1

Thanks, Rom.....there is one more value of k......

 

If the discriminant is 0, we also have a rational solution  (root)

 

So

 

10^2 -  4k^2  = 0

 

100 = 4k^2

 

25  = k^2

 

5 = k 

 

 

cool cool cool   

 Mar 21, 2019

6 Online Users