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

IneedHALP Mar 21, 2019

#1**+1 **

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

.Rom Mar 21, 2019