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

 Mar 21, 2019
 #1
avatar+6046 
+1

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

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 Mar 21, 2019
 #2
avatar+106515 
+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

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