### Instruction: For which positive integer values of $k$ does $kx^2+20x+k=0$ have rational solutions? Express your answers separated by commas and in increasing order.### Response: To find the values of $k$ that make the equation have rational solutions, I need to use the discriminant of the quadratic formula.
The discriminant is the part under the square root: $b^2-4ac$.
In this case, $a=k$, $b=20$, and $c=k$.
So the discriminant is $20^2-4k^2$.
For the equation to have rational solutions, the discriminant must be a perfect square, since otherwise the square root would be irrational.
So I need to find the values of $k$ that make $20^2-4k^2$ a perfect square.
I can try some small values of $k$ and see if they work.
If $k=1$, then $20^2-4k^2=400-4=396$, which is not a perfect square.
If $k=2$, then $20^2-4k^2=400-16=384$, which is also not a perfect square.
If $k=3$, then $20^2-4k^2=400-36=364$, which is still not a perfect square.
If $k=4$, then $20^2-4k^2=400-64=336$, which is not a perfect square either.
If $k=5$, then $20^2-4k^2=400-100=300$, which is not a perfect square.
If $k=6$, then $20^2-4k^2=400-144=256$, which is a perfect square, since $256=16^2$.
So $k=6$ is one value that works.
I wonder if there are any larger values of $k$ that work.
I notice that as $k$ increases, $20^2-4k^2$ decreases, and it must remain positive for the equation to have real solutions.
So the largest possible value of $k$ is when $20^2-4k^2=0$.
This happens when $k=10$.
If $k=10$, then $20^2-4k^2=0$, which is a perfect square, since $0=0^2$.
So $k=10$ is another value that works.
There might be some values of $k$ between $6$ and $10$ that work, so I need to check them.
If $k=7$, then $20^2-4k^2=400-196=204$, which is not a perfect square.
If $k=8$, then $20^2-4k^2=400-256=144$, which is a perfect square, since $144=12^2$.
So $k=8$ is another value that works.
If $k=9$, then $20^2-4k^2=400-324=76$, which is not a perfect square.
So the only values of $k$ that make the equation have rational solutions are $k=6$, $k=8$, and $k=10$.