### 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 out when the quadratic equation has rational solutions, I need to use the quadratic formula and check when the discriminant is a perfect square.
The discriminant is $b^2-4ac$, where $a=k$, $b=20$, and $c=k$.
So the discriminant is $20^2-4k^2=400-4k^2$.
Now I need to find the values of $k$ that make this a perfect square.
I can start by listing some perfect squares that are less than or equal to 400: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, and 400.
I can try to match each of these to the expression $400-4k^2$ and see if I get an integer value for $k$.
For example, if I try 0, I get $400-4k^2=0$, which implies that $k=\pm 10$.
That means $k=10$ and $k=-10$ are both possible values, but I only want positive values, so I keep $k=10$ and discard $k=-10$.
I can continue this process for the other perfect squares, but I can also use some shortcuts.
For instance, I know that $k$ has to be even, because otherwise $4k^2$ would be odd and $400-4k^2$ would not be even.
Also, I know that $k$ has to be less than or equal to 10, because otherwise $4k^2$ would be greater than 400 and $400-4k^2$ would be negative.
So I only need to check the even values of $k$ from 2 to 10.
For $k=2$, I get $400-4k^2=400-16=384$, which is not a perfect square.
For $k=4$, I get $400-4k^2=400-64=336$, which is not a perfect square.
For $k=6$, I get $400-4k^2=400-144=256$, which is a perfect square.
So $k=6$ is another possible value.
For $k=8$, I get $400-4k^2=400-256=144$, which is a perfect square.
So $k=8$ is another possible value.
Finally, for $k=10$, I already know that it works, so I don't need to check it again.
Therefore, the values of $k$ that make the equation have rational solutions are 6, 8, and 10.