### 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: I notice that this is a quadratic equation, and I know that a quadratic equation has rational solutions if and only if its discriminant is a perfect square.
The discriminant is the part under the square root in the quadratic formula, so I can write it as $D = 20^2-4k^2$.
To make $D$ a perfect square, I need to find values of $k$ that make $20^2-4k^2$ a perfect square.
I can rewrite this as $400-4k^2$, and then factor out a 4 to get $4(100-k^2)$.
Now, I need $100-k^2$ to be a perfect square, say $n^2$.
This means that $100-k^2 = n^2$, or equivalently, $100 = k^2 + n^2$.
This is a Pythagorean equation, and I know that the only positive integer solutions are $(k,n) = (6,8)$ or $(8,6)$.
Therefore, the only positive integer values of $k$ that make $D$ a perfect square are $k = 6$ or $k = 8$.