Question

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.d

Answer 1

When you solve this equation using the quadratic formula, you will get
`x = (-20\pm √(400-4k^2))/(2k)`. The only way for this number to be irrational is for
`√(400-4k^2)` to be irrational. The square root of any number that is not a perfect square is irrational*, so the solutions of the quadratic are rational if and only if
`400-4k^2` is a perfect square. We can factor out the 4 (which is already a perfect square), which means that
`100-k^2` must be a perfect square. This occurs exactly when k is equal to one of the following:
`√(100),√(99),√(96),√(91),√(84),√(75),√(64),√(51),√(36),√(19), √(0)`.

Of these, the only positive integer values of k are:
`√(100), √(64), √(36)`, or simply 6, 8, and 10.

* This is quite simple to show: Take any rational number, a/b. Without loss of generality, we can assume that a/b is in reduced form, that is, a and b have no common factors. (a/b)^2 is a^2/b^2, and since a and b have no common factors, neither do a^2 and b^2. Therefore, a^2/b^2 cannot be an integer. In the event that a/b is an integer, b would equal 1, and this proof would not hold.