When you solve this equation using the quadratic formula, you will get [tex]x = \frac{-20\pm \sqrt{400-4k^2}}{2k}[/tex]. The only way for this number to be irrational is for [tex]\sqrt{400-4k^2}[/tex] 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 [tex]400-4k^2[/tex] is a perfect square. We can factor out the 4 (which is already a perfect square), which means that [tex]100-k^2[/tex] must be a perfect square. This occurs exactly when k is equal to one of the following:[tex]\sqrt{100},\sqrt{99},\sqrt{96},\sqrt{91},\sqrt{84},\sqrt{75},\sqrt{64},\sqrt{51},\sqrt{36},\sqrt{19}, \sqrt{0}[/tex].
Of these, the only positive integer values of k are: [tex]\sqrt{100}, \sqrt{64}, \sqrt{36}[/tex], 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.
