For how many integer values of a does the equation \(x^2 + ax + 5a = 0\) have integer solutions for x?

Guest Jan 11, 2019

#1**+1 **

using the quadratic formula, the discriminant is a^2-20a, which must be an perfect square. it factors as a(a-20). settings this equal to some perfect squares and repeating the quadratic formula, you can get the new discriminant, which is 400+(the perfect square). this is also a perfect square, so you can write it as 400+(perfect square)=(another perfect square). subtracting, you get a difference of squares where the difference is 400. the perfect square on the left side can be 99^2, 48^2, 21^2, 15^2, and 0^2. plugging it back in, you get the equations

a^2-20a-99^2=0

a^2-20a-48^2=0

a^2-20a-21^2=0

a^2-20a-15^2=0

a^2-20a-0^2=0

since a is an integer, the square root of the discriminant is an integer. therefore you can take out the first equation, the third equation, and the fourth equation. solving the remaining two, you get the values of a as -16, 0, 20, and 36 and the answer is 4.

HOPE THIS HELPED!!!!

also, to check, you can plug in the values and solve for x.

asdf335 Jan 12, 2019