Recent content by addzy94

There can only be two integers whose multiplication can result in -1. That is 1 and -1. In my previous post I substitute 1 and minus -1 alternatively for the two integer values (x+a) and (x+1991) which are integer values themselves, thus proving it.

I don't know how that would work. It's alright I got the answer. As both a and x are integers (x + a)(x+1991) = -1 Thus (x + a) = 1 or -1 (x+1991) = -1 or 1 So we get two integral values for a. I would also like to know how to do it your way.

Homework Statement The number of integral values of 'a' for which the quadratic equation (x + a) (x + 1991) + 1 = 0 has integral roots are: Homework Equations D = b² - 4ac The Attempt at a Solution What I did was simplify the given equation and I got: x² + (1991 + a)x + (1991a + 1) = 0...