The discussion revolves around finding integral values of 'a' in the quadratic equation (x + a)(x + 1991) + 1 = 0 that yield integral roots. Participants are exploring the implications of the discriminant and the conditions for the roots to be integers.
The discussion is active, with participants providing different perspectives on the problem. Some suggest that the original poster's reasoning may not fully address the requirements for integral roots, while others assert that there are only two valid integral values for 'a'. There is no explicit consensus on the correct interpretation of the conditions needed for the roots.
Participants note that the condition derived from the discriminant must be further examined, particularly whether it should equate to the square of an integer. There is also mention of the assumption that both 'a' and 'x' are integers, which influences the discussion.
addzy94 said: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
Now: (1991+a)² > 4(1991a + 1)
What do I do after this: Please help, I am not even sure that I am on the correct path.
HallsofIvy said:I think you are misunderstanding the question. Isn't it obvious that there will be an infinite number of such a?
Your "(1991+a)² > 4(1991a + 1)" is simply the condition that the roots be real numbers.
pasmith said:There are indeed only two integer values of a which ensure that x is an integer, but the OP's argument in post #3 doesn't actually prove that.