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Find Integral values in a Quadratic equation

  1. Jul 21, 2014 #1
    1. The problem statement, all variables and given/known data

    The number of integral values of 'a' for which the quadratic equation (x + a) (x + 1991) + 1 = 0
    has integral roots are:

    2. Relevant equations

    D = b² - 4ac

    3. 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.
     
  2. jcsd
  3. Jul 21, 2014 #2
    Assume that ##b## and ##c## are the integer roots. Then,
    $$(x+a)(x+1991)+1=(x+b)(x+c)$$
    Substitute ##x=-b##, you should be able to take it from here.
     
  4. Jul 22, 2014 #3
    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.
     
  5. Jul 22, 2014 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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.
     
  6. Jul 22, 2014 #5

    pasmith

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    Homework Helper

    There are indeed only two integer values of [itex]a[/itex] which ensure that [itex]x[/itex] is an integer, but the OP's argument in post #3 doesn't actually prove that.


    Instead the condition the OP needs is that [itex](1991 + a)^2 - 4(1991 a + 1)[/itex] must be the square of an integer.
     
  7. Jul 22, 2014 #6
    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.
     
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