Solution

Question: For how many integer values of x is (2x^2- 10x - 4)/(x^2 - 4x - 3) an integer? Solution: (2x^2- 10x - 4)/(x^2 - 4x - 3) = {2(x^2 - 4x - 3) - 2x + 2}/(x^2 - 4x - 3) = {2(x^2 - 4x - 3) - 2(x - 1)}/(x^2 - 4x - 3) = 2 - [{2(x - 1)}/(x^2 - 4x - 3)] Now it is clear that in the above expression, the first part of the expression is 2, which is an integer. So for the entire expression to be an integer, the expression [{2(x - 1)}/(x^2 - 4x - 3)] also has to be an integer. This is possible only when the following are true: Case I: x^2 - 4x - 3 = 1 or, x^2 - 4x - 4 = 0 But from here we do not get any rational value of x and therefore, we do not get any integer solutions. Case II: x^2 - 4x - 3 = -1 or, x^2 - 4x - 2 = 0 But from here we do not get any real value of x and therefore, we do not get any integer solutions. Case III: x^2 - 4x - 3 = 2 or, x^2 - 4x - 5 = 0 or, (x - 5)(x + 1) = 0 or, x = - 1, 5 (For these values of x, that is, for x = - 1 and x = 5, the values of the given expression (2x^2- 10x - 4)/(x^2 - 4x - 3) are 4 and - 2 respectively). Case IV: x^2 - 4x - 3 = - 2 or, x^2 - 4x - 1 = 0 Again from here we do not get any real value of x and therefore, we do not get any integer solutions. Case V: x - 1 = 0 or, x = 1 (For this value of x, that is, for x = 1, the value of the given expression (2x^2- 10x - 4)/(x^2 - 4x - 3) is 2). (Note: May be I have missed out on any other possibility or may be there is a much better method to solve this problem. I have just posted what came to my mind first. Sorry in case of any calculation errors or any other mistakes). Thank You Ravi Raja