namratakher
New member
For how many integer values of x is (2x^2-10x-4)/(x^2-4x-3) an integer?
number system ques.
- Thread starter namratakher
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jitenmazee
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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).
(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).