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Isaiah Farrell 2022-05-23 Answered
Solve 1 x 2 1 1 x + 1
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Answers (2)

rideonthebussp
Answered 2022-05-24 Author has 10 answers
Use the difference of two squares to get
1 ( x + 1 ) ( x 1 ) 1 x + 1
Case 1: If x + 1 > 0 x > 1, we have:
( x 1 , 1 ) 1 x 1 1
The x 1 , 1 comes from the denominator ( x + 1 ) ( x 1 ), which is undefined when x = 1 , 1.
If x 1 > 0 x > 1, then multiplying both sides by x 1 gives 1 x 1 x 2. If x 1 < 0, then because x > 1, there are no solutions for this case. Hence the intersection of x > 1 , x > 1, and x 2 imply 1 < x 2.
Case 2: If x + 1 < 0 x < 1, we have:
( x 1 , 1 ) 1 x 1 1
x 1 > 0 is not possible in this case because x < 1. Thus x 1 < 0, or just x < 1, and there are no more conditions.
So the solution to the inequality is x < 1, 1 < x 2 for x R.
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Thomas Hubbard
Answered 2022-05-25 Author has 2 answers
We note on the side that x 2 1 and multiply by x 2 1, so need to distinguish two cases:
{ x 2 1 > 0 1 x 1 , x 2 1 < 0 1 x 1.
The second case is impossible and we are left with
x < 1 1 < x 2.
You can also work in systematic way, using a table of sign variations,
1 1 2 x + 1 0 + + + + + x 1 0 + + + 2 x + + + + + 0 + | | + 0
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