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solve Absolute Value Inequalities|x+1/x|<4

|x+1/x|<4  .
By: Guest
Date: Sun-Oct-14-2012-
Response
0
This is one of the VERY tricky questions as it is inequalities AND absolute values AND quadratic equations all-in-one.

First a couple of observations:

x cannot be zero (1/x would be undefined).

x+1/x is positive for x > 0 and negative for x < 0

First we would have to separate this into two distinct cases for x > 0 and for x < 0, i.e. the two cases for x+1/x positive and negative.

For x > 0:

|x+1/x| < 4 becomes x+1/x < 4

Multiply by x on both sides

(x+1/x) * x < 4x

Reduce

x^2 + 1 < 4x

Subtract 4x from both sides

x^2 - 4x + 1 < 0

Solve x^2 - 4x + 1 = 0

x = 2 - sqrt(3) or x = 2 + sqrt(3)

x^2 - 4x + 1 < 0 for 2 - sqrt(3) < x < 2 + sqrt(3)

For x < 0:

|x+1/x| < 4 becomes -x-1/x < 4

Multiply by -x on both sides (do NOT to flip the inequality, x is less than zero so -x is positive!)

(-x-1/x) * (-x) < -4x

Reduce

x^2 + 1 < -4x

Add 4x to both sides

x^2 + 4x + 1 < 0

Solve x^2 + 4x + 1 = 0

x = -2 - sqrt(3) or x = -2 + sqrt(3)

x^2 + 4x + 1 < 0 for -2 - sqrt(3) < x < -2 + sqrt(3)

The total solution then becomes (are you ready?):

-2 - sqrt(3) < x < -2 + sqrt(3) OR 2 - sqrt(3) < x < 2 + sqrt(3)

Let us test with some values.

x less than or equal to -2 - sqrt(3):

x = -4 should not be OK: |-4+1/(-4)| = |-4 1/4| = 4 1/4 (not < 4, as expected)

x in the range (-2 - sqrt(3), -2 + sqrt(3)):

x = -2 should be OK: |-2+1/(-2)| = |-2 1/2| = 2 1/2 (< 4, as expected)

x in the range [-2 + sqrt(3), 0):

x = -.25 should not be OK: |-.25+1/(-.25)| = |-4 1/4| = 4 1/4 (not < 4, as expected)

x in the range (0, 2 - sqrt(3)]:

x = .25 should not be OK: |.25+1/.25| = |4 1/4| = 4 1/4 (not < 4, as expected)

x in the range (2 - sqrt(3), 2 + sqrt(3)):

x = 2 should be OK: |2+1/2| = |2 1/2| = 2 1/2 (< 4, as expected)

x greater than or equal to 2 + sqrt(3):

x = 4 should not be OK: |4+1/4| = |4 1/4| = 4 1/4 (not < 4, as expected)

[d] By: Guest
Date: Sun-Oct-14-2012
Response
What is 1 + 100



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