Post by Admin on Sept 7, 2009 10:05:29 GMT
Ok ABS for BunBuns:
1)
|x+1| = 5
First off, test where x+1 is originally +
// x+1 = 5
// x = 4
Then test where x+1 is -
//-x-1 = 5
//-x = 6
//x = -6
therefore answer is x = 4, -6
2)
|x+1| > 5
First off, test where x+1 is originally +
// x+1 > 5
// x > 4
Then test where x+1 is -
//-x-1 > 5
//-x > 6
//x < -6
therefore answer is x > 4, x <-6
note: you may want to sub in a point to test.
3)
|x+1| > |2x|
Test all possibilities:
1// |x+1| > |2x|
2// -|x+1| > |2x|
3// |x+1| > -|2x|
4// -|x+1| > -|2x|
1//x+1 > 2x
//-x+1 > 0
//-x > -1
//x < 1
2//-x-1 > 2x
//-3x-1 > 0
//-3x > 1
//x < -1/3
As you can see 2// is exactly the same as 3// except three is negative version of 2//
Try multiplying the equation 3// by negative
|x+1| > -|2x|
Negative: -|x+1| < |2x|
Therefore 3// is exactly the same as 2// EXCEPT the sign changes.
The answer to 2// is:
x < -1/3
therefor as 3// is 2// except sign changes,
answer to 3// is x > -1/3
This is also the same as 4//, 4// is the same as 1// EXCEPT sign changes.
Therefore answer to 4// is x > 1
NOW YOU HAVE THE FOUR SOLUTIONS!
x < 1
x < -1/3
x > -1/3
x > 1
Now, placing these results on a number line you see which of the solutions are the absolute boundaries. In this case it would be:
-1/3 > x < 1
Testing for points is the easy way to find out where your boundaries belong. All you have to do is find the two points of equality (in this case 1 and -1/3) then substitute points.
ANOTHER METHOD TO DO THIS:
Square both sides, as an ABS is always +ve, then squaring should make no difference.
|x+1| > |2x|
x[2] + 2x + 1 > 4x[2]
-3x[2] + 2x +1 > 0
3x[2] -2x -1 < 0
(3x+1)(x-1) < 0
Note: [2] 'means to the power of 2' i.e. 'Squared'
Points of equality are -1/3 and 1
then sub points in.
1)
|x+1| = 5
First off, test where x+1 is originally +
// x+1 = 5
// x = 4
Then test where x+1 is -
//-x-1 = 5
//-x = 6
//x = -6
therefore answer is x = 4, -6
2)
|x+1| > 5
First off, test where x+1 is originally +
// x+1 > 5
// x > 4
Then test where x+1 is -
//-x-1 > 5
//-x > 6
//x < -6
therefore answer is x > 4, x <-6
note: you may want to sub in a point to test.
3)
|x+1| > |2x|
Test all possibilities:
1// |x+1| > |2x|
2// -|x+1| > |2x|
3// |x+1| > -|2x|
4// -|x+1| > -|2x|
1//x+1 > 2x
//-x+1 > 0
//-x > -1
//x < 1
2//-x-1 > 2x
//-3x-1 > 0
//-3x > 1
//x < -1/3
As you can see 2// is exactly the same as 3// except three is negative version of 2//
Try multiplying the equation 3// by negative
|x+1| > -|2x|
Negative: -|x+1| < |2x|
Therefore 3// is exactly the same as 2// EXCEPT the sign changes.
The answer to 2// is:
x < -1/3
therefor as 3// is 2// except sign changes,
answer to 3// is x > -1/3
This is also the same as 4//, 4// is the same as 1// EXCEPT sign changes.
Therefore answer to 4// is x > 1
NOW YOU HAVE THE FOUR SOLUTIONS!
x < 1
x < -1/3
x > -1/3
x > 1
Now, placing these results on a number line you see which of the solutions are the absolute boundaries. In this case it would be:
-1/3 > x < 1
Testing for points is the easy way to find out where your boundaries belong. All you have to do is find the two points of equality (in this case 1 and -1/3) then substitute points.
ANOTHER METHOD TO DO THIS:
Square both sides, as an ABS is always +ve, then squaring should make no difference.
|x+1| > |2x|
x[2] + 2x + 1 > 4x[2]
-3x[2] + 2x +1 > 0
3x[2] -2x -1 < 0
(3x+1)(x-1) < 0
Note: [2] 'means to the power of 2' i.e. 'Squared'
Points of equality are -1/3 and 1
then sub points in.
