|Sep26-12, 08:17 AM||#1|
Absolute Value inequalities
I was hoping someone could give a little more insight, or perhaps enlighten me to a better way of approaching solving these seemingly simple Algebra 2 inequalities.
I did some google searching but I was not able to find the answers I seek.
The problem came up when a friend of mine had an equation that looked like this:
|x-3| > -2 + |2x + 8|
Question 1: What is the generally accepted best way to solve this inequality.
My method of solving involved setting up 4 equalities:
1) x-3 = -2 + 2x + 8
2) -(x-3) = -2 + 2x + 8
3) x-3 = -2 - (2x + 8)
4) -(x-3) = -2 - (2x + 8)
These yielded the following values for x:
1) x= -9
2) x = -1
3) x = -7/3
4) x = -13
With each of these x values I then went back and plugged into the original inequality. My goal was to check for equality.
Only 2) and 4) resulted in equality so I threw the other answers out.
Once I had these two solutions I tested the points 0, -5, and -20 in the original inequality. And since -5 was the only value that yielded a true statement, I declared the solution:
(technically I had checked -7/3 and -9, checking -5 was unnecessary I realize now)
-13 < x < -1
Question 2: Can someone give me more information on why there are extraneous solutions?
As I understand things, when you have:
|x| = A
You need to solve x = A and x = -A
When you have:
|x| = |y|
You need to solve x = y, x = -y, -x = y and -x = -y, but because x = -y, -x = y are the same equation and x = y, -x = -y are the same equation you only need to solve two equations.
When you have:
|x| = A + |y|
You need to solve x = A + y, x = A - y, -x = A + y, and -x = A - y, since these are all different, you need to solve all four. Essentially the prior case (|x| = |y|) is a special case of this, just two of the equations you need to solve happen to be the same.
Question 3: Is there a more rigorous way of handling these equations? Should I be using |x| = Sqrt(x^2) here?
Thank you in advance.
|Sep26-12, 08:43 AM||#2|
Your method does not scale well. For n abs value terms you would have 2n cases. Better to list out the critical x values (-4 and 3 in your first example), then consider the n+1 ranges these generate: <-4, -4 to 3, >3.
|Sep26-12, 08:53 AM||#3|
I do not fully understand.
I understand that my method will result in 2n cases. That is clear to me.
I understand how to find what you call critical x values for each set of absolute value quantities. This is also clear.
I do not understand what it means to "consider" these ranges. How would I go about solving once I have the ranges: <-4, -4 to 3, >3?
|Sep26-12, 09:00 AM||#4|
Absolute Value inequalities
|Sep26-12, 09:21 AM||#5|
Please allow me to continue for my own benefit so that I may understand completely.
For x < -4 we have |x-3|=3-x and |2x+8|=-2x-8
For -4 < x < 3 we have |x-3|=3-x and |2x+8|= 2x+8
For x > 3 we have |x-3|= x - 3 and |2x+8|= 2x+8
Thus I must solve 3 equations, with constraints, as you predicted n + 1 = 3 in this case.
For x < -4 since |x-3|=3-x and |2x+8|=-2x-8
We must solve 3-x = -2 -2x-8. This yields x = -13 which is a valid answer since we require x < -4.
For -4 < x < 3 since|x-3|=3-x and |2x+8|= 2x+8
We must solve 3-x = -2 + 2x+8
This yields x = -1 which is a valid answer since we require -4 < x < 3.
For x > 3 since |x-3|= x - 3 and |2x+8|= 2x+8
We must solve x - 3 = -2 + 2x + 8
This yields x = -9. Which is NOT valid because we require x > 3.
Then by solving just 3 equations I can arrive at the critical values -1 and -13. I can then do my tests between -1 and -13 and beyond -1 and -13 to determine where the original inequality is true and where it is false.
Is this correct?
|Sep26-12, 02:21 PM||#6|
|Sep28-12, 07:14 AM||#7|
For -4 <= x <= 3 we have 3-x > -2 + 2x+8 ie x < -1, So [-4, -1) is valid.
And for x>= 3 we have x - 3 > -2 + 2x + 8 ie x < -9 So no solutions added in this case.
Thus (-13, -1) is our solution set. This way I don't have to do all that business with testing points in between....
Thank you haruspex for your help.
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