Absolute Value inequalities

Diffy

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.

haruspex

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.

Diffy

Thank you very much for your response.

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?

Thanks.

haruspex

Within each of those ranges, every abs value term has a known simplification. For x<-4, |x-3|=3-x and |2x+8|=-2x-8. For -4<x<3, |x-3|=3-x and |2x+8|=2x+8, etc.

Diffy

Thank you.

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?

Thank you.

haruspex

No, for this range (actually, x<=-4) you have 3-x > -2 + (-2x-8), i.e x>-13. So (-13,-4] is an entire interval of valid solutions. Proceed similarly with the other two cases.

Diffy

Ah yes. It took me a while to understand what you are saying here. Very clever.

So for
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.