Question 39425: I am having trouble solving absolute value inequalities which involves exponents.
1. |x - 1| = |x to the second power - 2x +1|
2. |x to the second power + x - 2| < x + 3
Your help will be appreciated.
Answer by kev82(151)   (Show Source):
You can put this solution on YOUR website! Hi,
There are quite a few ways to solve these, and I don't know which you prefer, so we'll do the first one geometrically, and the second one with a bit of algebra.
You may not know this but the 'absolute value' operator as you call it, is actually a distance operator. The way to think about it is like this
|A-B|=D(A, B)
The distance from A to B is |A-B|. For example |5-3|=|3-5|=2, if you draw a number line and examine the distance from 3 to 5, or from 5 to 3 you will see that it is 2.
The equation |x|=1, can be written as |x-0|=1 so it really says the distance between x and 0 is 1. There are two points on the number line that are a distance of 1 from 0, and they are 1, and -1. Which as we both know are the solutions to |x|=1.
Anyway, what we are dealing with is the equation |x-1|=|x^2-2x+1|. We can factorise x^2-2x+1 as (x-1)(x-1). So we have |x-1|=|(x-1)^2| or (because abs can move through products) |x-1|=|x-1|^2.
As I said above |x-1| is the distance between x and 1. So let's call it d. We have the equation d=d^2. The solutions to this are clearly d=0 and d=1. The only point that is a distance of 0 from 1 is one itsself, so x=1 is a solution. The points which are a distance of 1 from 1 are 0, and 2. So the solutions to this are x=0, 1, and 2.
Hopefully I didn't lose you in all of that. If it helps, try drawing number lines.
Now let's get into some real algebra and solve the inequality.
If we were give something like |f|=g, then we know that we can write that as two equations f=g, and -f=g. The same is true with inequalities, so  produces two inequalities  and 
Doing the same with this question gives the two inequalities
Now, quadratic inequalities are a bit of a pain, because the sign changes if we multiply/divide by a negative number. To get past this we rewrite the inequality and consider sign changes.
The first inequality can be written as
 , so has sign changes at  We have to consider the truthfulness of the inequality ecah side of the sign changes. If we take a number way bigger than  say for example x=10, we can see that  is  which is false. Similarly if we choose a number way below  say -10, we can again see that the inequality doesn't hold. Choosing a number in between and like 0 we can see that the inequality does hold  . So  .
We also need to check the other inequality. Rearraging that gives
^2>0)
This is always true for any (real) value of x, so doesn't matter.
So we finally have or more simply 
Again, I hope I didn't lose you in all of that. The only hard part there is solving quadratic inequalities, which you should have learnt before attempting such inequalities so you should be ok.
If you want any more help with this or just a clarification, please write back.
Hope that helps,
Kev
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