Inequalities and absolute value

  • #1
highcontrast
13
0

Homework Statement


1) x^5 > x^2
2) 7| x + 2 | + 5 > 4
3) 3 - | 2x + 4 | <= 1

Homework Equations





The Attempt at a Solution


1)
x5 - x2 > 0
x2(x3 - 1) > 0
x2(x - 1)(x2 + x + 1) > 0
Im not too sure what to do next. I can't factor it any further, at least I don't think so. Which leads me to ask how exactly am I suppose to find the numbers to check what the solution is?

2)

7| x + 2 | + 5 > 4
7| x + 2 | > -1
|x + 2 | > -1/7
Can this be correct? The absolute value must always equal 0, or a positive number, right? How would I go about solving this? Or should I say the solutions do not exist?

3)
3 - | 2x + 4 | <= 1
- | 2x + 4 | <= -2
| 2x + 4 | => 2
2x + 4 => 2
2x => -2
x => -1
or
2x + 4 <= -2
2x <= -6
x <= -3
is this the right answer?

Thanks for your time
 

Answers and Replies

  • #2
Bohrok
867
0
For 1), now that you've factored it, find where the graph crosses the x-axis to get some intervals between those points. Each interval will be either above or below the x-axis.

[Edit] I must have thought the the inequality sign for 2) was the other way...
The absolute value of a real number is always ≥0, so |x + 2| > -1/7 is always true, for any real x.

3 seems correct.
 
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  • #3
Pagan Harpoon
93
0
When at |x + 2 | > -1/7, recall that this is an inequality, not an equation, it doesn't say that |x+2| is less than 0, it says that it is greater than -1/7. No value for x would make this untrue, so x can be any real number.
 
  • #4
hominid
10
0
x5 - x2 > 0
x2(x3 - 1) > 0
x2(x - 1)(x2 + x + 1) > 0

Solving an inequality would mean to express the solution as a union of intervals. In this case, which values of x will result in a value greater than 0 when plugged into the inequality.
 
  • #5
highcontrast
13
0
When at |x + 2 | > -1/7, recall that this is an inequality, not an equation, it doesn't say that |x+2| is less than 0, it says that it is greater than -1/7. No value for x would make this untrue, so x can be any real number.

So, I should go about solving the equation then?

Such as,

x + 2 > -1/7
x > -1/7 - 2
x > -15/7
or
x + 2 < 1/7
x < -13/7

It seems these answers conflict, though. How can x be greater than -15/7, and less than
-13/7.

I'm rather confused about absolute value because they have drilled it into my head that they always must be positive, or 0. So, when I saw an absolute inequality with it saying > -1/7, I assumed that the absolute value, while greater than 1/7, was still a negative. Does this mean in the cases of absolute values and inequalities, it doesn't matter if there is a negative value after one of the <,> signs?

Thanks Again
 
Last edited:
  • #6
hominid
10
0
Think of an absolute value as a distance in that a distance is going to be positive. The statement is true because since you know |x + 2| is always positive, you know |x + 2| is greater than -1/7 no matter what value of x you plug in. Remember it is not an equation, so it even if it said |x + 2| > -100,000 it would still be true.
 
  • #7
highcontrast
13
0
I understand. Was the posted solution to that question correct? The answers left me confused.
 
  • #8
hominid
10
0
Don't think of plugging those values of x into |x + 2 | > -1/7. You're trying to find out which values of x make this statement true: 7| x + 2 | + 5 > 4. Try plugging your solution into the inequality for x and then seeing if that proves true.
 
  • #9
highcontrast
13
0
I plugged them, and they work. I was just concerned because the textbook asks for me to solve the question also in a graph form.
 
  • #10
hominid
10
0
Since you are confused that the answers seem to overlap, think about what that means. It means that all real numbers are included.
 
  • #11
highcontrast
13
0
That makes sense!
Thanks for your help.
 

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