Inequalities and absolute value

AI Thread Summary
The discussion revolves around solving three inequalities involving absolute values and polynomial expressions. For the first inequality, x^5 > x^2, the factorization leads to determining intervals where the expression is positive. The second inequality, 7|x + 2| + 5 > 4, is clarified to show that since absolute values are always non-negative, the inequality is always true for any real x. The third inequality, 3 - |2x + 4| <= 1, is correctly solved to find two possible solutions for x, indicating the values that satisfy the inequality. Overall, the key takeaway emphasizes understanding absolute values in the context of inequalities, confirming that they cannot be negative.
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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
 
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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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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.
 
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.
 
Pagan Harpoon said:
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
 
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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.
 
I understand. Was the posted solution to that question correct? The answers left me confused.
 
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.
 
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
Since you are confused that the answers seem to overlap, think about what that means. It means that all real numbers are included.
 
  • #11
That makes sense!
Thanks for your help.
 
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