Question

Need urgent help:
Solve the following inequality:
(x^2 - 1)|x| > 0

Answer 1

So are they not cases to this quesiton
(ie. |x| =
{ -x, if x < 0
x, if x > 0?

Answer 2

either
x^2-1>0
x^2>1
if x^a>a^2 then x>a or x <-a
so, x <-1 or x>1
or.
|x| >0
-x, if x < 0
x, if x > 0?

Answer 3

i think this is the solution ,isnt it?

Answer 4

If x is positive, then x(x^2-1) > 0 would mean that x^2-1>0, and hence that x^2>1, or that x > 1.
If x is negative, then x(x^2-1) > 0 would mean that x^2-1<0, and hence that x^2<1, or that x < -1.

Thus, x is either less than -1 or greater than 1.

Answer 5

yea the final answer should be x>1

Answer 6

Ok this is what I think:
x^2-1(x) > 0, <-- Case A
x^2 - 1(-x) > 0, <-- Case B

Now what do I do from here?

Answer 7

In general, \( ab > 0 \) means that either
- both terms positive, \( a > 0, b > 0 \); or
- both terms negative, \( a < 0, b < 0 \)

Hence \( (x^2 - 1)|x| > 0 \) means that either both terms are positive or negative. But \( |x| \) is always not negative, so both must be positive. Hence
\[ (x^2 - 1)|x| > 0 \ \ \implies |x| > 0 \ \hbox{ and } \ x^2 - 1 > 0 \]

Now this last equation means that \( x^2 > 1 \) which implies \( |x| > 1 \)

Thus \( (x^2 - 1)|x| > 0 \) implies \( |x| > 0 \) and \( |x| > 1 \). Which means that in fact \( |x| > 1 \).

Answer 8

Why is absolute x always not negtative?

Answer 9

By definition.
|x| = x, if x > 0
= 0, if x = 0
= -x, if x < 0

In every case, \( |x| \geq 0 \)

Answer 10

Yeah that means that my original solution is correct.

Answer 11

I am sorry but I don't understand. I thought I am suppose to use the chart method for:
Case A --> (x^-1)(-x) > 0
Case B --> (x^-1)(x) > 0

Answer 12

@xactxx. Yes, it was. Or in interval notation,
\[ x \in (-\infty,-1) \cup (1,\infty) \]

@qrawarrior. You can do it however you like.

Answer 13

Ok I did Case A using the chart method for this, and I got:
(-infinity, -1), (-1,0), (0,1) <-- union or intersection?

For Case B I got:
(-1,0), (1,infinity) <--union or intersection?

Answer 14

Wait before I set up the chart for the case A:
(x^2-1)(-x) > 0, should I first flip the sign and get rid of the negative on the LHS?

Answer 15

If you're going to do it your way, be careful.

Case A. Let x > 0. Then we have
(x^2-1)x > 0
=> x^2 - 1 > 0
=> x^2 > 1.
But as x > 0, that implies x > 1.

Case B. Let x < 0. Then we have
(x^2-1)(-x) > 0,
i.e., (x^2 - 1)x < 0
=> x^2 - 1 > 0 because x is negative
=> x^2 > 1
=> x < -1, again because x is negative.

Therefore either x > 1 or x < 1. The same answer as we obtained earlier.

Answer 16

=> x^2 - 1 > 0
=> x^2 > 1.
^ This is where I lost you.

How did you get from: (x^2-1)x > 0
to the next linesa bove?

Answer 17

So => means implies

Now
(x^2-1)x > 0
=> x^2 - 1 > 0

because if x is a positive number, we can divide both sides by it and preserve the direction of the inequality.

Answer 18

Why would we want to get rid of x?

Answer 19

because we're trying to find a simple expression for x that characterizes when the inequality is true.

Answer 20

So lowering the power of x in the expression by getting rid of it certainly helps.

Answer 21

Anyway. You have everything you need now in our responses for your answer. Read them again, work them on your own paper. Think about them.

Then work them again on a blank piece of paper without looking. When you can do that, then you understand this problem.

Answer 22

But look if you keep (x^2-1)(x) > 0, you have this:
(x-1)(x+1)(x) > 0
So now, we have -1,+1,0

When you do chart method, you make intervals out from those 3 x-values. I tried this and this is what I keep getting: intervals from this three x-values

Answer 23

That relation is true if we know already that x > 0. So there's only two zones to consider. x > 1 and 0 < x < 1. In the first of those, x > 1, the relation holds. In the other, it doesn't.