I am having trouble with this inequality solution. I did not understand the rephrasing from the practice guide.

#154

Is x negative?

(1) x^3 (1-x^2) < 0

(2) x^2 -1 <0


When I factor I get x^3 (1+x)(1-x) <0

then I solve:

x^3 < 0 = x < cube root of 0 = x<0

1+x < 0 = x < -1
-1 -1

1-x < 0 = 1 < x
+x +x

so I get: -1>x>1 x<0

But the guide has it as:

-1 < x < 0 or x > 1 I am also not sure why there is an "or" involved.

In statement 2

(2) x^2 -1 <0

I factor and get

(x + 1) (x-1) < 0

x + 1 < 0 = x < -1
-1 -1

x-1 < 0 = x < 1
+1 +1

So I get 1> x and -1 > x, but the book has it as

-1 < x < 1



Could you explain how this solution is supposed to work?

Thank you

Ok -

Is x negtive?

(1) x^3(1-x^2)<0

Good job factoring to:

x^3(1-x)(1+x)<0

Here's the tricky bit - this is not an equation, rather it's an inequality. So you can't solve for each factor by itself being less than zero. The whole expression will be less than zero only if exactly one factor is negative while the other two are positive or if all three factors are negative. In the other cases (all positive factors, or two negative factors and a positive factor) the product would be greater than zero.

Since each of the three factors is pretty simple, I found it v. helpful to sketch a quick graph with each factor on it (x^3, (1-x), (1+x)). From the graph, it's pretty easy to see that there is no region where all three are negative and two regions where exactly one is negative and theother two are posive: -1<x<0 and x>1.

So (1) by itself is not sufficient to answer whether x is negative and you can rule out A&D.

Now consider (2):

x^2-1<0
factor:
(x+1)(x-1)<0

The inequality above will be true only if exactly one of the factors is negative while the other is positive. Just like above, I'd suggest sketching a quick graph of the two factors and you'll see the only region where exactly one factor is negative while the other is positive is -1<x<1. By itself, this is not sufficient to answer whether x is negative so you can rule out B.

Now consider the two together: -1<x<0 or x>1 and -1<x<1. the only region where both are satisified is -1<x<0 so x is def. negative and the answer is C.

The take away from this problem, other than that the test continues to get harder, is that inequalities have their own rules and procedures and can't be dealt with exactly like equations.

Jeff

#154

Is x negative? (not much of a rephrasal needed)

(1) x^3 (1-x^2) < 0

(2) x^2 -1 <0


(1) Multiply: x^3 (1-x^2) = x^3 - x^5 < 0
= x^5 > x^3

Not sufficient, x could be positive, or x could be a negative fraction ( i.e. - 1/2)

(2) x^2 -1 <0
= x^2 <1
x^2 is always positive, if x^2 is less than 1, then x is a proper fraction -- x could be a negative or a positive "proper fraction".
So (2) is not sufficient

Looking at statement 1 and 2 in tandem -- both solutions contain the concept that x is a negative "proper fraction" - i.e. negative fraction between -1 and 0.

Consider this as an alternate method to solving without getting into the complicated inequality algebra - that I usually try to avoid.


MGMAT Instructors - Is this a viable way to solve the problem?

Jeff,

I was under the impression you can solve the inequalities like that of an equation.

For example: x + 3 < 0.






so i thought

x < -3?

Mikek -

Yup, you can solve simple in equalities much like equations.

2x+3<11
2x<8
x<4

It gets more complicated when you have products. Consider the equation:

x*(x-5)*(x+4)=0

This equation is satisfied as long as at least on of the factors is zero (x=0,x=5,x=-4 all work).

Now consider the inequality:

x*(x-5)*(x+4)<0

When will this inequality hold? It's not enough for at least one factor to be less than zero. Because if two of them are negative and the remaining one positive the product would be positive. You need an odd number (1 or 3) negtive factors. For the inequality above this happens in the region (0<x<5) and (x<-4).

Jeff

This is a creative and viable way to solve the problem. The only thing I would change is your analysis of x^5 > x^3. We can conclude from x^5 > x^3 that EITHER x>1 OR -1<x<0.

-Jad

Jeff - thanks for all your contributions. Your solution makes a lot of sense.

For people who don't like to make graphs, it's possible to save some time by following GMATPaduan's solution (in this thread) or at least by not factoring (1 - x^2) and (x^2 - 1).

In interpreting statement (1), we could say:

If x^3 (1 - x^2) < 0,

then either (a) x^3 is negative and (1-x^2) is positive, or (b) x^3 is positive and (1-x^2) is negative.

In scenario (a) x^3 < 0, which means x<0, AND (1-x^2) >0 , so 1>x^2, so -1<x<1. Combining the two boldface statements, we can say that scenario (a) means -1<x<0

In scenario (b) x^3 > 0, which means x>0, AND (1-x^2) <0 , so 1<x^2, so x>1 or x<-1. Combining the two boldface statements, we can say that scenario (b) means x>1.

Summarizing our interpretation of statement (1), we see that it means EITHER (a) -1<x<0 OR (b) x>1. This is insufficient to show whether x is negative.


In interpreting statement (2), we could say:

If x^2 - 1 < 0 then x^2<1 so -1<x<1. This is also insufficient to show whether x is negative.


To combine statements (1) and (2), we can say that (-1<x<1) AND (EITHER (a) -1<x<0 OR (b) x>1). This means -1<x<0, which is sufficient to establish that x is negative.

-Jad

Another approach to solving this problem more intuitively:

Given that (1) can be factored to x^3(1-x)(1+x)<0, you can see that this function crosses the x-axis at (and only at) -1,0,1. So there are only two possible "shapes" to the curve. The first would be to approach the x-axis from below when x<-1, then by greater than 0 when -1<x<0, less than zero again when 0<x<1 and greater than zero when x>1. The other possible shape just goes the other way, being greater than zero when x<-1. You can see which one you're dealing with here by plugging in, e.g., -2. This gives you (-2)^3(3)(-1)=24. So you can see that for (1) the function will be less than zero when -1<x<0 and x>1.

Sorry if the explanation above is confusing...much easier to see/explain on a whiteboard - and actually a pretty quick technique in practice.

Jeff