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Inequality Question

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Inequality Question

by barrelbowl » Sun Dec 18, 2011 2:56 pm
Hey guys - sorry to flood the forum with questions, but I just wanted to resolve some issues I've had before my test on Wed.

I just wanted to resolve a basic misunderstanding I've had regarding inequalities.

For example:

If x^3 < 16x which of the following includes at least some of the possible solutions for x, but no values that are not solutions?

A) |x|<4
B) X <4
C) X>4
D) X<-4
E) X>0

The way I always attempt to solve these inequalities algebraically is by moving terms to the same side and solving

x^3-16x < 0
x(x^2-16) <0

So if (x^2-16) is <0 then x>0 and if (x^2 - 16) then x<0.

For the second part if x<0 then (x^2-16)<0 so (x+4)(x-4) < 0 which means x<-4 or X<4.
On the other hand, if x>0 then (x^2-16)>0 so (x+4)(x-4)>0 which means x>-4 or x>4.

If my understanding is correct then would it be the case where EITHER:

1) x>0, (x>-4, OR x>-4)

OR

2) x<0, (x<-4 OR x<4).

I guess my inherent confusion lies in the distinction between the "OR" between the two sets, and the "OR" within the sets.
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by barrelbowl » Sun Dec 18, 2011 2:56 pm
OA is D btw.
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by GmatMathPro » Sun Dec 18, 2011 3:41 pm
barrelbowl wrote:Hey guys - sorry to flood the forum with questions, but I just wanted to resolve some issues I've had before my test on Wed.

I just wanted to resolve a basic misunderstanding I've had regarding inequalities.

For example:

If x^3 < 16x which of the following includes at least some of the possible solutions for x, but no values that are not solutions?

A) |x|<4
B) X <4
C) X>4
D) X<-4
E) X>0

The way I always attempt to solve these inequalities algebraically is by moving terms to the same side and solving

x^3-16x < 0
x(x^2-16) <0

So if (x^2-16) is <0 then x>0 and if (x^2 - 16) then x<0.

For the second part if x<0 then (x^2-16)<0 so (x+4)(x-4) < 0 which means x<-4 or X<4.
On the other hand, if x>0 then (x^2-16)>0 so (x+4)(x-4)>0 which means x>-4 or x>4.

If my understanding is correct then would it be the case where EITHER:

1) x>0, (x>-4, OR x>-4)

OR

2) x<0, (x<-4 OR x<4).

I guess my inherent confusion lies in the distinction between the "OR" between the two sets, and the "OR" within the sets.
There are a lot of problems here, so let me just suggest a better way of solving it.

x^3<16x

x^3-16x<0 (Start the same way, bringing everything to the left)

x(x-4)(x+4)<0 (Factor completely)

Now, note the values of x that make the expression equal to zero. This will be x=-4,0,4. Think of these values as dividing up the number line into four sections: x<-4, -4<x<0, 0<x<4, x>4

Now, plug in values of x from each of these sections to test whether they make the expression x^3-16x positive or negative. Because our inequality is x^3-16x<0, we only want to include values of x that make this expression negative:

1. x>4
Choose any value of x that is greater than 4, let's say x=5, and plug it into x(x-4)(x+4)--->5(5-4)(5+4)...note that you don't have to get an exact value. Just do enough work to see if the answer is going to be positive or negative. Here it is clear that all the factors will be positive, so the overall answer will also be positive, and we DON'T want to include x>4 in our solution set.

2. 0<x<4

Let x=3: x(x-4)(x+4)---->3(3-4)(3+4) This will give us two positive factors and one negative factor so the overall product will be negative, so we DO want to include this in our solution set

3. -4<x<0

Let x=-1: x(x-4)(x+4)---->-1(-1-5)(-1+4) This will give us two negatives and a positive, so the product is positive, so we DON'T want to include it in our solution set

4. x<-4

Let x=-5: x(x-4)(x+4)---->-5(-5-4)(-5+4) This will give you three negatives, so the product is negative, so we DO want to include it in our solution set.

Combining all of these, we say our solution set is x<-4 or 0<x<4. This means all values of x that are EITHER less than -4 OR between 0 and 4 satisfy the original inequality. Only D satisfies the question.

There are some shortcuts available here, but if you're taking the test on Wednesday, I wouldn't worry about trying to learn when they apply. Just do a few of these to get comfortable with the method. You made several mistakes using your method, so I'd probably abandon that if I were you. For example, (x+4)(x-4)<0 doesn't mean that x<-4 or x<4, and this expression doesn't even really make sense, which may be why it caused you confusion. Let me know if you have any questions about any of this.
Pete Ackley
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by neelgandham » Sun Dec 18, 2011 3:50 pm
If x^3 < 16x which of the following includes at least some of the possible solutions for x, but no values that are not solutions?

A) |x|<4
B) X <4
C) X>4
D) X<-4
E) X>0

I will try to solve the question in the simplest way(known to me!) and let me know if you are still have doubts.

x^3 < 16x.
x(x^2-16)<0, so the conditions
x>0 and x^2-16<0 ORx<0 and x^2-16>0 satisfy the inequation.

Condition 1 - (Read and = Intersection)
x>0 and x^2-16<0
x>0 and x^2<16
x>0 and -4<x<4
0<x<4 satisfies the inequation

Condition 2 - (Read and = Intersection)
x<0 and x^2-16>0
x<0 and x^2>16
x<0 and (x<-4 or x>4)
x<-4 satisfies the inequation

so 0<x<4 UNION x<-4 satisfies the inequation x^3 < 16x

A) |x|<4, implies -4<x<4. Values of x, where -4<x<0 DONOT satisfy the Inequation - Not the answer
B) X<4, Values of x, where -4<x<0 DONOT satisfy the Inequation - Not the answer
C) X>4, No where in the picture - Not the answer
D) X<-4 - hmm all values of x, where x<-4 satisfy the Inequation
E) X>0 - Forget it ;)

Answer D

Now the GMAT way ;)
A) |x|<4 - let x = 0, x^3 < 16x = 0 < 0 Oops! doesn't satisfy the inequation - Incorrect
B) X<4 - let x = 0, x^3 < 16x = 0 < 0 Oops! doesn't satisfy the inequation - Incorrect
C) X>4 - let x = 5, x^3 < 16x = 125 < 80 Oops again doesn't satisfy the inequation - Incorrect and also eliminates Option E
D) X<-4 - Left over = correct
E) X>0 - let x = 5, x^3 < 16x = 125 < 80 Oops again doesn't satisfy the inequation - Incorrect

Hope it helps !
Anil Gandham
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by barrelbowl » Sun Dec 18, 2011 4:25 pm
Thanks a lot guys very useful information.

GMATMathPro that explanation of inequalities breaking up the number line (and then checking) makes a whole lot more sense.
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