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Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?

x^3 – 4x^5 < 0 x^3(1-4x^2) < 0 (1-4x^2) < 0 1 < 4x^2 √1 < √4x^2 (when you take the square root of 4x^2 you take the square root of a square so...) 1 < |2x|

1<(2x) 1/2 < x OR 1<-2x -1/2>x

I am still a bit confused as to how we get 0. I see how it is done with the "root" method but my way of solving was just a bit different. Any thoughts?

The step in red above is your problem. How did you get rid of x^3? Can you divide both sides by x^3 when you have an inequality? You don't know whether x^3 is positive or negative. If you divide both sides by x^3 and x^3 is negative, the sign will flip. So you must retain the x^3 and that will give you 3 transition points (-1/2, 0 , 1/2) Even in equations, it is not a good idea to cancel off x from both sides. You might lose a solution in that case x = 0 e.g. x(x - 1) = 0 (x - 1) = 0 x = 1 (Incomplete)

Re: Which of the following represents the complete range of x [#permalink]

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09 Jul 2013, 22:10

Thank you, I figured that was my problem and you confirmed it!

VeritasPrepKarishma wrote:

WholeLottaLove wrote:

Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?

x^3 – 4x^5 < 0 x^3(1-4x^2) < 0 (1-4x^2) < 0 1 < 4x^2 √1 < √4x^2 (when you take the square root of 4x^2 you take the square root of a square so...) 1 < |2x|

1<(2x) 1/2 < x OR 1<-2x -1/2>x

I am still a bit confused as to how we get 0. I see how it is done with the "root" method but my way of solving was just a bit different. Any thoughts?

The step in red above is your problem. How did you get rid of x^3? Can you divide both sides by x^3 when you have an inequality? You don't know whether x^3 is positive or negative. If you divide both sides by x^3 and x^3 is negative, the sign will flip. So you must retain the x^3 and that will give you 3 transition points (-1/2, 0 , 1/2) Even in equations, it is not a good idea to cancel off x from both sides. You might lose a solution in that case x = 0 e.g. x(x - 1) = 0 (x - 1) = 0 x = 1 (Incomplete)

Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?

x^3 – 4x^5 < 0 x^3(1-4x^2) < 0 (1-4x^2) < 0 1 < 4x^2 √1 < √4x^2 (when you take the square root of 4x^2 you take the square root of a square so...) 1 < |2x|

1<(2x) 1/2 < x OR 1<-2x -1/2>x

I am still a bit confused as to how we get 0. I see how it is done with the "root" method but my way of solving was just a bit different. Any thoughts?

You cannot reduce by x^3 in the red part.
_________________

Re: Which of the following represents the complete range of x [#permalink]

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09 Jul 2013, 22:43

I didn't. I was trying to show that x^3 < 0 and (1-4x^2) < 0 (to get the check points) but I forgot to show for x^3!

Thank you, though!

Bunuel wrote:

WholeLottaLove wrote:

Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?

x^3 – 4x^5 < 0 x^3(1-4x^2) < 0 (1-4x^2) < 0 1 < 4x^2 √1 < √4x^2 (when you take the square root of 4x^2 you take the square root of a square so...) 1 < |2x|

1<(2x) 1/2 < x OR 1<-2x -1/2>x

I am still a bit confused as to how we get 0. I see how it is done with the "root" method but my way of solving was just a bit different. Any thoughts?

This might be a difficult question to answer, but here it is:

I understand the methodology in how the correct answer was arrived at (thanks, Bunuel) but my question is, how do I know to use that methodology with this particular question?

Also, could I solve for this problem using x^3(1-4x^2)<0 as opposed to (1+2x)*x^3*(1-2x)<0?

As always, thanks to the community for all of your help.

When you have linear factors and inequalities, think of this method. Since this method is useful for linear factors, you need to split the quadratic (1 - 4x^2) into (1-2x)*(1+2x). Some quadratic or higher powers may not be a problem (e.g. (x + 1)^2, (x^2 + 1) etc are always positive) so they can be ignored. For more, check: http://www.veritasprep.com/blog/2012/07 ... s-part-ii/ _________________

Re: Which of the following represents the complete range of x [#permalink]

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02 Aug 2013, 00:21

2

This post received KUDOS

VeritasPrepKarishma wrote:

lesnin wrote:

Hi All,

Could I conclude that for this case i.e (1+2x)*x^3*(1-2x)<0 even if one of the terms <0, that does not necessarily mean that the entire product of the 3 terms <0. Cause like if the eq was (1+2x)*x^3*(1-2x)= 0 ....I could have safely concluded that However in this case for the entire product <0.. either 1 terms or 2 terms or even all 3 terms can be - ve.

When you have product of two or more terms, the product will be negative only when odd number of terms are negative i.e. either only one term is negative and rest are positive or only 3 terms are negative and rest are positive or only 5 terms are negative and rest are positive. (-)(+)(+) = (-) (-)(-)(+) = (+) (-)(-)(-) = (-)

According to me answer should be

x^3(1-4x^2)<0 x^3(1-2x)(1+2x)<0

+ (-1/2) - (0) + (1/2) -

If less than 0, select (-) curves.

Answer: -1/2 < x < 0 or 1/2 < x ==> C

I solved all the the difficult inequality questions using this technqiue.I guess i am missing something.

Now the factors are of the form required and it is clear that the transition points are -1/2, 0, 1/2.

The required range is x > 1/2 or -1/2 < x< 0

Dear Karishma

So they always have to be in (x-a)(x-B) form??? also, the red part above can be written as \((x+1/2)*x^3*(x-1/2)>0??\)... we can divide both sides by 4 right?? that wont affect the problem na?
_________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

Now the factors are of the form required and it is clear that the transition points are -1/2, 0, 1/2.

The required range is x > 1/2 or -1/2 < x< 0

Dear Karishma

So they always have to be in (x-a)(x-B) form??? also, the red part above can be written as \((x+1/2)*x^3*(x-1/2)>0??\)... we can divide both sides by 4 right?? that wont affect the problem na?

Yes, always put them in (x - a)(x - b) format. That way, there will be no confusion.

And yes, you can divide both sides by 4. Think why it doesn't affect our inequality - We have: Expression > 0. If it were 4*Expression, that would be positive too since Expression is positive. If it were Expression/4, that would be positive too since Expression is positive. So positive constants don't affect the inequality.
_________________

Hi Bunuel, sorry for this noob question but, can you explain how do you find the sign for the equality roots - (I know how to find the roots but not able to understand how do we equate to the roots) \(-\frac{1}{2}<x<0\) or \(x>\frac{1}{2}\).

Hi Bunuel, sorry for this noob question but, can you explain how do you find the sign for the equality roots - (I know how to find the roots but not able to understand how do we equate to the roots) \(-\frac{1}{2}<x<0\) or \(x>\frac{1}{2}\).

Please read the whole thread and follow the links given in experts posts. You can benefit a lot from this approach.

Which of the following represents the complete range of x over which x^3 - 4x^5 < 0?

(A) 0 < |x| < ½ (B) |x| > ½ (C) –½ < x < 0 or ½ < x (D) x < –½ or 0 < x < ½ (E) x < –½ or x > 0

Responding to a pm: The problem is the same here. How do you solve this inequality: \((1+2x)*x^3*(1-2x)<0\)

Again, there are 2 ways - The long algebraic method: When is \((1+2x)*x^3*(1-2x)\) negative? When only one of the terms is negative or all 3 are negative. There will be too many cases to consider so this is painful.

The number line method: Multiply both sides of \((1+2x)*x^3*(1-2x)<0\) by -1 to get \((2x + 1)*x^3*(2x - 1)>0\) Take out 2 common to get \(2(x + 1/2)*x^3*2(x - 1/2)>0\) [because you want each term to be of the form (x + a) or (x - a)] Now plot them on the number line and get the regions where this inequality holds. Basically, you need to go through this entire post: inequalities-trick-91482.html

Responding to a pm:

Quote:

Why we meed to multiply the both sides by -1? What if the question is x^3 ( 2x+1) ( 1-2x )<0 or >0 do

we need in this caee to multiply the both sides by -1?

We need to bring the factors in the (x - a)(x - b) format instead of (a - x) format.

So how do you convert (1 - 2x) into (2x - 1)? You multiply by -1.

Say, if you have 1-2x < 0, and you multiply both sides by -1, you get -1*(1 - 2x) > (-1)*0 (note here that the inequality sign flips because you are multiplying by a negative number)

-1*(1 - 2x) > (-1)*0 -1 + 2x > 0 (2x -1) > 0

So you converted the factor to x - a form.

In case you have x^3 ( 2x+1) ( 1-2x )<0, you will multiply both sides by -1 to get x^3 ( 2x+1) ( 2x - 1 ) > 0 (inequality sign flips)
_________________

When reducing an inequality by a variable, we must know it's sign: if it's positive the sign of the inequality stays the same but if it's negative the sign of the inequality flips.
_________________

Which of the following represents the complete range of x [#permalink]

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19 Oct 2015, 13:21

gmatpapa wrote:

Which of the following represents the complete range of x over which x^3 – 4x^5 < 0?

A. 0 < |x| < ½ B. |x| > ½ C. –½ < x < 0 or ½ < x D. x < –½ or 0 < x < ½ E. x < –½ or x > 0

1. \(x^3(1-4x^2)=x^3*(1-2x)(1+2x)<0\) 2. Find critical points (zero points): x=0, x=1/2, x=-1/2 3. Pick some easy numbers to test each range (see attachment) and insert it in the inequality to find the final sign of it: -1: - + - + -1/4: - + + - 1/4: + ++ + 1: + - + -

To answer this question we just need describe the range with a -ve sign from the attachment: \(-\frac{1}{2}< X < 0\) and x > \(\frac{1}{2}\)

Attachments

Inequality.png [ 2.5 KiB | Viewed 528 times ]

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