Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

\(-ve --- -1/2---- +ve--- 0----- -ve-----1/2--- +ve\) Could you please explain where I'm wrong ?

Regards, Subhash

Even though your question is directed to Bunuel, I will give a quick explanation.

The concept of the rightmost section being positive is applicable when every term is positive in the rightmost region. This is the case whenever the expressions involved are of the form (x - a) or (ax - b) etc. When you have a term such as (1-2x), the rightmost region becomes negative. So either, as Bunuel mentioned, check for an extreme value of x or convert (1-2x) to (2x - 1) and flip the sign to >.
_________________

A question - what is the best way u use to know if the "good" area is above or below?

i mean - what was the best way for u to know that its between -1/2 to 0

i used numbers ex. 1/4 but it consumes time! is there any better technique?

thanks.

Check the link in my previous post. There are beautiful explanations by gurpreetsingh and Karishma.

General idea is as follows:

We have: \((1+2x)*x^3*(1-2x)<0\) --> roots are -1/2, 0, and 1/2 (equate the expressions to zero to get the roots and list them in ascending order), this gives us 4 ranges: \(x<-\frac{1}{2}\), \(-\frac{1}{2}<x<0\), \(0<x<\frac{1}{2}\) and \(x>\frac{1}{2}\) --> now, test some extreme value: for example if \(x\) is very large number than first two terms ((1+2x) and x) will be positive but the third term will be negative which gives the negative product, so when \(x>\frac{1}{2}\) the expression is negative. Now the trick: as in the 4th range expression is negative then in 3rd it'll be positive, in 2nd it'l be negative again and finally in 1st it'll be positive: + - + -. So, the ranges when the expression is negative are: \(-\frac{1}{2}<x<0\) (2nd range) or \(x>\frac{1}{2}\) (4th range).

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 _________________

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

Show Tags

01 Nov 2012, 16:06

5

This post received KUDOS

Bunuel Thanx a trillion for your post on solving inequalities using graph You know i paid over 300$ to test prep institutes but got nothing out of it.......when i asked such basic question the tutor got frustrated and insulted me.....But hats off to you... MAx wat will i give 1 kudo...... Wat an expeirence it has been with GMAt club

Thanx a lot Bunuel

Trillion kudos to you and Hats off to you for addressing problems with patience..............I cant express myself how satisfied i am feeling.

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

Show Tags

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.

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. (-)(+)(+) = (-) (-)(-)(+) = (+) (-)(-)(-) = (-)
_________________

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)

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)
_________________

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

i solved it using similar approach but what i learned from this link : inequalities-trick-91482.html is that we should plot the number line starting with +ve on the right most segment and then changing the alternatively as we go from right to left but......

using this approach i'm getting

-1/2<x<0 , x>1/2

but there's no option like that its just the opp. please help!!

You are right about starting with a positive sign in the rightmost segment. But all factors should be of the form (ax+b) or (ax - b). Note that you have 1 - 2x which should be converted to 2x - 1. For that you multiply both sides by -1 and hence the inequality sign flips. You will get the correct answer.
_________________

In this case we have points -1/2 ,0, 1/2 So the sequence of signs should be -+-+ So the range should be x<-1/2 or 0<x<1/2 But the OA is different. where did i go wrong ?

Recall that if you are going to start with a positive sign from the rightmost region, the factors should be in the form (x - a) etc

(a - x) changes the entire thing.

x^3(1−2x)(1+2x)<0 has ( 1- 2x) which is 2*(1/2 - x). This is of the form (a - x).

You need to multiply the inequality by -1 here to get

x^3 * (2x - 1) * (1 + 2x) > 0

Now you will get the correct answer.
_________________

Karishma I flipped the sign before. So I got x^3(2x-1)(2x-1) > 0

2 cases - both +ve or both -ve

case 1 ------- x > 0 and |x| > 1/2. Hence x > 1/2

case 2 ------ x < 0 and 4x^2 - 1 < 0 x < 0 and -1/2 < x < 1/2 Taking the most restrictive value- -1/2 < x < 0

I hope this is correct. Btw this is 750 level in 2 mins.

VeritasPrepKarishma wrote:

subhashghosh wrote:

Hi Bunuel

I'm getting D as answer :

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

\(-ve --- -1/2---- +ve--- 0----- -ve-----1/2--- +ve\) Could you please explain where I'm wrong ?

Regards, Subhash

Even though your question is directed to Bunuel, I will give a quick explanation.

The concept of the rightmost section being positive is applicable when every term is positive in the rightmost region. This is the case whenever the expressions involved are of the form (x - a) or (ax - b) etc. When you have a term such as (1-2x), the rightmost region becomes negative. So either, as Bunuel mentioned, check for an extreme value of x or convert (1-2x) to (2x - 1) and flip the sign to >.

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...