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Re: Everything is Less Than Zero [#permalink]
13 Feb 2011, 21:05

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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 >. _________________

Re: Everything is Less Than Zero [#permalink]
10 Feb 2011, 02:28

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144144 wrote:

Thanks Bunuel. +1

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).

Re: Which of the following represents the complete range of x [#permalink]
01 Nov 2012, 15:06

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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: Everything is Less Than Zero [#permalink]
20 Jun 2012, 21:42

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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

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]
01 Aug 2013, 23:21

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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.

Re: Which of the following represents the complete range of x [#permalink]
15 Nov 2012, 18:03

1

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Expert's post

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

Re: Which of the following represents the complete range of x [#permalink]
09 Jul 2013, 20:08

1

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Expert's post

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)

Re: Everything is Less Than Zero [#permalink]
02 Mar 2011, 17:40

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 >.

Re: Which of the following represents the complete range of x [#permalink]
15 Nov 2012, 12:58

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.

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