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


A. \(0 < |x| < \frac{1}{2}\)

B. \(|x| >\frac{1}{2}\)

C. \(–\frac{1}{2} < x < 0\) or \(\frac{1}{2} < x\)

D. \(x < –\frac{1}{2}\) or \(0 < x < \frac{1}{2}\)

E. \(x < –\frac{1}{2}\) or \(x > 0\)
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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.

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).

Hope its clear.
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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 >.
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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
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x > 0 and |x| > 1/2. Hence x > 1/2

case 2
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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.

|x| > 1/2 means that x<-1/2 or x>1/2.

The range you wrote is wrong also because x<-1/2 and x < 0 doesn't makes any sense.

Check Walker's post on absolute values for more: math-absolute-value-modulus-86462.html
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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.

Using the amazing technique:

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

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

If less than 0, select (-) curves.

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

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?