Which of the following represents the complete range of x

hi Karshima and Bunuel,
One part of inequality is:(1+2x)<0

How did you guys convert into the answer part: x >-1/2, I am getting x <-1/2...would really appreciate if you could reply. Thanks in advance

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!!

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 ?

Hi Bunuel VeritasKarishma , I solved this question in this manner:-


1. x^3 - 4x^2<0
= x^3<4x^3

Now, when x>0

dividing both sides by x^3
1<4x^2 which means x>(1/2)

in the same way, when x<0

dividing both sides by x^3
1>4x^2 which means x<(1/2) but since x is negative, the range shall be x<0

from both these ranges, my final answer was x>(1/2) and x<0

Could you please tell me what is wrong in my approach?
Thank you!

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 >.

KarishmaB
I have been tying hard to find a detailed explanation of this concept on google to understadn the mechanics intuitively however I have not been been able to. Could you pls share the relavant link explaining this in detail. COuld be a blog or video. Also will appreciate if you can help with other queries too.

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.

Hi,

While I partially understand the wavy line method to solve inequalities, can you help me understand why do we equate the expressions to zero to get the roots? This is an inequality right? And since it is <0, one of them needs to be negative and 2 of them positive. Please help me fill the gap here.

Bunuel chetan2u KarishmaB

Basically we are asked to find the range of \(x\) for which \(x^3-4x^5<0\) is true.

\(x^3-4x^5<0\);

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

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

"Roots" are -1/2, 0, and 1/2: \(-\frac{1}{2}<x<0\) or \(x>\frac{1}{2}\).

Answer: C.

Check this for more: https://gmatclub.com/forum/inequalities ... 91482.html