vdadwal wrote:

If x is an integer, is |x|>1.

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

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

Can somebody please explain this question?

Thanks

Vikram

This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach":

x2-4x-94661.html#p731476If x is an integer, is |x| > 1?First of all: is

|x| > 1 means is

x<-1 (-2, -3, -4, ...) or

x>1 (2, 3, 4, ...), so for YES answer

x can be any integer but -1, 0, and 1.

(1) (1 - 2x)(1 + x) < 0 --> rewrite as

(2x-1)(x+1)>0 (so that the coefficient of x^2 to be positive after expanding): roots are

x=-1 and

x=\frac{1}{2} --> "

>" sign means that the given inequality holds true for:

x<-1 and

x>\frac{1}{2}.

x could still equal to 1, so not sufficient.

(2) (1 - x)(1 + 2x) < 0 --> rewrite as

(x-1)(2x+1)>0: roots are

x=-\frac{1}{2} and

x=1 --> "

>" sign means that the given inequality holds true for:

x<-\frac{1}{2} and

x>1.

x could still equal to -1, so not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is

x<-1 and

x>1. Sufficient.

Answer: C.

This question is also discussed here:

m14-72785.htmlSolving inequalities:

x2-4x-94661.html#p731476 (

check this one first)

inequalities-trick-91482.htmldata-suff-inequalities-109078.htmlrange-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535everything-is-less-than-zero-108884.html?hilit=extreme#p868863Hope it helps.

I have a question regarding the above solution let's say in statement 1, when you solve the inequality why do you say that x<-1 AND x >1/2

why is this an AND condition ....why not OR? If this were a quadratic equation x (can be) = 1/2 OR -1 OR both