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
For inequality why is the same thing an AND as opposed to OR?
close
64% (04:03) correct
