Confusing Inequality.

**Seth** · 11-27-2006, 03:41 PM

Is |x-2|+|x+2|<4?
1). x<0
2). x is within (-2, 2)

SPOILER: D

I disagree with Official Answer, IMO it's

SPOILER: B

What do you think?

opps Sorry, it's wrong forum.

**JustMe123** · 11-27-2006, 06:45 PM

D is right.

1) is sufficient b/c
for x < 0, x can be such that
a) -2 < x < 2 so |x-2|+|x+2|=-(x-2)+x+2==x+2+x+2 = 4
b) x < -2 so |x-2|+|x+2|= -(x-2)+-(x+2)= -2x > 4 (remember x is < -2)
Since it's either = 4 or > 4 is never less than 4.

**r1234** · 11-27-2006, 09:48 PM

I think the Official Answer is wrong. It should be B. I think just123 can help here.

just123,

can you please explain this?

b) x < -2 so |x-2|+|x+2|= -(x-2)+-(x+2)= -2x > 4 (remember x is < -2)

If I pluggin -4 for x I get,

|-4 - 2| + |-4+2| => |-6| + |-2| => 6+2 => 8 which is not less than 4.

Thanks!

**JustMe123** · 11-28-2006, 01:01 AM

Originally Posted by r1234

b) x < -2 so |x-2|+|x+2|= -(x-2)+-(x+2)= -2x > 4 (remember x is < -2)

If I pluggin -4 for x I get,

|-4 - 2| + |-4+2| => |-6| + |-2| => 6+2 => 8 which is not less than 4.

Thanks!

Hi,
Since it's a yes/no problem, statement 1 (x<0) is considered sufficient if we can show that for all x < 0, the answer to this statement "Is |x-2|+|x+2|<4?" is either always "yes" or always "no." In this case, we've shown |x-2|+|x+2| can never be less than 4 (i.e. either 4 or greater). So this statement is considered sufficient.

Question: Is |x-2|+|x+2|<4?
condition: x < 0
x = -1/2, |-1/2-2|+|-1/2+2| = 4 <-- answer = no
x = -1, |-1-2|+|-1+2| = 4 <-- answer = no
x = -2, |-2-2|+|-2+2| = 4 <-- answer = no
x = -3, |-3-2|+|-3+2| = 6 <-- answer = no
x = -4, |-4-2|+|-4+2| = 8 <-- answer = no
...
x = -8, |-8-2|+|-8+2| = 16 <-- answer = no

**r1234** · 11-28-2006, 05:24 AM

Great! Thanks justme123. Appreciate your reply.

**Seth** · 11-28-2006, 08:19 AM

Ummm, it's my careless.

Anyway, thanks a lot, JustMe123

**nhy02** · 12-01-2006, 09:46 PM

hi just me123
2 is sufficient since it means x>-2 and x<2
x>-2 ==>x+2>0==>absolute x+2=x+2
x<2==>x-2<0==>abs x-2= -x+2
so when we add the absolute values ==> we get -x+2+x+2=4
which is not less than 4 so B is correct

**JustMe123** · 12-01-2006, 10:00 PM

Originally Posted by nhy02

hi just me123
2 is sufficient since it means x>-2 and x<2
x>-2 ==>x+2>0==>absolute x+2=x+2
x<2==>x-2<0==>abs x-2= -x+2
so when we add the absolute values ==> we get -x+2+x+2=4
which is not less than 4 so B is correct

Hello. Read my answer. I picked D.
D means either is sufficient by itself - A is sufficient, so is B sufficient by itself.

**profiler01** · 12-01-2006, 10:19 PM

I think the best strategy is picking numbers.
1) x < 0. So x can be -0.5 or -1 or -2 or - 3 per example.
-0.5 ==> |-0.5-2| + |-0.5+2| = 2.5 + 1.5 = 4
-1 ==> |-1-2| + |-1+2| = 3 + 1 = 4
-2 ==> |-2-2| + |-2+2| = 4 + 1 = 4
-3 ==> |-3-2| + |-3+2| = 5 + 1 = 6
So x <0 mean |x-2|+|x+2| >= 4. Answer is NO. Thus 1 is ok

2) x is within (-2, 2)
-0.5 ==> |-0.5-2| + |-0.5+2| = 2.5 + 1.5 = 4
-1 ==> |-1-2| + |-1+2| = 3 + 1 = 4
-2 ==> |-2-2| + |-2+2| = 4 + 1 = 4
1 ==> |1-2| + |1+2| = 1 + 3 = 4
-2 ==> |2-2| + |2+2| = 0 + 1 = 4
So x is within (-2, 2) mean |x-2|+|x+2| >= 4. Answer is NO. Thus 2 is ok

So answer = D

**nhy02** · 12-02-2006, 01:00 AM

yes the answer is D,that is why one has to continue till the end always

trying to sollve it mathematical way

x<0==>result will be -x+2+abs(x+2)
if -2<x<0 then it will give 4
if x<-2 then it will give -2x>4
so it will always give 4 or above
vey tricky but nice