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# 4 < x^2 + 2x + 1 < 16

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Intern
Joined: 26 Feb 2017
Posts: 28
Location: Brazil
GMAT 1: 610 Q45 V28
GPA: 3.11
4 < x^2 + 2x + 1 < 16 [#permalink]

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23 Oct 2017, 17:25
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Difficulty:

85% (hard)

Question Stats:

49% (02:08) correct 51% (01:23) wrong based on 176 sessions

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How many integers satisfy this inequality?
4 < x^2 + 2x + 1 < 16

A - 4
B - 3
C - 2
D - 1
E - 0
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Aug 2009
Posts: 5647
4 < x^2 + 2x + 1 < 16 [#permalink]

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23 Oct 2017, 19:07
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vitorpteixeira wrote:
How many integers satisfy this inequality?
4 < x^2 + 2x + 1 < 16

A - 4
B - 3
C - 2
D - 1
E - 0

hi..

so lets solve the equation..
$$4 < x^2 + 2x + 1 < 16......4<(x+1)^2<16......$$
we are looking for integer values..
(x+1) can take two values (x+1)=3 or x+1=-3......
so x=3-1=2 and x=-3-1=-4

so two values : 2 and -4
C
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

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Joined: 02 Apr 2014
Posts: 392
Re: 4 < x^2 + 2x + 1 < 16 [#permalink]

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27 Jan 2018, 01:42
4 < $$x^2 + 2x + 1$$ < 16
4 < $$(x+1)^2$$ < 16
taking square root,
2 < |x+1| < 4

if x > -1, then |x+1| = x + 1
2 < x+1 < 4 => only x = 2 satisfies

if x < -1, then |x+1| = -x - 1
2 < -x - 1 < 4 => 3 < -x < 5 => only x = -4 satisfies

so two possible integers of x {2,-4} => (C)
Re: 4 < x^2 + 2x + 1 < 16   [#permalink] 27 Jan 2018, 01:42
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