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

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Intern
Joined: 26 Feb 2017
Posts: 29
Location: Brazil
GMAT 1: 610 Q45 V28
GPA: 3.11
4 < x^2 + 2x + 1 < 16  [#permalink]

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23 Oct 2017, 18:25
1
16
00:00

Difficulty:

85% (hard)

Question Stats:

51% (01:54) correct 49% (01:51) wrong based on 307 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
Math Expert
Joined: 02 Aug 2009
Posts: 8331
4 < x^2 + 2x + 1 < 16  [#permalink]

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23 Oct 2017, 20:07
1
2
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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Re: 4 < x^2 + 2x + 1 < 16  [#permalink]

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27 Jan 2018, 02:42
1
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)
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Joined: 12 Apr 2019
Posts: 365
Re: 4 < x^2 + 2x + 1 < 16  [#permalink]

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13 Jun 2019, 07:26
First things first, I think the options here mean 4,3, 2, 1, and 0 IMHO and not -4,-3,-2,-1 and -0 (there's no meaning to -0, that's a clear clue). The hyphens could have been left out!

This is not a difficult question, conceptually. Since both the extremities of the inequality are positive numbers, we can take the square root directly.
But, because the term in the middle is a perfect square, we need to consider both positive and negative values.

$$x^2$$ + 2x + 1 = $$(x+1)^2$$. Therefore, we can say, 4 < $$(x+1)^2$$ < 16. Taking the square root of all terms of the inequality,

2 < (x+1) < 4 OR -4 < (x+1) < -2.

Subtracting 1 from both sides of the inequality, we have 1 < x < 3 OR -5 < x < -3. So, there are two values that x can take i.e. x = 2 OR x = -4.

An easy way of verifying your answer is to plug these values back into the inequality given in the question statement and see if it satisfies.

The correct answer option is C.

Hope this helps!
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Re: 4 < x^2 + 2x + 1 < 16  [#permalink]

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23 Nov 2019, 01:02
we get 2 equations:

x^2 + 2x - 3 >0 & X^2 + 2x - 15 <0

Solving these 2 equations,

We get, X<-3, X>1 & (-5,3).

The integers that satisfy this equation are -4 & 2.

Ans C
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Re: 4 < x^2 + 2x + 1 < 16  [#permalink]

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23 Nov 2019, 01:20
vitorpteixeira wrote:
How many integers satisfy this inequality?
4 < x^2 + 2x + 1 < 16

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

4 < x^2 + 2x + 1 < 16
4< (x+1)^2 < 16
Taking square root

2< |x+1 | < 4

Combined solution set is (-5,-3) and (1,3). Only possible solution are -2, 2
Hence C is correct.
Re: 4 < x^2 + 2x + 1 < 16   [#permalink] 23 Nov 2019, 01:20
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