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

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Intern
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Joined: 26 Feb 2017
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GMAT 1: 610 Q45 V28
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4 < x^2 + 2x + 1 < 16  [#permalink]

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New post 23 Oct 2017, 18:25
1
13
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

52% (01:55) correct 48% (01:56) wrong based on 250 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
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Joined: 02 Aug 2009
Posts: 7949
4 < x^2 + 2x + 1 < 16  [#permalink]

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New post 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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New post 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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Re: 4 < x^2 + 2x + 1 < 16  [#permalink]

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New post 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] 13 Jun 2019, 07:26
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