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# inite range of values of "x"

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Joined: 06 Apr 2010
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inite range of values of "x" [#permalink]  03 Sep 2010, 06:09
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Question Stats:

71% (02:17) correct 29% (00:00) wrong based on 7 sessions
Which of the following inequalities have a finite range of values of "x" satisfying them?
A. x^2 + 5x + 6 > 0
B. |x + 2| > 4
C. 9x - 7 < 3x + 14
D.x^2 - 4x + 3 < 0
E. (B) and (D)
[Reveal] Spoiler: OA
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Re: inite range of values of "x" [#permalink]  03 Sep 2010, 06:53
A. X^2 + 5X >-6 you can see that X can be all positive including 0 so not finite
B. this shows X has to be >3 and <-7 so again not finite
C. move terms around gets 6x < 21 so x can be any negative value and couple of positives so infinite
D. x^2 - 4x <-3 You can readily see that X cant be negative since the -4X and x^2 results in positives thus invalidates the inequality. Try plugin couple of numbers - 0 doesnt work since 0<-3 is invalid. X=1 gets you -3<-3 so no. x=2 gets you 4-8 = -4<-3 which works x=3 gets 9-12=-3 <-3 so no. Stop here and you see x can be only 2

so D
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Re: inite range of values of "x" [#permalink]  03 Sep 2010, 06:53
Lets get started with this, choice by choice:

Choice I: x^2 + 5x + 6 > 0
Break this one: (x+2)(x+3) > 0. So either x>-2 or x<-3
So infinite values of x.

Choice II:|x + 2| > 4
Break this one: either x+2>4 or x+2<-4 i.e x>2 or x<-6
So infinite values of x.

Choice III:9x - 7 < 3x + 14
Break this one: x< 3.33
So infinite values of x.

Choice IV:x^2 - 4x + 3 < 0
Break this one: (x-3)(x-1)<0. So x is between 1<x<3
So here x have finite values.

Re: inite range of values of "x"   [#permalink] 03 Sep 2010, 06:53
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