GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 29 Mar 2020, 13:51

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

(x^2+6x-7)/|x+4| < 0

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Intern
Intern
avatar
Joined: 30 Jun 2013
Posts: 4
(x^2+6x-7)/|x+4| < 0  [#permalink]

Show Tags

New post Updated on: 06 Dec 2017, 20:01
5
32
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

69% (02:38) correct 31% (02:48) wrong based on 465 sessions

HideShow timer Statistics

\(\frac{(x^2+6x-7)}{|x+4|} < 0\)

A) -7<x<-5 and -4<x<1
B) -7<x<-5 and -4<x<0
C) -7<x<-4 and -4<x<1
D) -7<x<-6 and -4<x<1

I am looking for a good way to approach inequality problems that involve absolute value ratios, as given above. Any help is greatly appreciated.

SOURCE: this is not from GMAT prep book; this problem is from a GRE practice book, which have many problems of this kind. I went through M-GMAT book inequalities chapter but haven't seen anything similar to this.

Thanks!

_________________
“Not everyone who works hard is rewarded, however all those who succeed have worked hard!” -Coach Kamogawa

Originally posted by Juggernaut23 on 19 Aug 2013, 21:23.
Last edited by Bunuel on 06 Dec 2017, 20:01, edited 4 times in total.
RENAMED THE TOPIC.
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62289
Re: (x^2+6x-7)/|x+4| < 0  [#permalink]

Show Tags

New post 20 Aug 2013, 01:30
9
12
\(\frac{(x^2+6x-7)}{|x+4|} < 0\)

A) -7<x<-5 and -4<x<1
B) -7<x<-5 and -4<x<0
C) -7<x<-4 and -4<x<1
D) -7<x<-6 and -4<x<1

Absolute value is always non-negative, thus \(\frac{(x^2+6x-7)}{|x+4|}=\frac{x^2+6x-7}{positive} < 0\) to hold true the numerator must be negative, keeping in mind that x cannot be -4 (to avoid division by zero).

\(x^2+6x-7<0\) --> \((x+7)(x-1)<0\).

The roots are -7 and 1 (check here: x2-4x-94661.html#p731476) --> "<" sign indicates that the solution lies between the roots: \(-7<x<1\) --> since x cannot be -4, then finally we have that \(-7<x<-4\) and \(-4<x<1\).

Answer: C.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html
inequations-inequalities-part-154664.html
inequations-inequalities-part-154738.html

Hope it helps.
_________________
General Discussion
MBA Section Director
User avatar
V
Affiliations: GMAT Club
Joined: 21 Feb 2012
Posts: 7498
City: Pune
GMAT ToolKit User
Re: (x^2+6x-7)/|x+4| < 0  [#permalink]

Show Tags

New post 19 Aug 2013, 22:42
6
6
Answer Should be C

First of all, equation within Modulus can not be negative. So |x+4| has to be positive.

Now if \(\frac{a}{b}\) is negative and b is positive, then a must be negative. i.e. \(x^2 + 6x - 7 < 0\)

So we have that \(x^2 + 6x - 7 < 0\) --------> (x+7)(x-1) < 0 -------> Now we get three ranges of x
1) x < -7
2) -7 < x < 1
3) x > 1
Here we can see that the range mentioned in serial number 2 satisfies the inequality (x+7)(x-1) < 0. So -7 < x < 1

Now let's work on denominator. We know that |x+4| > 0 --------> x > -4 or x < -4 that means x is not equal to -4 -------[ |x-a| > r then x > a+r or x < a-r ]

Consider both the ranges together
-7 < x < 1 and x is not equal to -4
So -7 < x < -4 and -4 < x < 1

Should you want to review basics concepts of Inequations and Modules, Visit my this Article. And this too.

Hope that helps
_________________
2020 MBA Applicants: Introduce Yourself Here!

MBA Video Series - Video answers to specific components and questions about MBA applications.

2020 MBA Deadlines, Essay Questions and Analysis of all top MBA programs
Manager
Manager
avatar
Joined: 21 Aug 2012
Posts: 97
Re: (x^2+6x-7)/|x+4| < 0  [#permalink]

Show Tags

New post 19 Aug 2013, 23:23
Narenn wrote:
Answer Should be C

First of all, equation within Modulus can not be negative. So |x+4| has to be positive.

Now if \(\frac{a}{b}\) is negative and b is positive, then a must be negative. i.e. \(x^2 + 6x - 7 < 0\)

So we have that \(x^2 + 6x - 7 < 0\) --------> (x+7)(x-1) < 0 -------> Now we get three ranges of x
1) x < -7
2) -7 < x < 1
3) x > 1
Here we can see that the range mentioned in serial number 2 satisfies the inequality (x+7)(x-1) < 0. So -7 < x < 1

Now let's work on denominator. We know that |x+4| > 0 --------> x > -4 or x < -4 that means x is not equal to -4 -------[ |x-a| > r then x > a+r or x < a-r ]

Consider both the ranges together
-7 < x < 1 and x is not equal to -4
So -7 < x < -4 and -4 < x < 1

Should you want to review basics concepts of Inequations and Modules, Visit my this Article. And this too.

Hope that helps


Hi Narenn,

Here we can see that the range mentioned in serial number 2 satisfies the inequality (x+7)(x-1) < 0. So -7 < x < 1

I have a doubt.

Now, (x+7)(x-1) < 0
CASE 1: x+7 is +ve and x-1 is negative.

x+7 >0 and x -1 <0
x > -7 and x< 1

CASE 2: x+7 is -ve and x-1 is +ve

x+7 <0 and x-1 >0
x< -7 and x> 1

Why haven't you consider the second case???

Thanks,
Jai
_________________
MODULUS Concept ---> http://gmatclub.com/forum/inequalities-158054.html#p1257636
HEXAGON Theory ---> http://gmatclub.com/forum/hexagon-theory-tips-to-solve-any-heaxgon-question-158189.html#p1258308
MBA Section Director
User avatar
V
Affiliations: GMAT Club
Joined: 21 Feb 2012
Posts: 7498
City: Pune
GMAT ToolKit User
Re: (x^2+6x-7)/|x+4| < 0  [#permalink]

Show Tags

New post 20 Aug 2013, 09:16
1
jaituteja wrote:
Narenn wrote:
Answer Should be C

First of all, equation within Modulus can not be negative. So |x+4| has to be positive.

Now if \(\frac{a}{b}\) is negative and b is positive, then a must be negative. i.e. \(x^2 + 6x - 7 < 0\)

So we have that \(x^2 + 6x - 7 < 0\) --------> (x+7)(x-1) < 0 -------> Now we get three ranges of x
1) x < -7
2) -7 < x < 1
3) x > 1
Here we can see that the range mentioned in serial number 2 satisfies the inequality (x+7)(x-1) < 0. So -7 < x < 1

Now let's work on denominator. We know that |x+4| > 0 --------> x > -4 or x < -4 that means x is not equal to -4 -------[ |x-a| > r then x > a+r or x < a-r ]

Consider both the ranges together
-7 < x < 1 and x is not equal to -4
So -7 < x < -4 and -4 < x < 1

Should you want to review basics concepts of Inequations and Modules, Visit my this Article. And this too.

Hope that helps


Hi Narenn,

Here we can see that the range mentioned in serial number 2 satisfies the inequality (x+7)(x-1) < 0. So -7 < x < 1

I have a doubt.

Now, (x+7)(x-1) < 0
CASE 1: x+7 is +ve and x-1 is negative.

x+7 >0 and x -1 <0
x > -7 and x< 1

CASE 2: x+7 is -ve and x-1 is +ve

x+7 <0 and x-1 >0
x< -7 and x> 1

Why haven't you consider the second case???

Thanks,
Jai


We are looking for the value(s) of x that would satisfy the the inequality (x+7)(x-1)<0

Now we know that (x+7)(x-1) will be negative when any one of them be negative and other one be positive. For this to happen we should find the value(s) of x that would simultaneously make one positive and other one negative.

We got two critical points (or check points, or reference points, whatever you call to it) as -7 and 1. We should carry out our search for the right value considering these two reference points.

-Infinity,........-13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, .......... Infinity

Consider the right side of the number line. You will notice that every value beyond critical point 1 makes both the equations positive. So the values beyond 1 do not agree with the inequality and hence can not be the solution of it.

Similarly every value beyond -7 towards negative side makes both the equations negative, thereby making the multiplication (x+7)(x-1) positive. So these values also can not be the solution of the inequality.

Whereas every value between -7 and 1 makes one of the equation positive and at the same time other one negative. So these values can satisfy the inequality and hence are the solution.

You will also notice that the reverse is true when the inequation has greater than(>) sign. That means in such case values between the critical points will not satisfy the inequality and those beyond the critical points would.

You can generalize this as for the quadratic inequation of the form ax^2 + bx + c < 0 the solution lies between the roots and for the quadratic inequation of the form ax^2 + bx + c > 0 the solution lies beyond the roots
(Note :- There is one exception for this rule that is in quadratic equation ax^2 + bx + c < or > 0, if discriminant i.e. b^2 - 4ac is equal to zero then the quadratic equation will give only one root (Critical Point). All values of x except for root(critical point) will satisfy the inequality ax^2 + bx + c > 0 and the inequality ax^2 + bx + c < 0 will not have any solution. Check this example.

Further to understand quadratic inequalities in depth try these methods proposed by some legendary members
BUNUEL
Veritas Karishma
Zarrolou

Hope that Helps!
_________________
2020 MBA Applicants: Introduce Yourself Here!

MBA Video Series - Video answers to specific components and questions about MBA applications.

2020 MBA Deadlines, Essay Questions and Analysis of all top MBA programs
Intern
Intern
avatar
Joined: 30 Jun 2013
Posts: 4
Re: (x^2+6x-7)/|x+4| < 0  [#permalink]

Show Tags

New post 20 Aug 2013, 11:09
Thank you all. The replies are helpful..I will review the posts quoted here first before solving more problems.
_________________
“Not everyone who works hard is rewarded, however all those who succeed have worked hard!” -Coach Kamogawa
Manager
Manager
User avatar
Status: folding sleeves up
Joined: 26 Apr 2013
Posts: 120
Location: India
Concentration: Finance, Strategy
GMAT 1: 530 Q39 V23
GMAT 2: 560 Q42 V26
GPA: 3.5
WE: Consulting (Computer Hardware)
Re: (x^2+6x-7)/|x+4| < 0  [#permalink]

Show Tags

New post 26 Mar 2014, 09:22
4
I think this question can be solved easily with diagram

HTH
Attachments

image.jpg
image.jpg [ 552.47 KiB | Viewed 8599 times ]

Director
Director
User avatar
G
Joined: 23 Jan 2013
Posts: 512
Schools: Cambridge'16
Re: (x^2+6x-7)/|x+4| < 0  [#permalink]

Show Tags

New post 05 May 2016, 03:19
(x+7)*(x-1)<0

x<-7 AND x>1=Nothing

x>-7 AND x<1=-7<x<1

test denominator for 0: only -4 gives |x+4|=0

so, -7<x<-4 AND -4<x<1
Intern
Intern
avatar
Joined: 01 May 2015
Posts: 36
Re: (x^2+6x-7)/|x+4| < 0  [#permalink]

Show Tags

New post 16 Jun 2016, 05:40
I have a question. Since we know that |x+4| is always positive, why can't we multiply both sides of the inequality by |x+4|, to arrive at the following:

x^2 + 6x - 7 < 0
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62289
Re: (x^2+6x-7)/|x+4| < 0  [#permalink]

Show Tags

New post 16 Jun 2016, 05:53
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 14448
Re: (x^2+6x-7)/|x+4| < 0  [#permalink]

Show Tags

New post 11 Jun 2019, 19:26
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Bot
Re: (x^2+6x-7)/|x+4| < 0   [#permalink] 11 Jun 2019, 19:26
Display posts from previous: Sort by

(x^2+6x-7)/|x+4| < 0

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne