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

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(x^2+6x-7)/|x+4| < 0 [#permalink] New post 19 Aug 2013, 21:23
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70% (02:57) correct 30% (02:03) wrong based on 152 sessions
\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!
[Reveal] Spoiler: OA

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Last edited by Bunuel on 27 Mar 2014, 07:40, edited 3 times in total.
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Re: Inequality with absolute values [#permalink] New post 19 Aug 2013, 22:42
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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
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Re: Inequality with absolute values [#permalink] 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
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MODULUS Concept ---> inequalities-158054.html#p1257636
HEXAGON Theory ---> hexagon-theory-tips-to-solve-any-heaxgon-question-158189.html#p1258308

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Re: (x^2+6x-7)/|x+4| < 0 [#permalink] New post 20 Aug 2013, 01:30
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\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.
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Re: Inequality with absolute values [#permalink] New post 20 Aug 2013, 09:16
Expert's post
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!
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Re: Inequality with absolute value problem. [#permalink] 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.
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Re: Inequality with absolute value problem. [#permalink] New post 26 Mar 2014, 09:22
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I think this question can be solved easily with diagram

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Re: Inequality with absolute value problem.   [#permalink] 26 Mar 2014, 09:22
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