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# Find the range of values of x such that ((x+6) (3x-4)^3)/(x+2)^2 > 0

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Find the range of values of x such that ((x+6) (3x-4)^3)/(x+2)^2 > 0  [#permalink]

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Updated on: 31 Aug 2018, 01:47
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Solving inequalities- Number Line Method - Practice Question #4

Find the range of values of x such that $$\frac{((x+6) (3x-4)^3)}{(x+2)^2} >0$$

A. x > $$\frac{4}{3}$$
B. (-6, 4/3)
C. (-inf, 4/3) U (6, +inf) – {-2}
D. (-inf, -6 ) U(4/3, +inf)
E. (-inf, -6) U (4/3, +inf) – {-2}

To read the article: Solving inequalities- Number Line Method

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Originally posted by EgmatQuantExpert on 23 Aug 2018, 04:31.
Last edited by EgmatQuantExpert on 31 Aug 2018, 01:47, edited 7 times in total.
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Find the range of values of x such that ((x+6) (3x-4)^3)/(x+2)^2 > 0  [#permalink]

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Updated on: 31 Aug 2018, 01:48

Solution

Given:
An inequality,$$\frac{(x+6) (3x-4)^3}{(x+2)^2}$$>0

To find:
The range of x, that satisfies the above inequality

Approach and Working:
The first step is to multiply the inequality by $$(x+2)^2$$
Thus, we get,
$$\frac{(x+2)^2 (x+6) (3x-4)^3}{(x+2)^2} >0$$
Which implies, $$(x + 6) * (3x - 4)^3 > 0$$ and the denominator, x + 2 ≠ 0, implies, x≠ -2
This can be written as $$(x + 6) * (3x - 4) * (3x - 4)^2 > 0$$
Since, $$(3x-4)^2$$, is always positive, expect for x = 4/3
We can write the inequality as, (x + 6) * (3x - 4) > 0
The zero points of this inequality are, x = -6 and x = 4/3
Plot these points on the number line.
Since, 4/3 > -6, the inequality will be positive, for all the points to the right of 4/3, on the number line
And, it is negative, in the region, between -6 and 4/3
It is again positive for all the values less than -6

Therefore, the range of x is x > 4/3 and x < -6 and x ≠ -2

Hence, the correct answer is option D.

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Originally posted by EgmatQuantExpert on 23 Aug 2018, 04:32.
Last edited by EgmatQuantExpert on 31 Aug 2018, 01:48, edited 2 times in total.
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Re: Find the range of values of x such that ((x+6) (3x-4)^3)/(x+2)^2 > 0  [#permalink]

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23 Aug 2018, 06:21
EgmatQuantExpert wrote:
Solving inequalities- Number Line Method - Practice Question #4

Find the range of values of x such that $$\frac{((x+6) (3x-4)^3)}{(x+2)^2} >0$$

A. x > $$\frac{4}{3}$$
B. (-6, 4/3)
C. (-inf, 4/3) U (6, +inf) – {2}
D. (inf, -6 ) U(4/3, +inf)
E. (-inf, -6) U (4/3, +inf) – {2}

$$\frac{((x+6) (3x-4)^3)}{(x+2)^2} >0$$

Critical points:- -6, 4/3
Points that to be excluded from the range of x: -2

Using wavy curve method:
Range of x: x < -6 or x>4/3
Range of x in interval form:- (-inf, -6 ) U(4/3, +inf) (Note: x=-2 is already excluded)

Ans. (D)

Seems option (D) is a typo error. It should be -inf instead of inf.
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Re: Find the range of values of x such that ((x+6) (3x-4)^3)/(x+2)^2 > 0  [#permalink]

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23 Aug 2018, 11:25
Director
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Re: Find the range of values of x such that ((x+6) (3x-4)^3)/(x+2)^2 > 0  [#permalink]

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28 Aug 2018, 21:17
EgmatQuantExpert wrote:
Thanks PKN,
I have updated the options.

Regards,
Ashutosh
e-GMAT

Amendment done in the answer options:-
1) {+2} has been replaced with {-2} in answer options C & E.
2) No change in answer option D, it must be (-inf, -6 ) U(4/3, +inf), which makes sense.

Now correct answer option can't be E. Reason:-
1) As per official explanation the range of x: x > 4/3 and x < -6 and x ≠ -2

We always subtract or exclude points(or positions or ranges) when those points of a range are subset of a bigger range of points.
For example , If I say , set of real numbers except -2. It can be written as (-inf,+inf)-{-2}, since -2 is already present in {-inf,inf}, and we want the interval excluding 2.

In line with the above reasoning, option E is to be discarded.(how to exclude -2 where the interval (-inf, -6) U (4/3, +inf) doesn't possess the point -2. It's already excluded.)

I hope with a small correction in option D, we would arrive at the correct answer option. Please correct me.

Thanking you.

Hi EgmatQuantExpert,
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Re: Find the range of values of x such that ((x+6) (3x-4)^3)/(x+2)^2 > 0  [#permalink]

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31 Aug 2018, 01:52
Thanks PKN,
You are correct.
I apologise for typo in the answer options.

Regards,
Ashutosh
e-GMAT
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Joined: 10 Sep 2018
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Re: Find the range of values of x such that ((x+6) (3x-4)^3)/(x+2)^2 > 0  [#permalink]

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19 Sep 2018, 01:41
PKN wrote:
EgmatQuantExpert wrote:
Solving inequalities- Number Line Method - Practice Question #4

Find the range of values of x such that $$\frac{((x+6) (3x-4)^3)}{(x+2)^2} >0$$

A. x > $$\frac{4}{3}$$
B. (-6, 4/3)
C. (-inf, 4/3) U (6, +inf) – {2}
D. (inf, -6 ) U(4/3, +inf)
E. (-inf, -6) U (4/3, +inf) – {2}

$$\frac{((x+6) (3x-4)^3)}{(x+2)^2} >0$$

Critical points:- -6, 4/3
Points that to be excluded from the range of x: -2

Using wavy curve method:
Range of x:x < -6or x>4/3
Range of x in interval form:- (-inf, -6 ) U(4/3, +inf) (Note: x=-2 is already excluded)

Ans. (D)

Seems option (D) is a typo error. It should be -inf instead of inf.

Hi, can you please tell me how x will be less than -6?
I think x should be greater than -6. Can you tell me what I'm missing?
Director
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Find the range of values of x such that ((x+6) (3x-4)^3)/(x+2)^2 > 0  [#permalink]

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19 Sep 2018, 01:44
srijnasingh wrote:
PKN wrote:
EgmatQuantExpert wrote:
Solving inequalities- Number Line Method - Practice Question #4

Find the range of values of x such that $$\frac{((x+6) (3x-4)^3)}{(x+2)^2} >0$$

A. x > $$\frac{4}{3}$$
B. (-6, 4/3)
C. (-inf, 4/3) U (6, +inf) – {2}
D. (inf, -6 ) U(4/3, +inf)
E. (-inf, -6) U (4/3, +inf) – {2}

$$\frac{((x+6) (3x-4)^3)}{(x+2)^2} >0$$

Critical points:- -6, 4/3
Points that to be excluded from the range of x: -2

Using wavy curve method:
Range of x:x < -6or x>4/3
Range of x in interval form:- (-inf, -6 ) U(4/3, +inf) (Note: x=-2 is already excluded)

Ans. (D)

Seems option (D) is a typo error. It should be -inf instead of inf.

Hi, can you please tell me how x will be less than -6?
I think x should be greater than -6. Can you tell me what I'm missing?

Request to furnish your explanation on x>-6 so that I can share my reasoning with you.

I have used wavy curve method to determine the intervals.
_________________

Regards,

PKN

Rise above the storm, you will find the sunshine

Intern
Joined: 10 Sep 2018
Posts: 3
Re: Find the range of values of x such that ((x+6) (3x-4)^3)/(x+2)^2 > 0  [#permalink]

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19 Sep 2018, 02:06
PKN wrote:
EgmatQuantExpert wrote:
Solving inequalities- Number Line Method - Practice Question #4

Find the range of values of x such that $$\frac{((x+6) (3x-4)^3)}{(x+2)^2} >0$$

A. x > $$\frac{4}{3}$$
B. (-6, 4/3)
C. (-inf, 4/3) U (6, +inf) – {2}
D. (inf, -6 ) U(4/3, +inf)
E. (-inf, -6) U (4/3, +inf) – {2}

$$\frac{((x+6) (3x-4)^3)}{(x+2)^2} >0$$

Critical points:- -6, 4/3
Points that to be excluded from the range of x: -2

Using wavy curve method:
Range of x: x < -6 or x>4/3
Range of x in interval form:- (-inf, -6 ) U(4/3, +inf) (Note: x=-2 is already excluded)

Ans. (D)

Seems option (D) is a typo error. It should be -inf instead of inf.

I used the following reasoning:

(x+6) * (3x-4)^3 > 0

This means that both the terms, (x+6) and (3x-4)^3 should be positive.

so, x+6>0 and 3x-4>0 (Because only cube of a positive number can be positive)

so, x>-6 and 3x>4

or x>-6 and x>4/3

Is this reasoning incorrect?
Director
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Posts: 852
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Re: Find the range of values of x such that ((x+6) (3x-4)^3)/(x+2)^2 > 0  [#permalink]

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19 Sep 2018, 02:30
1
srijnasingh wrote:

I used the following reasoning:

(x+6) * (3x-4)^3 > 0

This means that both the terms, (x+6) and (3x-4)^3 should be positive.

so, x+6>0 and 3x-4>0 (Because only cube of a positive number can be positive)

so, x>-6 and 3x>4

or x>-6 and x>4/3

Is this reasoning incorrect?

Hi srijnasingh,

a*b>0 ,
There are 2 cases:
b) a<0 , b<0
We have to find out the common interval in which both the cases (a) and (b) are valid.

You may check, if x>-6 (say x=-5) then the expression $$(x+6)*(3x-4)^3$$=(+ve)*(-ve)=(-ve), which contradicts question stem.

You may try wavy-curve method explained thru below link:-
https://gmatclub.com/forum/wavy-line-me ... 24319.html
_________________

Regards,

PKN

Rise above the storm, you will find the sunshine

Re: Find the range of values of x such that ((x+6) (3x-4)^3)/(x+2)^2 > 0 &nbs [#permalink] 19 Sep 2018, 02:30
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