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Find the range of values of x such that ((x+6) (3x4)^3)/(x+2)^2 > 0
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Updated on: 31 Aug 2018, 01:47
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Solving inequalities Number Line Method  Practice Question #4Find the range of values of x such that \(\frac{((x+6) (3x4)^3)}{(x+2)^2} >0\) 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} Previous Question To read the article: Solving inequalities Number Line MethodTo read all our articles:Must Read articles to reach Q51
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Find the range of values of x such that ((x+6) (3x4)^3)/(x+2)^2 > 0
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Updated on: 31 Aug 2018, 01:48
Solution Given: An inequality,\(\frac{(x+6) (3x4)^3}{(x+2)^2}\)>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,
\(\frac{(x+2)^2 (x+6) (3x4)^3}{(x+2)^2} >0\) 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, \((3x4)^2\), 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. Answer: D
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Re: Find the range of values of x such that ((x+6) (3x4)^3)/(x+2)^2 > 0
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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 \(\frac{((x+6) (3x4)^3)}{(x+2)^2} >0\)
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}
\(\frac{((x+6) (3x4)^3)}{(x+2)^2} >0\) 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) (3x4)^3)/(x+2)^2 > 0
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23 Aug 2018, 11:25



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Re: Find the range of values of x such that ((x+6) (3x4)^3)/(x+2)^2 > 0
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28 Aug 2018, 21:17
EgmatQuantExpert wrote: Thanks PKN, I have updated the options. Regards, Ashutosh eGMATAmendment 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 ≠ 2We 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, Please guide.
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Re: Find the range of values of x such that ((x+6) (3x4)^3)/(x+2)^2 > 0
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31 Aug 2018, 01:52



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Re: Find the range of values of x such that ((x+6) (3x4)^3)/(x+2)^2 > 0
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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 \(\frac{((x+6) (3x4)^3)}{(x+2)^2} >0\)
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}
\(\frac{((x+6) (3x4)^3)}{(x+2)^2} >0\) 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?



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Find the range of values of x such that ((x+6) (3x4)^3)/(x+2)^2 > 0
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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 \(\frac{((x+6) (3x4)^3)}{(x+2)^2} >0\)
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}
\(\frac{((x+6) (3x4)^3)}{(x+2)^2} >0\) 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.
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Re: Find the range of values of x such that ((x+6) (3x4)^3)/(x+2)^2 > 0
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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 \(\frac{((x+6) (3x4)^3)}{(x+2)^2} >0\)
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}
\(\frac{((x+6) (3x4)^3)}{(x+2)^2} >0\) 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) * (3x4)^3 > 0 This means that both the terms, (x+6) and (3x4)^3 should be positive. so, x+6>0 and 3x4>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?



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Re: Find the range of values of x such that ((x+6) (3x4)^3)/(x+2)^2 > 0
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19 Sep 2018, 02:30
srijnasingh wrote: I used the following reasoning:
(x+6) * (3x4)^3 > 0
This means that both the terms, (x+6) and (3x4)^3 should be positive.
so, x+6>0 and 3x4>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: a) a>0, b>0 (Your reasoning) 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 \((x+6)*(3x4)^3\)=(+ve)*(ve)=(ve), which contradicts question stem. You may try wavycurve method explained thru below link: https://gmatclub.com/forum/wavylineme ... 24319.html
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Re: Find the range of values of x such that ((x+6) (3x4)^3)/(x+2)^2 > 0
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