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Wavy Line Method Application  Exercise Question #7
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Updated on: 07 Aug 2018, 03:55
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Re: Wavy Line Method Application  Exercise Question #7
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26 Aug 2016, 10:17
EgmatQuantExpert wrote: Wavy Line Method Application  Exercise Question #7Find the product of the integer values of x that satisfy the inequality \(\frac{(x^24)^3}{(x5)^5 (x^2  9)^4} < 0\) Wavy Line Method Application has been explained in detail in the following post:: http://gmatclub.com/forum/wavylinemethodapplicationcomplexalgebraicinequalities224319.htmlDetailed solution will be posted soon. The inequality is \(\frac{(x^24)^3}{(x5)^5 (x^2  9)^4} < 0\) or \(\frac{(x2)^3 (x+2)^3}{(x5)^5 (x  3)^4 (x + 3)^4} < 0\) Now, we have (x3) and (x+3) with EVEN powers, so we will not move the curve direction to the upper side. So, the zero points of x will be 3,2,2,3 and 5 But we cannot have x = 5,3 or 3 as they are at the denominator. So, the range of x will be (infinity, 2) U (2,5). Please correct me if I am missing anything.
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Re: Wavy Line Method Application  Exercise Question #7
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26 Oct 2016, 18:09
My answer is {3<x<2} U {2<x<3}U{x>5}.
Not sure if I did it correctly. Please correct me if I am wrong



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Re: Wavy Line Method Application  Exercise Question #7
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29 Oct 2016, 11:32
EgmatQuantExpert wrote: Wavy Line Method Application  Exercise Question #7Find the product of the integer values of x that satisfy the inequality \(\frac{(x^24)^3}{(x5)^5 (x^2  9)^4} < 0\) Wavy Line Method Application has been explained in detail in the following post:: http://gmatclub.com/forum/wavylinemethodapplicationcomplexalgebraicinequalities224319.htmlDetailed solution will be posted soon. [(x2)(x+2)]^3 / (x5)^5 ( x^29)^4 critical values are 2,2,5, the inequality is valid in ranges x<3 and 2<x<5 , x not equal 3 , thus +ve integer value of x is 4 , and the ve integer values belong to (infinity,4) how can i get the product then or where am i going wrong??



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Re: Wavy Line Method Application  Exercise Question #7
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18 Nov 2016, 03:33
SolutionHey Everyone, Please find below, the solution of the given problem. Rewriting the inequality to easily identify the zero pointsWe know that \((x^2 – 4) = (x+2)(x2)\) And similarly, \((x^29) = (x+3)(x3)\) So, \((x^24)^3/((x – 5)^5 (x^29)^4 )>0\) can be written as \(((x+2)^3 (x2)^3)/((x – 5)^5 (x+3)^4 (x3)^4 )>0\) (x + 3) and (x3) are the terms with even power. So, the wavy line should bounce back at the corresponding zeropoints x = 3 and x = 3. Plotting the zero points and drawing the wavy line:Required Range: 2< x < 2 or x > 5 Calculations: We need the product of all integers in this range. However, note that the integer 0 is a part of the range (2 < x < 2). Therefore, the product will be 0. Correct Answer: Option C
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Re: Wavy Line Method Application  Exercise Question #7
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27 Nov 2016, 11:51
EgmatQuantExpert wrote: SolutionHey Everyone, Please find below, the solution of the given problem. Rewriting the inequality to easily identify the zero pointsWe know that \((x^2 – 4) = (x+2)(x2)\) And similarly, \((x^29) = (x+3)(x3)\) So, \((x^24)^3/((x – 5)^5 (x^29)^4 )>0\) can be written as \(((x+2)^3 (x2)^3)/((x – 5)^5 (x+3)^4 (x3)^4 )>0\) (x + 3) and (x3) are the terms with even power. So, the wavy line should bounce back at the corresponding zeropoints x = 3 and x = 3. Plotting the zero points and drawing the wavy line:Required Range: 2< x < 2 or x > 5 Calculations: We need the product of all integers in this range. However, note that the integer 0 is a part of the range (2 < x < 2). Therefore, the product will be 0. Correct Answer: Option C just had a question to solidify my understanding of this  if the question was < instead of >, then there would be no ranges that satisfy this equation? Thank you.



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Re: Wavy Line Method Application  Exercise Question #7
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27 Nov 2016, 13:04
OreoShake wrote: EgmatQuantExpert wrote: SolutionHey Everyone, Please find below, the solution of the given problem. Rewriting the inequality to easily identify the zero pointsWe know that \((x^2 – 4) = (x+2)(x2)\) And similarly, \((x^29) = (x+3)(x3)\) So, \((x^24)^3/((x – 5)^5 (x^29)^4 )>0\) can be written as \(((x+2)^3 (x2)^3)/((x – 5)^5 (x+3)^4 (x3)^4 )>0\) (x + 3) and (x3) are the terms with even power. So, the wavy line should bounce back at the corresponding zeropoints x = 3 and x = 3. Plotting the zero points and drawing the wavy line:Required Range: 2< x < 2 or x > 5 Calculations: We need the product of all integers in this range. However, note that the integer 0 is a part of the range (2 < x < 2). Therefore, the product will be 0. Correct Answer: Option C just had a question to solidify my understanding of this  if the question was < instead of >, then there would be no ranges that satisfy this equation? Thank you. Hey, If the question was less than (<) instead of greater than(>), then we can still find the range of the range of x. Look at the wavy curve in the solution given above and think, if we had to find the range of x for which the expression is less than zero, what range will we consider? Hint: We had considered the positive range in the wavy curve to find the range of x for which the expression was more than zero. Thanks, Saquib Quant Expert eGMAT
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Re: Wavy Line Method Application  Exercise Question #7
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04 Dec 2016, 02:45
[/quote] Hey, If the question was less than (<) instead of greater than(>), then we can still find the range of the range of x. Look at the wavy curve in the solution given above and think, if we had to find the range of x for which the expression is less than zero, what range will we consider? Hint: We had considered the positive range in the wavy curve to find the range of x for which the expression was more than zero. Thanks, Saquib Quant Expert eGMAT[/quote] Hi Saquib, Thank you for your prompt response. If we look at the graph, the for negatives we will get the ranges  x<3, 2>x>3, 3>x>2, and 5>x>3. Here the product of the integer values cannot be calcuated. Would these be correct ranges if the question asked for negative? Please confirm. Thanks a ton. Oreo



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Re: Wavy Line Method Application  Exercise Question #7
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04 Dec 2016, 05:27
EgmatQuantExpert wrote: Find the product of the integer values of x that satisfy the inequality \(\frac{(x^24)^3}{(x5)^5 (x^2  9)^4} > 0\)
We could solve this question easily with the tool called: factor table with sign\(\frac{(x^24)^3}{(x5)^5 (x^2  9)^4} > 0 \iff \frac{(x2)^3(x+2)^3}{(x5)^5(x3)^4(x+3)^4} > 0\) First, find the value that all factors are equal to 0 \(\begin{split} x2=0 &\iff x=2 \\ x+2=0 &\iff x=2\\ x5=0 &\iff x=5\\ x3=0 &\iff x=3\\ x+3=0 &\iff x=3 \end{split}\) Now, make a factor table with sign like this Attachment:
Capture table 1.PNG [ 12.77 KiB  Viewed 2835 times ]
In line (1), we review all possible values of \(x\) that make each fator equal to 0. The order is ascending (from lowest to highest) Next, from line (2) to (6), we review the sign of each raw factor without power that based on the range value of \(x\). Character "" means that we no need to care about the specific value of them, just the sign positive (+) or negative (). Next, from line (7) to (11), we review the sign of each fator with power that exists in the expression. Finally, in line (12), we could quickly review the sign of the expression. Hence, we have \(\frac{(x2)^3(x+2)^3}{(x5)^5(x3)^4(x+3)^4} > 0 \iff x \in (2,2) \cup (5, +\infty) \) You could find more about this tool in this link: fatortablewithsigntheusefultooltosolvepolynomialinequalities229988.html#p1771368
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Re: Wavy Line Method Application  Exercise Question #7
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08 Oct 2017, 10:04
Given inequality \(\frac{(x^24)^3}{(x5)^5 (x^2  9)^4} > 0\) Since only expressions with odd powers play role in determining the sign of the equation, 1.) we can remove the even power expressions and 2.) make all odd powers equivalent to 1. We also note that x cannot be equal to 5, 3, 3 as these will make the numerator '0'. Thus, we can solve for below inequality instead. \(\frac{(x^24)}{(x5)} > 0\) Or, \({(x^24)(x5)} > 0\) Points are : 2, 2 and 5. Thus, Ans: (2,2) U (5,infinity). KS



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Re: Wavy Line Method Application  Exercise Question #7
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19 Oct 2017, 09:51
EgmatQuantExpert wrote: SolutionHey Everyone, Please find below, the solution of the given problem. Rewriting the inequality to easily identify the zero pointsWe know that \((x^2 – 4) = (x+2)(x2)\) And similarly, \((x^29) = (x+3)(x3)\) So, \((x^24)^3/((x – 5)^5 (x^29)^4 )>0\) can be written as \(((x+2)^3 (x2)^3)/((x – 5)^5 (x+3)^4 (x3)^4 )>0\) (x + 3) and (x3) are the terms with even power. So, the wavy line should bounce back at the corresponding zeropoints x = 3 and x = 3. Plotting the zero points and drawing the wavy line:Required Range: 2< x < 2 or x > 5 Calculations: We need the product of all integers in this range. However, note that the integer 0 is a part of the range (2 < x < 2). Therefore, the product will be 0. Correct Answer: Option C Hi for satisfying the inequality (x^24)^3 and ((x – 5)^5 should have the same sign so, they could be both negative or positive the range will be different am I wrong?



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Re: Wavy Line Method Application  Exercise Question #7
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22 Oct 2017, 00:09
soodia wrote: EgmatQuantExpert wrote: SolutionHey Everyone, Please find below, the solution of the given problem. Rewriting the inequality to easily identify the zero pointsWe know that \((x^2 – 4) = (x+2)(x2)\) And similarly, \((x^29) = (x+3)(x3)\) So, \((x^24)^3/((x – 5)^5 (x^29)^4 )>0\) can be written as \(((x+2)^3 (x2)^3)/((x – 5)^5 (x+3)^4 (x3)^4 )>0\) (x + 3) and (x3) are the terms with even power. So, the wavy line should bounce back at the corresponding zeropoints x = 3 and x = 3. Plotting the zero points and drawing the wavy line:Required Range: 2< x < 2 or x > 5 Calculations: We need the product of all integers in this range. However, note that the integer 0 is a part of the range (2 < x < 2). Therefore, the product will be 0. Correct Answer: Option C Hi for satisfying the inequality (x^24)^3 and ((x – 5)^5 should have the same sign so, they could be both negative or positive the range will be different am I wrong? Hi soodiathe range of \(x\) is \(2<x<2\) & \(x>5\). In this range the expression \(\frac{(x^24)^3}{(x5)^5}\) will always be greater than \(0\). you can test this by using some values from the range let \(x=1\) then \(\frac{(x^24)^3}{(x5)^5} = \frac{(14)^3}{(15)^5}>0\) if \(x=6\) then \(\frac{(x^24)^3}{(x5)^5}=\frac{(6^24)^3}{(65)^5}>0\) Hi EgmatQuantExpertwe know that \((x^29)^4>0\) and \(x≠3\),\(3\) & \(5\) because in that case denominator will be \(0\) which is not possible. So isn't our equation be simplified as \(\frac{(x^24)^3}{(x5)^5}>0\) (crossmultiplying \((x^29)^4\)) And the wavy line for this equation will be as below giving us the same range \(2<x<2\) & \(x>5\) Attachment:
inequality 1.jpg [ 38.38 KiB  Viewed 1586 times ]




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