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Find the range of values of x that satisfy the inequality (x^24)(x5)
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Updated on: 23 Dec 2018, 03:48
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77% (01:40) correct 23% (01:57) wrong based on 159 sessions
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Wavy Line Method Application  Exercise Question #5Find the range of values of x that satisfy the inequality \(\frac{(x^24)}{(x5)(x^29)} < 0\) A. x < 3 or 3 < x < 5 B. x < 3 or 2 < x < 2 C. 2 < x < 2 or 3 < x < 5 D. x < 3 or 2 < x < 2 or 3 < x < 5 E. x < 3
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Originally posted by EgmatQuantExpert on 26 Aug 2016, 02:43.
Last edited by Bunuel on 23 Dec 2018, 03:48, edited 5 times in total.



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Re: Find the range of values of x that satisfy the inequality (x^24)(x5)
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26 Aug 2016, 10:10
EgmatQuantExpert wrote: Wavy Line Method Application  Exercise Question #5Find the range of values of x that satisfy the inequality \(\frac{(x^24)}{(x5)(x^29)} < 0\) Wavy Line Method Application has been explained in detail in the following post:: http://gmatclub.com/forum/wavylinemethodapplicationcomplexalgebraicinequalities224319.htmlDetailed solution will be posted soon. Inequality is \(\frac{(x^24)}{(x5)(x^29)} < 0\) it can be written as \(\frac{(x2)(x+2)}{(x5)(x3)(x+3)} < 0\) So, we have the Zero pints of x as 3,2,2,3 and 5 But We cannot have x = 3,3 or 5. So, when drawing these values on the number line, we will get the range of x as (infinity,3)U(2,2)U(3,5) Please correct me if I am missing anything.
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Re: Find the range of values of x that satisfy the inequality (x^24)(x5)
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Updated on: 07 Aug 2018, 06:46
Solution: Hey 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, the given inequality can be written as: \((x+2)(x2)/(x – 5)(x+3)(x3)<0\) Plotting the zero points and drawing the wavy line:Required Range: x < 3 or 2 < x < 2 or 3 < x < 5 Correct Answer: Option D
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Re: Find the range of values of x that satisfy the inequality (x^24)(x5)
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10 Apr 2017, 00:40
Here is what i did in this question==>
Critical points => Equate the numerator and denominator terms to zero and obtain the critical points.
In here => x^24=0 => x=2 and x=2 Also x5=0 => x=5 And finally x^29=0=> x=3 and x=3
Critical points => 3,2,2,3,5
Now mark these on the number line and pick up a number in each boundary. If the value makes the original inequality true => Pick that boundary else discard
=> x=> (∞,3)U(2,2)U(3,5)
Personal Opinion> This question might be good for practise but is most certainly not a GMAT like Question.The wavy figure that is attached above makes it look scary.In reality we don't need that crazy figure. Just plug in numbers in a simple plain straight line by marking the critical points.
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Re: Find the range of values of x that satisfy the inequality (x^24)(x5)
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31 Jul 2017, 02:42
Does the range include 2 and 2 because the points are not circled on the number line like other points?



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Re: Find the range of values of x that satisfy the inequality (x^24)(x5)
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05 Aug 2017, 09:35
Shiv2016 wrote: Does the range include 2 and 2 because the points are not circled on the number line like other points? The range would have included any of the critical points 3, 2, 2, 3 and 5 if the equation read \(\frac{(x^24)}{(x5)(x^29)} <= 0\) (notice the equal sign at the right) So, to answer your question, it doesn't include 2 and 2.



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Re: Find the range of values of x that satisfy the inequality (x^24)(x5)
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17 Mar 2019, 10:40
EgmatQuantExpert wrote: Solution: Hey 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, the given inequality can be written as: \((x+2)(x2)/(x – 5)(x+3)(x3)<0\) Plotting the zero points and drawing the wavy line:Required Range: x < 3 or 2 < x < 2 or 3 < x < 5 Correct Answer: Option D Hi, since we have 5 zero points and no number with an even power, can we deduce that the must be 3 or 4 intervals for x? Since there is no answer with 4 intervals and only one with 3 intervals, this must be the right answer, true?



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Re: Find the range of values of x that satisfy the inequality (x^24)(x5)
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19 Mar 2019, 12:23
EgmatQuantExpert wrote: Solution: Hey 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, the given inequality can be written as: \((x+2)(x2)/(x – 5)(x+3)(x3)<0\) Plotting the zero points and drawing the wavy line:Required Range: x < 3 or 2 < x < 2 or 3 < x < 5 Correct Answer: Option D Can you please explain, how we have come to the required range?



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Find the range of values of x that satisfy the inequality (x^24)(x5)
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19 Mar 2019, 12:40
arorni wrote: Can you please explain, how we have come to the required range? This is exactly what she did I think you have to specify your question a bit. Which part did you not understand?



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Find the range of values of x that satisfy the inequality (x^24)(x5)
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19 Mar 2019, 12:56
mtambiev wrote: arorni wrote: Can you please explain, how we have come to the required range? This is exactly what she did I think you have to specify your question a bit. Which part did you not understand? I saw Payal's other post on wavy line https://gmatclub.com/forum/wavylineme ... 24319.html. It indeed cleared my doubt about this question, but I am a little confused about drawing wavy lines. Do we always have to start drawing the wavy line from the ve region (ie Is it the upper right corner)? How the wavy line of (x1)^2 will look, I got the idea that for even powers the line will remain in the same region but why we have started drawing from the ve region?



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Find the range of values of x that satisfy the inequality (x^24)(x5)
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19 Mar 2019, 13:11
arorni wrote: mtambiev wrote: arorni wrote: Can you please explain, how we have come to the required range? This is exactly what she did I think you have to specify your question a bit. Which part did you not understand? I saw Payal's other post on wavy line https://gmatclub.com/forum/wavylineme ... 24319.html. It indeed cleared my doubt about this question, but I am a little confused about drawing wavy lines. Do we always have to start drawing the wavy line from the ve region (ie Is it the upper right corner)? How the wavy line of (x1)^2 will look, I got the idea that for even powers the line will remain in the same region but why we have started drawing from the ve region? Ok i got you. As you can see in her approach, the largest zero point for x is 5 (from all the others:3,2,2,3,5). So its easy to check if the equation gives a positive or negative result, if you plug in numbers that are less than the smallest zero point (here: 3) or greater the largest zero point (5). So let's try all number greater the largest zero point, which is 5. For example, plug in \(x=10\). Then all terms in the brackets will be positive, so for all \(x>5\) the equation is \(>0\). Remember: \(+*+=+\) \(*+=\) \(*=+\) Further on, for even powers, the line will not cross the axis, but bounces back, for odd powers it does cross the axis.




Find the range of values of x that satisfy the inequality (x^24)(x5)
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19 Mar 2019, 13:11






