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Find the range of values of x such that (x5)^3 (24x) < 0
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Updated on: 23 Aug 2018, 21:25
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Solving inequalities Number Line Method  Practice Question #1Find the range of values of x such that \((x5)^3 (24x) < 0\) A. \(\frac{1}{2} < x < 5\) B. \(0 < x < \frac{1}{2}\) C. \(x >5\) D. \(\frac{1}{2} < x < 5\) E. \((x < \frac{1}{2}\)) and (\(x > 5\)) Next QuestionTo 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 (x5)^3 (24x) < 0
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Updated on: 27 Aug 2018, 02:05
Solution Given:• An inequality, \((x5)^3 (24x) < 0\) To find:• The range of x, that satisfies the above inequality Approach and Working: • So, if we observe carefully, the inequality given to us can be written as,
o \((x5)^2 * (x5) * (24x) < 0\) • We know that, any number of the form N2 is always positive, expect for N = 0.
o So, we can say that, \((x  5)^2\) is always positive, expect for x = 5, o Thus, the inequality can be written as (x  5) * (2  4x) < 0 • Now, let’s multiply the inequality by 1 to make the coefficient of x, positive
o And, note that the inequality sign must be changed as we are multiplying it by a negative number, 1 o Thus, the inequality becomes, (x  5) * (4x  2) > 0 • The zero points of this inequality are x = 5 and \(x = \frac{1}{2}\) • Plot these points on the number line.
o Since, both (x – 5) and (4x – 2) > 0, the inequality will be positive, for all the points to the right of 5 o And, it is negative, in the region, between \(\frac{1}{2}\) and 5, since (x – 5) is negative and (4x – 2) is positive in this region o It is again positive for all the values less than \(\frac{1}{2}\), since both (x – 5) and (4x – 2) are negative in this region Therefore, the range of x is \(x < \frac{1}{2}\) and x > 5 Hence, the correct answer is option E. Answer: E
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Re: Find the range of values of x such that (x5)^3 (24x) < 0
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23 Aug 2018, 05:38
EgmatQuantExpert wrote: Find the range of values of x such that \((x5)^3 (24x) < 0\)
A. \(\frac{1}{2} < x < 5\) B. \(0 < x < \frac{1}{2}\) C. \(x >5\) D. \(\frac{1}{2} < x < 5\) E. \((x < \frac{1}{2}\)) and (\(x > 5\))
\((x5)^3 (24x) < 0...........(x5)^3(12x)<0\) two cases 1) \(x5<0\) or \(x<5\), then \(12x>0\) or \(2x<1\) or \(x<\frac{1}{2}\).... thus \(x<\frac{1}{2}\) 2) \(x5>0\) or \(x>5\), then \(12x<0\) or \(2x>1\) or \(x>\frac{1}{2}\).... thus \(x>5\) combined \(x<\frac{1}{2}\) and \(x>5\) E
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Re: Find the range of values of x such that (x5)^3 (24x) < 0
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23 Aug 2018, 06:35
EgmatQuantExpert wrote: Solving inequalities Number Line Method  Practice Question #1
Find the range of values of x such that \((x5)^3 (24x) < 0\)
A. \(\frac{1}{2} < x < 5\) B. \(0 < x < \frac{1}{2}\) C. \(x >5\) D. \(\frac{1}{2} < x < 5\) E. \((x < \frac{1}{2}\)) and (\(x > 5\))
\((x5)^3 (24x) < 0\) Or, \(2\left(x5\right)^3\left(2x1\right)<0\) Or, \(2\left(x5\right)^3\left(2x1\right)>0\) Critical points: 5, 1/2 Using wavy curve method: range of x: \(x<\frac{1}{2}\) or \(x>5\) Ans. (E) P.S: In option (E) , OR sounds logical instead of AND.
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Re: Find the range of values of x such that (x5)^3 (24x) < 0
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05 Sep 2018, 02:29
EgmatQuantExpert wrote: Solution Given:• An inequality, \((x5)^3 (24x) < 0\) To find:• The range of x, that satisfies the above inequality[/list
Approach and Working:
• So, if we observe carefully, the inequality given to us can be written as,
o \((x5)^2 * (x5) * (24x) < 0\) • We know that, any number of the form N2 is always positive, expect for N = 0.
o So, we can say that, \((x  5)^2\) is always positive, expect for x = 5, o Thus, the inequality can be written as (x  5) * (2  4x) < 0 • Now, let’s multiply the inequality by 1 to make the coefficient of x, positive
o And, note that the inequality sign must be changed as we are multiplying it by a negative number, 1 o Thus, the inequality becomes, (x  5) * (4x  2) > 0 • The zero points of this inequality are x = 5 and \(x = \frac{1}{2}\) • Plot these points on the number line.
o Since, both (x – 5) and (4x – 2) > 0, the inequality will be positive, for all the points to the right of 5 o And, it is negative, in the region, between \(\frac{1}{2}\) and 5, since (x – 5) is negative and (4x – 2) is positive in this region o It is again positive for all the values less than \(\frac{1}{2}\), since both (x – 5) and (4x – 2) are negative in this region Therefore, the range of x is \(x < \frac{1}{2}\) and x > 5 Hence, the correct answer is option E. Answer: EHello EgmatQuantExpert, thanks for the wholesome explanation. I would like to know why after multiplying the whole inequality by 1, the sign in (x5) did no change only that of 24x changed. Thanks. Posted from my mobile device



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Re: Find the range of values of x such that (x5)^3 (24x) < 0
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06 Sep 2018, 19:23
EgmatQuantExpert wrote: Find the range of values of x such that \((x5)^3 (24x) < 0\)
A. \(\frac{1}{2} < x < 5\) B. \(0 < x < \frac{1}{2}\) C. \(x >5\) D. \(\frac{1}{2} < x < 5\) E. \((x < \frac{1}{2}\)) and (\(x > 5\))
\(?\,\,\,:\,\,\,{\left( {x  5} \right)^3}\left( {2  4x} \right) < 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\boxed{\left( {x  5} \right)\left( {1  2x} \right) < 0}\) \(\left( {x  5} \right)\left( {1  2x} \right) = 0\,\,\,\,\, \Leftrightarrow \,\,\,\,x = 5\,\,{\text{or}}\,\,x = \frac{1}{2}\) \(?\,\,:\,\,\,\left\{ {x < \frac{1}{2}} \right\}\,\,\, \cup \,\,\,\left\{ {x > 5} \right\}\) The solution above follows the notations and rationale taught in the GMATH method. Regards, fskilnik.
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Re: Find the range of values of x such that (x5)^3 (24x) < 0
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06 Sep 2018, 21:09
I thought the answer must be written in form like x < 1/2 or x > 5 as there is no way x can satisfy both the condition. The "And" here made me confused.



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Find the range of values of x such that (x5)^3 (24x) < 0
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06 Sep 2018, 21:34
Hungluu92vn wrote: I thought the answer must be written in form like x < 1/2 or x > 5 as there is no way x can satisfy both the condition. The "And" here made me confused. Hi, Hungluu92vn! Excellent observation. I have avoided to use the term "and", because EACH x cannot be simultaneously in ]infinity, 1/2[ and in ]5, +infinity[. That´s why, for the solution set of a inequation such as the one presented in the question stem, the proper word is OR , the proper symbol is U (reunion). Other experts used the term AND, specifically in this question, because the word "range" was presented in the question stem! Their interpretation: there are real values "x" that belong to ]infinity, 1/2[ AND there are (OTHER) real values "x" that belong to ]5, +infinity[. (It is as if "x" is a dummy variable: different "roles" each time it appears! This is not the classical way of looking into this "theme" but... you got the point!) Regards, fskilnik.
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Find the range of values of x such that (x5)^3 (24x) < 0
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09 Sep 2018, 05:21
Under 30sec approach
You directly see that x can be bigger than 5. Eliminate A,B and D. Left with C or E. > x can also be negative therefore eliminate C. Hence, E



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Re: Find the range of values of x such that (x5)^3 (24x) < 0
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09 Sep 2018, 06:13
Hello Egmatquantexpert why are multiplying by 1? What is the need for it?
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Re: Find the range of values of x such that (x5)^3 (24x) < 0 &nbs
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