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For what range of values of 'x' ? [#permalink]
 03 Sep 2010, 07:02
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For what range of values of 'x' will the inequality 15x - 2/x > 1? A. x > 0.4 B. x < 1/3 C. -1/3 < x < 0.4, x > 15/2 D. -1/3 < x < 0, x > 2/5 E. x < -1/3 and x > 2/5
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Re: For what range of values of 'x' ? [#permalink]
03 Sep 2010, 08:17
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15x - 2/x > 1
Case1 : x > 0
Multiply both sides by "x"
=> 15x^2 - 2 > x
=> 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5)
Since x > 0 , so x > 2/5
Case 2 : x < 0
Multiply both sides by "x"
=> 15x^2 - 2 < x
=> 15x^2 - x - 2 < 0
Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.
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Re: For what range of values of 'x' ? [#permalink]
03 Oct 2010, 13:15
I do not understand why 2 cases are considered in the above explanation ? Only if we had |x| should the 2 cases be considered, am I right ?
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Re: For what range of values of 'x' ? [#permalink]
03 Oct 2010, 13:23
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Whenever you multiply an inequality by X , you should check for the -ve and +ve values of X, as it inverts the sign.
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Re: For what range of values of 'x' ? [#permalink]
03 Oct 2010, 13:26
Thanks for the lightning quick reply, I was not aware of this.
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Re: For what range of values of 'x' ? [#permalink]
03 Oct 2010, 13:36
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Eden wrote: For what range of values of 'x' will the inequality 15x - 2/x > 1? A. x > 0.4 B. x < 1/3 C. -1/3 < x < 0.4, x > 15/2 D. -1/3 < x < 0, x > 2/5 E. x < -1/3 and x > 2/5 devashish wrote: I do not understand why 2 cases are considered in the above explanation ?
Only if we had |x| should the 2 cases be considered, am I right ? anshumishra considered 2 cases for x in order to multiply both parts of the inequality by x and simplify it by doing so. One could also do as follows: 15x-\frac{2}{x}>1 --> \frac{15x^2-x-2}{x}>0 --> \frac{(x+\frac{1}{3})(x-\frac{2}{5})}{x}>0: Both denominator and nominator positive --> x>\frac{2}{5}; Both denominator and nominator negative --> -\frac{1}{3}<x<0; So inequality holds true in the ranges: -\frac{1}{3}<x<0 and x>\frac{2}{5}. Answer: D.
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Re: For what range of values of 'x' ? [#permalink]
03 Oct 2010, 13:45
Bunuel wrote: One could also do as follows:
15x-\frac{2}{x}>1 --> \frac{15x^2-x-2}{x}>0 --> \frac{(x+\frac{1}{3})(x-\frac{2}{5})}{x}>0:
Both denominator and nominator positive --> x>\frac{2}{5};
Both denominator and nominator negative --> -\frac{1}{3}<x<0;
So inequality holds true in the ranges: -\frac{1}{3}<x<0 and x>\frac{2}{5}.
Answer: D. Bunuel you are awesome
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Re: For what range of values of 'x' ? [#permalink]
03 Oct 2010, 19:00
Nice alternate solution.
Please note :
15x^2-x-2 -> x^2 - x/15 - 2/15 -> x^2 + x/3 - 2x/5 - 2/15 -> (x+1/3)(x-2/5)
This works well, however I would suggest to rely on quadratic equation formula: -b(+-)\sqrt{D}/2a, as you don't need to think, how to factorize an equation?
Try yourself to factorize the equation above vs using the formula (Unfortunately, if the equation doesn't has a fractional value (which I expect rarely could be the case in GMAT) , for example when D in above formula is irrational, then you might be trying to factorize the fraction forever.)
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Re: For what range of values of 'x' ?
[#permalink]
03 Oct 2010, 19:00
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