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# For what range of values of 'x' ?

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For what range of values of 'x' ? [#permalink]  03 Sep 2010, 06:02
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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

[Reveal] Spoiler:
D
[Reveal] Spoiler: OA
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Re: For what range of values of 'x' ? [#permalink]  03 Sep 2010, 07:17
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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, 12:15
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, 12:23
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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, 12:26
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, 12:36
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Expert's post
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

[Reveal] Spoiler:
D

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 $$x>\frac{2}{5}$$.

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Re: For what range of values of 'x' ? [#permalink]  03 Oct 2010, 12:45
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 $$x>\frac{2}{5}$$.

Bunuel you are awesome
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Re: For what range of values of 'x' ? [#permalink]  03 Oct 2010, 18:00
Nice alternate solution.

$$15x^2-x-2 -> x^2 - x/15 - 2/15 -> x^2 + x/3 - 2x/5 - 2/15 -> (x+1/3)(x-2/5)$$

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]  12 Aug 2013, 10:48
What I have learnt, if x is in denom, take LCM after bringing all element at one side.

If we have more than 2 roots, then we can use following method to know the range.

Attachment:

Solution.jpg [ 16.75 KiB | Viewed 2428 times ]

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Re: For what range of values of 'x' ? [#permalink]  14 Aug 2013, 22:13
hi,

can any one explain this with graphical explanation??

Rrsnathan.
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Re: For what range of values of 'x' ? [#permalink]  26 Nov 2013, 19:46
anshumishra wrote:
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.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.
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Re: For what range of values of 'x' ? [#permalink]  02 Mar 2014, 00:22
TooLong150 wrote:
anshumishra wrote:
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.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.

I have the same doubt. Shouldn't the answer be D?
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Re: For what range of values of 'x' ? [#permalink]  02 Mar 2014, 04:03
Expert's post
siriusblack1106 wrote:
TooLong150 wrote:
anshumishra wrote:
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.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.

I have the same doubt. Shouldn't the answer be D?

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Re: For what range of values of 'x' ? [#permalink]  02 Mar 2014, 17:02
Expert's post
TooLong150 wrote:
anshumishra wrote:
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.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.

After you determine boundary points, you still need to test the potential +ve/-ve ranges per the inequality
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Re: For what range of values of 'x' ? [#permalink]  07 Jun 2014, 16:21
Bunuel wrote:
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

[Reveal] Spoiler:
D

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 $$x>\frac{2}{5}$$.

Hi Bunuel, I thought that we shouldn't multiply an inequality with a variable whose range of value could be +ive and/or -ive.
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Re: For what range of values of 'x' ? [#permalink]  08 Jun 2014, 02:35
Expert's post
Enael wrote:
Bunuel wrote:
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

[Reveal] Spoiler:
D

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 $$x>\frac{2}{5}$$.

Hi Bunuel, I thought that we shouldn't multiply an inequality with a variable whose range of value could be +ive and/or -ive.

I didn't multiply.

$$15x-\frac{2}{x}>1$$ --> subtract 1: $$15x-\frac{2}{x}-1>0$$ --> put in common denominator: $$\frac{15x^2-x-2}{x}>0$$ --> factorize: $$\frac{(x+\frac{1}{3})(x-\frac{2}{5})}{x}>0$$.

Hope it's clear.
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Re: For what range of values of 'x' ? [#permalink]  10 Jun 2014, 07:51
anshumishra wrote:
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.

A graphical approach supplements this solution quite well: The roots are the blue dots, the yellow line separates the two cases, and the red lines are where the inequality holds true.

We assume two cases: x>0 and x<0.

For the first case, x>0, we want to see where 15x^2 -x-2 >0. Solving for the roots using the quadratic formula, we find roots when x = 2/5 and x = -1/3. Thus, for x>0, 15x^2 -x-2 >0 when x>2/5.

For the second case, x<0, we want to see where 15x^2 -x-2 <0. Again, using the roots, we know that the 15x^2 -x-2 < 0 when x > -1/3. We also assumed x<0 so we must include that in our solution.

Thus we have x>2/5 and -1/3<x<0. Answer: D
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Re: For what range of values of 'x' ? [#permalink]  15 Jul 2014, 04:31
Bunuel wrote:
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

[Reveal] Spoiler:
D

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 $$x>\frac{2}{5}$$.

Hi
Sorry for asking basic question. But do you have detail link to solve complicated quadratic equations like this: 15x2-x-2 ?
I don't know how to solve this????
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Re: For what range of values of 'x' ? [#permalink]  15 Jul 2014, 05:26
Expert's post
GGMAT760 wrote:
Bunuel wrote:
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

[Reveal] Spoiler:
D

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 $$x>\frac{2}{5}$$.

Hi
Sorry for asking basic question. But do you have detail link to solve complicated quadratic equations like this: 15x2-x-2 ?
I don't know how to solve this????

Hope this helps.
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Re: For what range of values of 'x' ? [#permalink]  23 Sep 2014, 01:50
PiyushK wrote:
What I have learnt, if x is in denom, take LCM after bringing all element at one side.

If we have more than 2 roots, then we can use following method to know the range.

Attachment:
Solution.jpg

Can you explain why before and after the "0" you change the sign?
What was the equation that got you to draw this?
Re: For what range of values of 'x' ?   [#permalink] 23 Sep 2014, 01:50
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