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Re: For what range of values of 'x' ? [#permalink]
03 Sep 2010, 07:17

2

This post received KUDOS

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.

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.)

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 1598 times ]

_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: WOULD: when to use? Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

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.

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.

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.

Re: For what range of values of 'x' ? [#permalink]
02 Mar 2014, 17:02

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

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.

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.

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.

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.

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????

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.

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????