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For what range of values of 'x' will the inequality 15x - 2/x > 1?

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Re: For what range of values of 'x' will the inequality 15x - 2/x > 1?  [#permalink]

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New post 17 Oct 2015, 18:44
Engr2012

Please tell me where am I going wrong. I took out the roots using the factor method and correctly deduced them but when I used the discriminant method, I am not getting the values.

15x^2-x-2>0

D=121 (b^2-4ac=1+120)

roots= (-(-1)+11)/30 and (-(-1)-11)/30

This gives 2/5 and 1/3 while the roots should be 2/5 and -1/3 :shock:
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For what range of values of 'x' will the inequality 15x - 2/x > 1?  [#permalink]

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New post 18 Oct 2015, 22:33
1
longfellow wrote:
Engr2012

Please tell me where am I going wrong. I took out the roots using the factor method and correctly deduced them but when I used the discriminant method, I am not getting the values.

15x^2-x-2>0

D=121 (b^2-4ac=1+120)

roots= (-(-1)+11)/30 and (-(-1)-11)/30

This gives 2/5 and 1/3 while the roots should be 2/5 and -1/3 :shock:


Hi longfellow,

I think you need to recheck your calculation:
(-(-1)-11)/30 = (1 - 11)/30 = -10/30 = -1/3

Does this help?
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Re: For what range of values of 'x' will the inequality 15x - 2/x > 1?  [#permalink]

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New post 19 Oct 2015, 00:08
TeamGMATIFY wrote:
longfellow wrote:
Engr2012

Please tell me where am I going wrong. I took out the roots using the factor method and correctly deduced them but when I used the discriminant method, I am not getting the values.

15x^2-x-2>0

D=121 (b^2-4ac=1+120)

roots= (-(-1)+11)/30 and (-(-1)-11)/30

This gives 2/5 and 1/3 while the roots should be 2/5 and -1/3 :shock:


Hi longfellow,

I think you need to recheck your calculation:
(-(-1)-11)/30 = (1 - 11)/30 = -10/30 = -1/3

Does this help?



Got it! Thanks! :-D
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For what range of values of 'x' will the inequality 15x - 2/x > 1?  [#permalink]

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New post 21 Oct 2015, 05:26
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 the first case I got x> 2/5 and x> -1/3 and x>0 , which results in x>2/5 so no problem here.

In the second case I have a problem. I get x<2/5 and x< -1/3 and x<0
This results in x < -1/3

So my answer comes out to be x<-1/3 or x>2/5 which is option E , but the answer is D. :?:

Can someone please explain what I am doing wrong here? Why should x be in the range -1/3<x<0 in the case when x<0 ?

A detailed explanation will be a great help because it seems something is wrong with my understanding of some fundamental inequality rules.
:cry:
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For what range of values of 'x' will the inequality 15x - 2/x > 1?  [#permalink]

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New post 21 Oct 2015, 05:59
1
nishantsharma87 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 the first case I got x> 2/5 and x> -1/3 and x>0 , which results in x>2/5 so no problem here.

In the second case I have a problem. I get x<2/5 and x< -1/3 and x<0
This results in x < -1/3

So my answer comes out to be x<-1/3 or x>2/5 which is option E , but the answer is D. :?:

Can someone please explain what I am doing wrong here? Why should x be in the range -1/3<x<0 in the case when x<0 ?

A detailed explanation will be a great help because it seems something is wrong with my understanding of some fundamental inequality rules.
:cry:



A common mistkae that people do on inequalitites is to multiply by a variable when you dont know the sign. You do not have to confuse yourself with the 2 cases above. A straightforward solution is:
15x-2/x>1 ---> 15x-2/x-1>0 ---> \(\frac{15x^2-2-x}{x}>0\) ---> \(\frac{15x^2-x-2}{x}>0\) ---> \(\frac{15x^2-6x+5x-2}{x}>0\)---> \(\frac{(3x+1)(5x-2)}{x}>0\)

---> \(\frac{(x+1/3)(x-2/5)}{x}>0\) ....(1)

Now, you have 3 roots at x=-1/3, 0 and 2/5.

So if you draw a numberline as shown in the attached figure, you will see that the expression in (1) will be true for

Attachment:
10-21-15 9-51-48 AM.jpg
10-21-15 9-51-48 AM.jpg [ 3.27 KiB | Viewed 1079 times ]


-1/3<x<0 and x>2/5, giving you D as the correct answer.

For your solution, the text in red is incorrect. After you get \(15x^2 - x - 2 <\)0, you can not get x<-1/3. It should be -1/3<x<0.

You can check it by assuming x<-1/3 or x=-2. You get \(15*(-2)^2-2-(-2) = 60-2+4 >0\) and NOT <0. Thus x<-1/3 DOES NOT satisfy the case 2 inequality of \(15x^2 - x - 2 < 0\).

Hope this helps.
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For what range of values of 'x' will the inequality 15x - 2/x > 1?  [#permalink]

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New post 21 Oct 2015, 06:03
longfellow wrote:
Engr2012

Please tell me where am I going wrong. I took out the roots using the factor method and correctly deduced them but when I used the discriminant method, I am not getting the values.

15x^2-x-2>0

D=121 (b^2-4ac=1+120)

roots= (-(-1)+11)/30 and (-(-1)-11)/30

This gives 2/5 and 1/3 while the roots should be 2/5 and -1/3 :shock:


I think TeamGMATIFY has answered your question but I would suggest to not use discriminant method to calculate the roots for such easier factorizations. You will end up spending more time than required.

Once you saw that you needed 2 numbers giving you a difference of 1 and when multiplied give you 30 (=15*2), it should have been easier to recognize that the pair can only be (6,5). Refer to my solution above (for-what-range-of-values-of-x-will-the-inequality-15x-2-x-100335-20.html#p1589956) for more detailed way of factorization the given inequality.

Hope this helps.
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Re: For what range of values of 'x' will the inequality 15x - 2/x > 1?  [#permalink]

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New post 21 Oct 2015, 08:34
Engr2012 wrote:
longfellow wrote:
Engr2012

Please tell me where am I going wrong. I took out the roots using the factor method and correctly deduced them but when I used the discriminant method, I am not getting the values.

15x^2-x-2>0

D=121 (b^2-4ac=1+120)

roots= (-(-1)+11)/30 and (-(-1)-11)/30

This gives 2/5 and 1/3 while the roots should be 2/5 and -1/3 :shock:


I think TeamGMATIFY has answered your question but I would suggest to not use discriminant method to calculate the roots for such easier factorizations. You will end up spending more time than required.

Once you saw that you needed 2 numbers giving you a difference of 1 and when multiplied give you 30 (=15*2), it should have been easier to recognize that the pair can only be (6,5). Refer to my solution above (for-what-range-of-values-of-x-will-the-inequality-15x-2-x-100335-20.html#p1589956) for more detailed way of factorization the given inequality.

Hope this helps.



Hi Engr2012

Even I think the factorization method is better but I was trying out if calculating D is a better way of solving quadratic and see how much time will be required for it.
But you are right. It is tedious and as seen in this case, a slight sign change can cost dearly.
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Re: For what range of values of 'x' will the inequality 15x - 2/x > 1?  [#permalink]

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New post 21 Oct 2015, 10:51
longfellow wrote:
Engr2012 wrote:
longfellow wrote:
Engr2012

Please tell me where am I going wrong. I took out the roots using the factor method and correctly deduced them but when I used the discriminant method, I am not getting the values.

15x^2-x-2>0

D=121 (b^2-4ac=1+120)

roots= (-(-1)+11)/30 and (-(-1)-11)/30

This gives 2/5 and 1/3 while the roots should be 2/5 and -1/3 :shock:


I think TeamGMATIFY has answered your question but I would suggest to not use discriminant method to calculate the roots for such easier factorizations. You will end up spending more time than required.

Once you saw that you needed 2 numbers giving you a difference of 1 and when multiplied give you 30 (=15*2), it should have been easier to recognize that the pair can only be (6,5). Refer to my solution above (for-what-range-of-values-of-x-will-the-inequality-15x-2-x-100335-20.html#p1589956) for more detailed way of factorization the given inequality.

Hope this helps.



Hi Engr2012

Even I think the factorization method is better but I was trying out if calculating D is a better way of solving quadratic and see how much time will be required for it.
But you are right. It is tedious and as seen in this case, a slight sign change can cost dearly.


For these question you can also plug in values and eliminate options.

Option A, eliminated with x=-0.1
Option B, eliminated with x=2
Option C, eliminated with x=2
Option D
Option E, eliminated with x=-2

So the correct answer is D.
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Re: For what range of values of 'x' will the inequality 15x - 2/x > 1?  [#permalink]

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New post 06 Nov 2015, 00:57
Engr2012 wrote:
nishantsharma87 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 the first case I got x> 2/5 and x> -1/3 and x>0 , which results in x>2/5 so no problem here.

In the second case I have a problem. I get x<2/5 and x< -1/3 and x<0
This results in x < -1/3

So my answer comes out to be x<-1/3 or x>2/5 which is option E , but the answer is D. :?:

Can someone please explain what I am doing wrong here? Why should x be in the range -1/3<x<0 in the case when x<0 ?

A detailed explanation will be a great help because it seems something is wrong with my understanding of some fundamental inequality rules.
:cry:



A common mistkae that people do on inequalitites is to multiply by a variable when you dont know the sign. You do not have to confuse yourself with the 2 cases above. A straightforward solution is:
15x-2/x>1 ---> 15x-2/x-1>0 ---> \(\frac{15x^2-2-x}{x}>0\) ---> \(\frac{15x^2-x-2}{x}>0\) ---> \(\frac{15x^2-6x+5x-2}{x}>0\)---> \(\frac{(3x+1)(5x-2)}{x}>0\)

---> \(\frac{(x+1/3)(x-2/5)}{x}>0\) ....(1)

Now, you have 3 roots at x=-1/3, 0 and 2/5.

So if you draw a numberline as shown in the attached figure, you will see that the expression in (1) will be true for

Attachment:
10-21-15 9-51-48 AM.jpg


-1/3<x<0 and x>2/5, giving you D as the correct answer.

For your solution, the text in red is incorrect. After you get \(15x^2 - x - 2 <\)0, you can not get x<-1/3. It should be -1/3<x<0.

You can check it by assuming x<-1/3 or x=-2. You get \(15*(-2)^2-2-(-2) = 60-2+4 >0\) and NOT <0. Thus x<-1/3 DOES NOT satisfy the case 2 inequality of \(15x^2 - x - 2 < 0\).

Hope this helps.


Thanks a lot Engr2012! This numberline-approach is a great time-saving approach for finding range of such inequalities. Could you please inform more such questions so I could practice this numberline approach to make it my unconscious way of approaching all questions involving such inequality range calculations ?

Thanks a ton again!
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For what range of values of 'x' will the inequality 15x - 2/x > 1?  [#permalink]

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New post 06 Nov 2015, 03:56
1
nishantsharma87 wrote:


Thanks a lot Engr2012! This numberline-approach is a great time-saving approach for finding range of such inequalities. Could you please inform more such questions so I could practice this numberline approach to make it my unconscious way of approaching all questions involving such inequality range calculations ?

Thanks a ton again!


The number line approach is the most common approach for tackling inequalities. Most inequalty questions that ask about range of a particular variable will invariably want you to use the number line. There is no way around it.

You can try the questions at search.php?search_id=tag&tag_id=189 and search.php?search_id=tag&tag_id=184

Alternately, for such questions, you can also plug in the values of use the options to your advantage. These techniques will be time saving techniques.

Hope this helps.
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Re: For what range of values of 'x' will the inequality 15x - 2/x > 1?  [#permalink]

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New post 06 Nov 2015, 04:04
1
nishantsharma87 wrote:
Thanks a lot Engr2012! This numberline-approach is a great time-saving approach for finding range of such inequalities. Could you please inform more such questions so I could practice this numberline approach to make it my unconscious way of approaching all questions involving such inequality range calculations ?

Thanks a ton again!


Check links below:
Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach
Inequality tips

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

Hope it helps.
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Re: For what range of values of 'x' will the inequality 15x - 2/x > 1?  [#permalink]

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