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# Solve the Inequality

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Senior Manager
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03 Aug 2009, 06:48
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X > 1/X

Kindly help to solve ...
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03 Aug 2009, 06:56
It would be helpful to know what the rest of the question is as well as the answer choices. Is this a data sufficiency?

As a statement, this is only true if X is positive. If X is negative, then that statement is false.
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03 Aug 2009, 08:57
1
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[EDIT - REMOVED BY USER]

Last edited by nplaneta on 17 Jun 2012, 17:43, edited 1 time in total.
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04 Aug 2009, 03:12
nplaneta wrote:
X > 1 or -1 < X < 0

hey can you kindly explain in steps ,how you arrived at the answer..
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04 Aug 2009, 03:37
crejoc wrote:
X > 1/X

Kindly help to solve ...

i will try and solve...Ian please comment

x>1/x
x-(1/x)>0
$$(x^2-1)/x > 0$$

so EITHER 1/x > 0 and $$x^2 - 1 > 0$$ OR 1/x<0 and $$x^2 - 1 < 0$$ [ since $$(x^2-1)/x > 0$$ ...either both are positive or both are negative]

Case 1:

1/x is greater than 0 then x > 0 and for $$x^2 - 1 > 0$$ ... x >1 or x>-1

so finally x>1

Cae 2:

1/x < 0 then x is less than 0 and for $$x^2 - 1 < 0$$ ..... x<-1 and x <1

so finally x < -1

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04 Aug 2009, 04:56
1
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apoorvasrivastva wrote:
crejoc wrote:
X > 1/X

Kindly help to solve ...

i will try and solve...Ian please comment

x>1/x
x-(1/x)>0
$$(x^2-1)/x > 0$$

so EITHER 1/x > 0 and $$x^2 - 1 > 0$$ OR 1/x<0 and $$x^2 - 1 < 0$$ [ since $$(x^2-1)/x > 0$$ ...either both are positive or both are negative]

Case 1:

1/x is greater than 0 then x > 0 and for $$x^2 - 1 > 0$$ ... x >1 or x>-1

so finally x>1

Cae 2:

1/x < 0 then x is less than 0 and for $$x^2 - 1 < 0$$ ..... x<-1 and x <1

so finally x < -1

apoorvasrivastva, your approach is good for case 1, however for case 2 it will be -1 < x < 0 as nplaneta posted earlier.

The reason being...
(x^2 - 1) / x > 0
When x < 0 then:
(x^2 - 1) < 0 [multiplying both sides by x (which is a -ve number, hence flipping the inequality)]
x^2 < 1
now as we are considering x < 0 then, when we take square root of both side:
sqrt(x^2) < sqrt(1)
x < 1 [but x is a negative number so we need to multiply both sides by -1]
-x > -1
x > -1 [left side became positive due to negative x and -1]
and since the case is for x < 0 , therefore, the solution is -1 < x < 0

The solution for x > 1/x being:
x>1 (when x>0) and -1 < x < 0 (when x<0)

For reference and review of inequality principles, see basic-mathematical-principles post by HongHu in this forum [sorry I can't post the direct link, as the forum isn't allowing me cos I am new member]
This exact problem is explained in this post as well.

~Cheers
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04 Aug 2009, 05:25
scarish wrote:
For reference and review of inequality principles, see basic-mathematical-principles post by HongHu in this forum

Actually i got this problem from the very link you mentioned of HongHu's post, as it is not explained in detail there , i came up posting it in a separate thread...

scarish wrote:
case 2 it will be -1 < x < 0 as nplaneta posted earlier.
When x < 0 then:
(x^2 - 1) < 0 [multiplying both sides by x (which is a -ve number, hence flipping the inequality)]
x^2 < 1
..........
x < 1 [but x is a negative number so we need to multiply both sides by -1]

I couldn't understand why you have to multiply x, you said to flip the inequality, but it is not flipped i think. Then why have to multiply by -1. sorry i don't get you. May be i am not comprehending it.. Could you explain clearly, the second case. i was also getting the same answer as apoorvasrivastva
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04 Aug 2009, 05:28
[EDIT - REMOVED BY THE USER]

Last edited by nplaneta on 17 Jun 2012, 19:13, edited 1 time in total.
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04 Aug 2009, 06:19
scarish wrote:
Cae 2:

1/x < 0 then x is less than 0 and for $$x^2 - 1 < 0$$ ..... x<-1 and x <1

so finally x < -1

apoorvasrivastva, your approach is good for case 1, however for case 2 it will be -1 < x < 0 as nplaneta posted earlier.

The reason being...
(x^2 - 1) / x > 0
When x < 0 then:
(x^2 - 1) < 0 [multiplying both sides by x (which is a -ve number, hence flipping the inequality)]
x^2 < 1
now as we are considering x < 0 then, when we take square root of both side:
sqrt(x^2) < sqrt(1)
x < 1 [but x is a negative number so we need to multiply both sides by -1]
-x > -1
x > -1 [left side became positive due to negative x and -1]
and since the case is for x < 0 , therefore, the solution is -1 < x < 0

The solution for x > 1/x being:
x>1 (when x>0) and -1 < x < 0 (when x<0)

For reference and review of inequality principles, see basic-mathematical-principles post by HongHu in this forum [sorry I can't post the direct link, as the forum isn't allowing me cos I am new member]
This exact problem is explained in this post as well.

~Cheers[/quote]

Thanks i realsed my mistake!!!Kudos given to u !!:)
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04 Aug 2009, 17:09
1
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scarish wrote:
case 2 it will be -1 < x < 0 as nplaneta posted earlier.
When x < 0 then:
(x^2 - 1) < 0 [multiplying both sides by x (which is a -ve number, hence flipping the inequality)]
x^2 < 1
..........
x < 1 [but x is a negative number so we need to multiply both sides by -1]

crejoc wrote:
I couldn't understand why you have to multiply x, you said to flip the inequality, but it is not flipped i think. Then why have to multiply by -1. sorry i don't get you. May be i am not comprehending it.. Could you explain clearly, the second case. i was also getting the same answer as apoorvasrivastva

Right, our equation was (x^2-1)/x > 0, so to get rid of x from the denominator you have to multiply both sides of the inequality by x, that is (x^2-1)*x/x > 0*x => (x^2-1)/x < 0 and the sign is flipped because x < 0. Multiplying both sides by x is what you do when we simply "cross-multiply" but we can only safely cross-multiply when it is equality but not in inequality.

Secondly, upon solving our equation we arrived at:
x < 1

Now x in this equation is -ve, since we are considering a case where x < 0. So if we leave it like x < 1 then the x in this case is not comparable with x in our solution for the case where x > 0. So in order to compare apples with apples, we need to get x to mean the same thing for both cases.
Therefore, we multiply both sides by -1 (which flips the sign again) to arrive at:
x > -1

Also, try picking few numbers in and outside the boundaries in the original equation, which may also help.

crejoc wrote:
Actually i got this problem from the very link you mentioned of HongHu's post, as it is not explained in detail there , i came up posting it in a separate thread...

On a separate note, I think if you have doubts on questions in a particular post then its better to post your doubts on that post itself, so as to keep all responses on the issue together, instead of starting the new post. I don't know if this is the policy on this forum or not, may be moderator can clarify?
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07 Aug 2009, 02:51
i think best approach in those 2 mins in GMAT will be to plug values.....just keep in mind that whenever it's about fraction....intervals should be 1<x; 0<x<1; -1<x<0; -1>x

we find only applicable are 1<x and -1<x<0

i hope this helps
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09 Aug 2009, 05:24
bhushan252 wrote:
i think best approach in those 2 mins in GMAT will be to plug values.....just keep in mind that whenever it's about fraction....intervals should be 1<x; 0<x<1; -1<x<0; -1>x

we find only applicable are 1<x and -1<x<0

i hope this helps

Yes the faster the better, thanks for your approach...
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03 Apr 2011, 13:27
X - (1/X) > 0

=> (X2 -1)/X >0

=> X^2>1 , X>0 or X^2<1 , X<0

=> X>1 or -1<x<0

Last edited by Spidy001 on 03 Apr 2011, 18:46, edited 1 time in total.
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03 Apr 2011, 18:24
@Spidy001, I don't think you can multiply by x with the inequality sign intact without knowing the sign of x.

See this solution :

=> x - 1/x > 0

=> (x^2 -1)/x > 0

So the sign of numerator and denominator are same and if x > 0 then x^2 - 1 > 0 and if x < 0 then x^2 - 1 < 0

(x+1)(x-1) > 0 and (x - 1) (x+1) < 0

so x > 1 Or x < -1 AND -1 < x < 1

Hnce x > 1 OR -1 < x < 0
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03 Apr 2011, 18:38
Agreed. Good catch.

subhashghosh wrote:
@Spidy001, I don't think you can multiply by x with the inequality sign intact without knowing the sign of x.

See this solution :

=> x - 1/x > 0

=> (x^2 -1)/x > 0

So the sign of numerator and denominator are same and if x > 0 then x^2 - 1 > 0 and if x < 0 then x^2 - 1 < 0

(x+1)(x-1) > 0 and (x - 1) (x+1) < 0

so x > 1 Or x < -1 AND -1 < x < 1

Hnce x > 1 OR -1 < x < 0
Re: Solve the Inequality   [#permalink] 03 Apr 2011, 18:38
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