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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5

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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 18 Jun 2012, 15:46
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Is x > 1?


(1) \((x+1)(|x| - 1) > 0\)

(2) \(|x| < 5\)

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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 18 Jun 2012, 16:15
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Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.


(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 18 Jun 2012, 16:30
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enigma123 wrote:
Thanks Bunuel. Can you please explain how did you get this?

If \(x>0\) then \(|x|=x\)


Check this: math-absolute-value-modulus-86462.html

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\).

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

Hope it helps.
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 12 Jun 2014, 00:44
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sevaro wrote:
Is x>1?

(1) (x+1)(|x|-1) > 0
(2) |x|<5

I got it right plugging in number. Any other options?

Thanks

This question is rated as hard by GMAC.


Question: Is x > 1?

(1) \((x+1)(|x|-1) > 0\)
For the left hand side to be positive, either both factors are positive or both are negative.

If both are positive,
x+1 > 0, x > -1
AND
|x|-1 > 0, |x|> 1 which means either x < -1 or x > 1
This is possible only when x > 1

If both are negative,
x+1 < 0, x < -1
AND
|x|-1 < 0, |x| < 1 which means -1 < x < 1
Both these conditions cannot be met and hence this is not possible.

This gives us only one solution: x > 1
So we can answer the question asked with "Yes".

(2) \(|x|<5\)
This implies that -5 < x < 5
x may be less than or more than 1. Not sufficient.

Answer (A)
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 27 Jan 2013, 23:17
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fozzzy wrote:
Is x>1

1) (x+1)(lxl-1) > 0
2) lxl < 5


Statement 1:
For (x+1)(lxl-1) > 0, we should have either (x+1)>0 and (lxl-1) > 0 or (x+1)<0 and (lxl-1) < 0
when (x+1)>0 and (lxl-1) > 0
(x+1)>0 => x>-1
(lxl-1) > 0 => x>1 or x <-1
From above two, possible solution is x>1
when (x+1)>0 and (lxl-1) < 0
(x+1)<0 => x<-1
(lxl-1) < 0 => -1<x<1
Both of these can not be satisfied by any value of x.
Hence we get only 1 solution, x>1. which is what we wanted to ascertain. Sufficient.

Statement 2:
|x| <5
=> -5<x<5
Clearly not sufficient to tell whether x>1 or not.

Ans A it is.
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 24 Jun 2013, 02:11
1
WholeLottaLove wrote:
So here is my question:

For #1 we have two cases: positive and negative

For x≥0
(x+1)(|x|-1)>0
(x+1)(x-1)>0
x^2-1>0
x^2>1

Of course, x could be 2 or negative two for all we know, so it seems like this is insufficient as we are testing for the range of x>1


x can never be -2, as because you have assumed that \(x\geq0\). Also, from the inequality you have correctly arrived at, i.e. \(x^2>1\) \(\to\) x>1 OR x<-1. As assumption was \(x\geq0\). thus only x>1 condition is valid. Also,as x>1 automatically makes \(x\geq0\), thus the correct range is x>1. Sufficient.
Quote:
For x≤0 No need to include equality with zero twice.
(x+1)(|x|-1)>0
(x+1)(-x-1)>0 This leads to \(-(x+1)^2>0\) and this is not possible for any real value of x. So,there is no solution for this.
-x^2-1>0
-x^2>-1


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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 22 Nov 2014, 07:30
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joseph0alexander wrote:
Bunuel,

Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?

1) x+1 > 0
2) x-1 > 0
3) -x-1 > 0

If correct, could you please demonstrate on how to proceed?


This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 06 Dec 2014, 14:52
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Is x > 1?

I would approach it the following way :

(1) (x+1) (|x| - 1) > 0

(+) * (+) > 0 or (-) * (-) > 0

For both of the parts to be positive we can see that x >1 . Just by trying few values you can figure this out. X cant be Zero as then the second part becomes - . X cant be 1 as then second part becomes 0 and hence the whole LHS becomes Zero.

For both of the parts to be negative we try any value less an Zero and see that no value will satisfy the equation. Hence X cannot be negative.. Hence A is Sufficient.

(2) |x| < 5

Clearly not sufficient

Answer is A.
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 24 Aug 2015, 16:02
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gmatcracker24 wrote:
Is X > 1 ?

1) (X+1) (|X| -1 ) >0

2) |X| <5


Starting with 2) -5 < X < 5 .hence clearly not sufficient .

However, i am not sure how to go about the 1st one .


Search for a question before posting a new topic. The question would have been discussed already.

As for the question,

Per statement 1, (x+1)(|x|-1) >0

Case 1: for x \(\geq\) 0 ---> |x| = x ----> (x+1)(x-1) > 0 ----> x>1 or x<-1 but as x \(\geq\) 0 ---> only possible case is x>1

Case 2: for x<0 ---> |x|=-x ---> (x+1)(-x-1)>0 ----> \(-(x+1)^2 > 0\) ---> \((x+1)^2<0\) . Now a square can never be <0 and thus x can not be negative.

Thus the only possible case from statement 1 is for x \(\geq\) 0 which gives a definite "yes" for x>1.

Hence A is the correct answer.
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 12 Mar 2016, 04:47
1
truongynhi wrote:
Hi Bunuel,

I have a question.

(x+1)(|x|-1)=0 has x=1 and x=-1 as its solution. However, x=-1 is a repeated root. Hence, on the number line, the sign will not change when it passes x=-1.

--------(-1)---------(1)+++++++

Therefore, only when x>1, (x+1)(|x|-1)>0.

Is my solution correct? I remember my highschool math teacher said something like 'the sign doesn't change when it passes a double root'. But, it's years ago and I just want to make sure that the approach is valid.

Thank you!


Hi,
the solution here is correct, But I am not sure if it has to do with Double root in this specific instance..

Double root is generally when a polynomial has two equal roots..
so when ever you are placing the polynomial as>0, you cannot do anything to the squared part as it is always positive so it will depend on the other roots..

example--
\((x-3)^2*(x-2)>0\)..
as can be seen \((x-3)^2\)will always be >0 except at x=3.. so we require to just look for x-2..
and x-2 will be positive only whenx>2..
so solution will be x>2 given that x is NOT equal to 3..

what is happening here --
(x+1)(|x|-1)>0.
x with a value <1 will make both (x+1) and (|x|-1) of oposite sign
a) x<-1
(x+1) is negtaive and (|x|-1) is positive
b) a) -1< x<1
(x+1) is positive and (|x|-1) is negative
so here (x+1)(|x|-1) will be negative and there will be no solutions for (x+1)(|x|-1)>0.

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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 18 Jun 2012, 16:26
Thanks Bunuel. Can you please explain how did you get this?

If \(x>0\) then \(|x|=x\)
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 19 Jun 2012, 12:22
Bunuel wrote:
Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.




(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.


Dear Bunuel,
i got (x+1)^2<0 .
and further solved to x+1<0 giving x <-1
and ended up in wrong answer E.
square of a number cannot be negative. but just confused for (x+1)^2 where x is a variable?
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 20 Jun 2012, 01:07
kashishh wrote:
Bunuel wrote:
Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.




(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.


Dear Bunuel,
i got (x+1)^2<0 .
and further solved to x+1<0 giving x <-1
and ended up in wrong answer E.
square of a number cannot be negative. but just confused for (x+1)^2 where x is a variable?


It doesn't matter that \(x\) is a variable, it's still some number and so is \(x+1\). So, \((x+1)^2\) is a square of that number and it cannot be negative.

Also your way of solving is not correct: \((x+1)^2<0\) does not mean \(x+1<0\) it means that \(|x+1|<0\). From that you could deduce the same: since absolute value cannot be negative then this equation has no solution.

Hope it's clear.
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 21 Jun 2012, 01:04
Hi Bunuel, I understood your solution but can you please tell me where am I going wrong?

When I see the expression, (x+1)(|x|-1)>0, I immediately think that these two brackets must be either positive or negative.

Hence, if that take both of them are positive, then x>-1 and and x>1 & x<-1

For these two ranges we realize when we start plugging in number that only for x>1 the equation hold true.--------(1)

Similarly, if we take both of them to be negative, then x<-1 and x<1 & x>-1

for x<-1 for example, -2, one of the brackets turn out to be positive, which does not satisfy our initial assumption that both must be negative.

for x<1 ex 0, one bracket is +ve and one is -ve <Not Desired>; for 0, the inequality is not > 0 <not desired>

if we go still less it follows the first case

for x>-1 pretty much follows like the above. Hence nothing desired.--------(2)

Hence, x>1

A is sufficient.
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 21 Jun 2012, 02:41
pavanpuneet wrote:
Hi Bunuel, I understood your solution but can you please tell me where am I going wrong?

When I see the expression, (x+1)(|x|-1)>0, I immediately think that these two brackets must be either positive or negative.

Hence, if that take both of them are positive, then x>-1 and and x>1 & x<-1

For these two ranges we realize when we start plugging in number that only for x>1 the equation hold true.--------(1)

Similarly, if we take both of them to be negative, then x<-1 and x<1 & x>-1

for x<-1 for example, -2, one of the brackets turn out to be positive, which does not satisfy our initial assumption that both must be negative.

for x<1 ex 0, one bracket is +ve and one is -ve <Not Desired>; for 0, the inequality is not > 0 <not desired>

if we go still less it follows the first case

for x>-1 pretty much follows like the above. Hence nothing desired.--------(2)

Hence, x>1

A is sufficient.


First of all I wouldn't recommend to solve this question the way you are doing.

Next, when you consider both multiples to be negative and get \(x<-1\) from the first one, then the second multiple automatically transformes to \((-x-1)\), since if \(x<-1<0\) then \(|x|=-x\). So, we have that \(-x-1<0\) must also be true or \(x>-1\), which contradicts the case for the first multiple (\(x<-1\)). So, both \(x+1\) and \(|x|-1\) can not be negative.

Also I think you got x<1 & x>-1 from |x|<1, and if yes, then it's not correct: \(|x|<1\) means that \(-1<x<1\). So, again \(x<-1\) (for the first multiple to be negative) and \(-1<x<1\) (for the second multiple to be negative) cannot simultaneously be true.


Hope it's clear.
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post Updated on: 28 Jan 2013, 00:25
so you just have to find the common region if its done on a number line and we ignore one of the cases, since there isn't a common region?

Originally posted by fozzzy on 28 Jan 2013, 00:21.
Last edited by fozzzy on 28 Jan 2013, 00:25, edited 1 time in total.
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 06 Feb 2013, 03:21
andrew40 wrote:
Bunuel wrote:
Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.


(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.


Sorry, I don't understand why we should consider the range where x>0. because of the absolute value?


We need to get rid of the modulus in the expression to solve it and this is the way to do that. Check here: is-x-134652.html#p1097668

Hope it helps.
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 18 May 2013, 05:56
Bunuel wrote:
Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.


(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.



Hi,

This is my first post so was little conscious to ask my doubt. In the above question, we took the roots as 1<x>1. However, in this question as mentioned below (unable to post the link as per new member rule)

is ((X-3)^2)^1/2 = 3-X ?

1) X does not = 3

2) -X|X| > 0

the roots are 3<x>3. Can you please explain the difference?
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 22 May 2013, 16:45
Please help,

When I try this question for statement 1:
(x+1) (|x| - 1) > 0
(x+1)|x|-(x+1)>0
(x+1)|x| >(x+1)
|x| > 1

x>1 ; x<-1
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5  [#permalink]

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New post 23 May 2013, 02:14
smartyman wrote:
Please help,

When I try this question for statement 1:
(x+1) (|x| - 1) > 0
(x+1)|x|-(x+1)>0
(x+1)|x| >(x+1)
|x| > 1

x>1 ; x<-1


Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign.

So you cannot reduce both parts of inequality (x+1)|x|>(x+1) by x+1 as you don't know the sign of x+1: if x+1>0 you should write |x|>1 BUT if x+1<0 you should write |x|<1 (flip the sign).

Hope it helps.
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Re: Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5   [#permalink] 23 May 2013, 02:14

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