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If x is positive, is x > 3

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If x is positive, is x > 3  [#permalink]

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If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4
(2) (x - 2)^2 > 9

Can someone point a mistake in my method?
(1)
Taking one of the equations:
(x - 1)^2 > 4
x^2 + 1 - 2x > 4
x^2 + 1 - 2x - 4 > 0
x^2 - 3x + 1x - 3 > 0
(x-3) (x+1) > 0
x > 3 and x > -1

while in the explanation given the ans is coming out to be x > 3 and x < -1

Please help ..

Originally posted by rohitgoel15 on 16 Apr 2012, 08:42.
Last edited by ENGRTOMBA2018 on 14 Oct 2015, 15:14, edited 1 time in total.
Formatted the question
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If x is positive, is x > 3  [#permalink]

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New post 16 Apr 2012, 10:58
9
18
rohitgoel15 wrote:
If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4
(2) (x - 2)^2 > 9

Can someone point a mistake in my method?
(1)
Taking one of the equations:
(x - 1)^2 > 4
x^2 + 1 - 2x > 4
x^2 + 1 - 2x - 4 > 0
x^2 - 3x + 1x - 3 > 0
(x-3) (x+1) > 0
x > 3 and x > -1

while in the explanation given the ans is coming out to be x > 3 and x < -1

Please help ..


If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> \((x+1)(x-3)>0\) --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<-1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient.

(2) (x - 2)^2 > 9 --> \((x+1)(x-5)>0\) --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<-1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient.

Answer: D.

Solving inequalities:
http://gmatclub.com/forum/x2-4x-94661.html#p731476 (check this one first)
http://gmatclub.com/forum/inequalities-trick-91482.html
http://gmatclub.com/forum/data-suff-ine ... 09078.html
http://gmatclub.com/forum/range-for-var ... me#p873535
http://gmatclub.com/forum/everything-is ... me#p868863


Another approach:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> since both sides of the inequality are non-negative then we can take square root from both parts: \(|x-1|>2\). \(|x-1|\) is just the distance between 1 and \(x\) on the number line. We are told that this distance is more than 2: --(-1)----1----3-- so, \(x<-1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient.

(2) (x - 2)^2 > 9 --> \(|x-2|>3\). The same here: \(|x-2|\) is just the distance between 2 and \(x\) on the number line. We are told that this distance is more than 3: --(-1)----2----5-- so, \(x<-1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient.

Answer: D.

Hope it helps.
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Re: If x is positive, is x > 3  [#permalink]

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New post 20 Jul 2013, 05:57
Hi bunuel,

Just wanted to clarify in the alternative approach you mentioned "non-negative" so if the other side of the inequality has a negative number, the only way to proceed with the problem is by expansion?
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Re: If x is positive, is x > 3  [#permalink]

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New post 20 Jul 2013, 06:32
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fozzzy wrote:
Hi bunuel,

Just wanted to clarify in the alternative approach you mentioned "non-negative" so if the other side of the inequality has a negative number, the only way to proceed with the problem is by expansion?


If it were (x - 1)^2 > -4, it would simply mean that x can take any value.

As for general rules for inequalities: taking the square root, squaring, ...

ADDING/SUBTRACTING INEQUALITIES:


You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

RAISING INEQUALITIES TO EVEN/ODD POWER:


A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

For multiplication check here: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html#p1242652

THEORY ON INEQUALITIES:


x2-4x-94661.html#p731476
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
inequations-inequalities-part-154664.html
inequations-inequalities-part-154738.html

QUESTIONS:


All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If x is positive, is x > 3  [#permalink]

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New post 05 Sep 2013, 02:47
Bunuel wrote:
rohitgoel15 wrote:
If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4
(2) (x - 2)^2 > 9

Can someone point a mistake in my method?
(1)
Taking one of the equations:
(x - 1)^2 > 4
x^2 + 1 - 2x > 4
x^2 + 1 - 2x - 4 > 0
x^2 - 3x + 1x - 3 > 0
(x-3) (x+1) > 0
x > 3 and x > -1

while in the explanation given the ans is coming out to be x > 3 and x < -1

Please help ..


If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> \((x+1)(x-3)>0\) --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<-1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient.

(2) (x - 2)^2 > 9 --> \((x+1)(x-5)>0\) --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<-1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient.

Answer: D.

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

Another approach:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> since both sides of the inequality are non-negative then we can take square root from both parts: \(|x-1|>2\). \(|x-1|\) is just the distance between 1 and \(x\) on the number line. We are told that this distance is more than 2: --(-1)----1----3-- so, \(x<-1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient.

(2) (x - 2)^2 > 9 --> \(|x-2|>3\). The same here: \(|x-2|\) is just the distance between 2 and \(x\) on the number line. We are told that this distance is more than 3: --(-1)----2----5-- so, \(x<-1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient.

Answer: D.


I used x= +6 and -6 ..Which is true in both the cases..it shud be E..
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Re: If x is positive, is x > 3  [#permalink]

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New post 05 Sep 2013, 02:52
SUNGMAT710 wrote:
Bunuel wrote:
rohitgoel15 wrote:
If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4
(2) (x - 2)^2 > 9

Can someone point a mistake in my method?
(1)
Taking one of the equations:
(x - 1)^2 > 4
x^2 + 1 - 2x > 4
x^2 + 1 - 2x - 4 > 0
x^2 - 3x + 1x - 3 > 0
(x-3) (x+1) > 0
x > 3 and x > -1

while in the explanation given the ans is coming out to be x > 3 and x < -1

Please help ..


If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> \((x+1)(x-3)>0\) --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<-1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient.

(2) (x - 2)^2 > 9 --> \((x+1)(x-5)>0\) --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<-1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient.

Answer: D.

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

Another approach:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> since both sides of the inequality are non-negative then we can take square root from both parts: \(|x-1|>2\). \(|x-1|\) is just the distance between 1 and \(x\) on the number line. We are told that this distance is more than 2: --(-1)----1----3-- so, \(x<-1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient.

(2) (x - 2)^2 > 9 --> \(|x-2|>3\). The same here: \(|x-2|\) is just the distance between 2 and \(x\) on the number line. We are told that this distance is more than 3: --(-1)----2----5-- so, \(x<-1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient.

Answer: D.


I used x= +6 and -6 ..Which is true in both the cases..it shud be E..


Stem says that x is a positive number, thus x cannot be -6.

Hope it's clear.
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Re: If x is positive, is x > 3  [#permalink]

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New post 27 Oct 2013, 21:43
Bunuel wrote:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> \((x+1)(x-3)>0\) --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<-1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient.

(2) (x - 2)^2 > 9 --> \((x+1)(x-5)>0\) --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<-1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient.

Answer: D.



I have a question if the sign was "<" instead of ">" what would the solution be? (x+1)(x-3) < 0
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If x is positive, is x > 3  [#permalink]

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New post 27 Oct 2013, 21:57
fozzzy wrote:
Bunuel wrote:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> \((x+1)(x-3)>0\) --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<-1\) or \(x>3\). Since given that \(x\) is positive then only one range is valid: \(x>3\). Sufficient.

(2) (x - 2)^2 > 9 --> \((x+1)(x-5)>0\) --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<-1\) or \(x>5\). Since given that \(x\) is positive then only one range is valid: \(x>5\). Sufficient.

Answer: D.



I have a question if the sign was "<" instead of ">" what would the solution be? (x+1)(x-3) < 0


\((x+1)(x-3)<0\) --> \(-1<x<3\).
\((x+1)(x-5)<0\) --> \(-1<x<5\).
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Re: If x is positive, is x > 3  [#permalink]

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New post 28 Dec 2015, 01:46
Key thing to note is that X is positive.

Statement 1 - If (x-1)^2 > 4, x has to be greater than 3. If x = 1, 2 or 3, then the expression will only be < or equal to 4. Hence 1 is sufficient.

Statement 2 - If (x - 2)^2 > 9, x has to be greater than 3. If x = 1, 2 or 3, then the expression will not be > 9. Hence 2 is also sufficient.

Answer - D.

Bunuel - Is this really a 700 level question ?
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Re: If x is positive, is x > 3  [#permalink]

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New post 05 Jul 2018, 23:41
rohitgoel15 wrote:
If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4
(2) (x - 2)^2 > 9




Given x > 0, asked is x > 3?

Statement 1: (x - 1)^2 > 4

Hence lx - 1l > 2, so we have x > 3 or x < -1

Since x > 0, there fore x > 3. Answer is YES.

Statement 1 alone is Sufficient.


Statement 2: (x - 2)^2 > 9

Hence lx - 2l > 3, so we have x > 5 or x < -1

Since x > 0, therefore x > 5 & hence x > 3. Answer is YES.

Statement 2 alone is Sufficient.


Answer D.


Thanks,
GyM
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Re: If x is positive, is x > 3  [#permalink]

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New post 22 Jul 2018, 05:46
Bunuel niks18 pushpitkc chetan2u KarishmaB amanvermagmat pikolo2510

I understood alternate approach by Bunuel since taking square root
of a squared variable and turning it in to absolute value came more naturally to me.

Where I faltered is doing this is less than 2 mins as per below steps:
|x-1| > 2
x-1 >2 or -x+1 > 2
x>3 or -x>1 ie x<-1

Similar painstaking calculations I had to do for St 2
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If x is positive, is x > 3  [#permalink]

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New post 22 Jul 2018, 06:41
adkikani

remember a small rule and this question can be done in 30 seconds.

See my approach.
Attachment:
IMG-20180722-WA0005.jpg
IMG-20180722-WA0005.jpg [ 84.46 KiB | Viewed 6028 times ]



adkikani wrote:
Bunuel niks18 pushpitkc chetan2u KarishmaB amanvermagmat pikolo2510

I understood alternate approach by Bunuel since taking square root
of a squared variable and turning it in to absolute value came more naturally to me.

Where I faltered is doing this is less than 2 mins as per below steps:
|x-1| > 2
x-1 >2 or -x+1 > 2
x>3 or -x>1 ie x<-1

Similar painstaking calculations I had to do for St 2

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Re: If x is positive, is x > 3  [#permalink]

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New post 22 Jul 2018, 07:24
Your approach is perfect :thumbup: .
Practice more of such questions so that the steps come like a reflex and save time.

adkikani wrote:
Bunuel niks18 pushpitkc chetan2u KarishmaB amanvermagmat pikolo2510

I understood alternate approach by Bunuel since taking square root
of a squared variable and turning it in to absolute value came more naturally to me.

Where I faltered is doing this is less than 2 mins as per below steps:
|x-1| > 2
x-1 >2 or -x+1 > 2
x>3 or -x>1 ie x<-1

Similar painstaking calculations I had to do for St 2

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Re: If x is positive, is x > 3  [#permalink]

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New post 18 Jan 2019, 02:23
f x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> since both sides of the inequality are non-negative then we can take square root from both parts: |x−1|>2|x−1|>2. |x−1||x−1| is just the distance between 1 and xx on the number line. We are told that this distance is more than 2: --(-1)----1----3-- so, x<−1x<−1 or x>3x>3. Since given that xx is positive then only one range is valid: x>3x>3. Sufficient.

(2) (x - 2)^2 > 9 --> |x−2|>3|x−2|>3. The same here: |x−2||x−2| is just the distance between 2 and xx on the number line. We are told that this distance is more than 3: --(-1)----2----5-- so, x<−1x<−1 or x>5x>5 . Since given that xx is positive then only one range is valid: x>5x>5. Sufficient.

Answer: D.

So, in the second approach, the answer for second statement comes out x>5. Which means x cannot be 4. Right? This will make the second statement incorrect.
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Re: If x is positive, is x > 3  [#permalink]

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New post 18 Jan 2019, 11:30
wali786 wrote:
If x is positive, is x > 3 ?

So, in the second approach, the answer for second statement comes out x>5. Which means x cannot be 4. Right? This will make the second statement incorrect.


It is unnecessary to ask yourself whether x can be 4. The question is YES/NO, "is x > 3?" If we know x > 5, the answer to the question is a definite YES --> Sufficient.


Side note: Read carefully! It's a good habit to be sure to write down, "x > 0" -- the most common error on these questions is rushing and ignoring "If x is positive". (The majority of the errors were E for this one)
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Re: If x is positive, is x > 3 &nbs [#permalink] 18 Jan 2019, 11:30
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