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# If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9

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If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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Updated on: 05 Jun 2019, 05:52
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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

Originally posted by rohitgoel15 on 16 Apr 2012, 09:42.
Last edited by Bunuel on 05 Jun 2019, 05:52, edited 2 times in total.
Renamed the topic and edited the question.
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Posts: 60605
Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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16 Apr 2012, 11:58
10
22
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

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.

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.

Hope it helps.
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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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20 Jul 2013, 06:57
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 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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20 Jul 2013, 07:32
1
8
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, ...

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: $$x^3<y^3$$.

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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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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05 Sep 2013, 03:47
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

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.

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.

I used x= +6 and -6 ..Which is true in both the cases..it shud be E..
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Posts: 60605
Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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05 Sep 2013, 03:52
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

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.

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.

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 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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27 Oct 2013, 22:43
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.

I have a question if the sign was "<" instead of ">" what would the solution be? (x+1)(x-3) < 0
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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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27 Oct 2013, 22:57
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.

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 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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28 Dec 2015, 02:46
1
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.

Bunuel - Is this really a 700 level question ?
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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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06 Jul 2018, 00:41
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.

Thanks,
GyM
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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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22 Jul 2018, 06:46
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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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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22 Jul 2018, 07:41
1
1

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

See my approach.
Attachment:

IMG-20180722-WA0005.jpg [ 84.46 KiB | Viewed 20397 times ]

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 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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18 Jan 2019, 03:23
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.

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 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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18 Jan 2019, 12:30
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 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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03 Mar 2019, 08:08
Hi,
I am confused
Dont we have to know the sign of x to answer this ?
in each cases, x can be negative and therefore the value will be <3 ?
x=3 works fine
But if x=-3, then answer is no
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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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07 Mar 2019, 21:31
hsn81960 wrote:
Hi,
I am confused
Dont we have to know the sign of x to answer this ?
in each cases, x can be negative and therefore the value will be <3 ?
x=3 works fine
But if x=-3, then answer is no

The question says "If x is positive".
This is a very common oversight and cause of mistakes. Read carefully!
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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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07 Dec 2019, 04:47
Experts Please suggest if this method is correct. I solved it as below. Request Bunuel, Veritaskarishma to suggest.

St1 (x-1)^2>4
x^2-2x+1>4
x^2-2x>3
x(x-2)>3
So x>3 or x-2>3
X>3 or x>5 Hence St 1 sufficient

St 2 (x-2)^2>9
x^2-4x+4>9
x^2-4x>5
x(x-4)>5
x>5 or x-4>5
x>5 or x>9

st 2 sufficient.

Ans Choice D
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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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07 Dec 2019, 06:38
Hi can we solve the equations in the following way

(x-1)^2>4
So on expanding we get x^2-2x+1>4;
x^2-2x>3
x(x-2)>3
so x>3 or x>5

Is this the correct way to solve this equation?
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If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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Updated on: 08 Dec 2019, 08:15
1
Kshah001 wrote:
Experts Please suggest if this method is correct. I solved it as below. Request Bunuel, Veritaskarishma to suggest.

St1 (x-1)^2>4
x^2-2x+1>4
x^2-2x>3
x(x-2)>3
So x>3 or x-2>3
X>3 or x>5 Hence St 1 sufficient

St 2 (x-2)^2>9
x^2-4x+4>9
x^2-4x>5
x(x-4)>5
x>5 or x-4>5
x>5 or x>9

st 2 sufficient.

Ans Choice D

Your mistake is in this step:
"x(x-2)>3
So x>3 or x-2>3"

You are most likely confusing this with when it is equal to zero -- for example:
x(x-2) = 0 ---> in this case, x = 0 OR x-2 = 0. (One of the two things we are multiplying on the left must be zero, to get zero on the right)
Important ---> This does not apply if it's not equal to zero.

Instead, we want to take the square root of both sides:

(1) $$(x - 1)^2$$ > 4
|x-1| > 2 (Important: when we take the square root of a square, we get the absolute value. It's a common mistake to forget this. For this problem, it doesn't affect the answer, because the question says "if x is positive", and the negative values don't apply. Another common mistake is to miss the "If x is positive" -- read carefully!)

Case A: If x-1 is positive, then x-1 > 2 ---> x > 3
Case B: If x-1 is negative, then the absolute value flips the signs on the left: -x+1 > 2 ---> x < -1 ---> this doesn't apply, because the question says "If x is positive".
Therefore, our answer to the question, "is x>3?", is YES, and (1) is sufficient.

We do the same process for statement 2:
(2) $$(x - 2)^2$$ > 9
|x-2| > 3
x > 5
Again, the negative case doesn't apply here. (If x-2 is negative, then the absolute value flips the signs on the left: -x+2 > 3 ---> x < -1 ---> this doesn't apply, because the question says "If x is positive")
Therefore, our answer to the question, "is x>3?", is YES, and (2) is sufficient.

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Originally posted by GMATCoachBen on 07 Dec 2019, 21:48.
Last edited by GMATCoachBen on 08 Dec 2019, 08:15, edited 1 time in total.
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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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08 Dec 2019, 05:11
rohitgoel15 wrote:
If x is positive, is x > 3 ?

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

x is positive

(1) |x-1| >2
x>3. Sufficient

(2) |x-2|>3
x>5 Sufficient

D is correct.
Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9   [#permalink] 08 Dec 2019, 05:11

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