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

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New post 16 Apr 2012, 11:58
10
20
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 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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New post 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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New post 20 Jul 2013, 07:32
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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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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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New post 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

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

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New post 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

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

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New post 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.

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

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New post 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.

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

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New post 28 Dec 2015, 02:46
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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 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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New post 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.


Answer D.


Thanks,
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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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New post 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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New post 22 Jul 2018, 07:41
1
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 12572 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 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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New post 22 Jul 2018, 08:24
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 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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New post 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.

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

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New post 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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New post 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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New post 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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New post 22 May 2019, 11:19
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



I have a very basic ( silly) query -

Why...
|x-1| > 2, and not |x-1| > |2|?


Thanks in advance :)

Regards,
Tamal
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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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New post 05 Jun 2019, 05:51
1
tamal99 wrote:
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



I have a very basic ( silly) query -

Why...
|x-1| > 2, and not |x-1| > |2|?


Thanks in advance :)

Regards,
Tamal


Those two are the same because |2| = 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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New post 21 Jul 2019, 08:28
Bunuel wrote:
tamal99 wrote:
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



I have a very basic ( silly) query -

Why...
|x-1| > 2, and not |x-1| > |2|?


Thanks in advance :)

Regards,
Tamal


Those two are the same because |2| = 2.



Let me extend the question a bit.

(x-1)^2 > 4 ...now, if you take square root of both sides, shouldn't we get => |x-1| > +/- 2?

This, I thought, should give us 4 conditions:

1. (x-1)>2
2. (x-1)>-2
3. -x+1<2 (or -x+1>2?)
4. -x+1<-2 (or -x+1>-2?)

Please correct me.
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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9   [#permalink] 21 Jul 2019, 08:28
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