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

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

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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adkikani

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

See my approach.



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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.

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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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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Experts Please suggest if this method is correct. I solved it as below. Request Bunuel, Veritaskarishma to suggest.

Also I am confused - why am i getting different range for x. Please help

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.

Answer is D
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x is positive

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

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

D is correct.

is the only way to deduce your comment on X must be to the *left of the smaller root and right of the greater root on the x domain* by graphical interpretation since its a parabola? at first I was only able to test out the numbers to show that the solution only lies in the above * section by plugging in numbers because the expansion illustrated the number you plug in must result in x ">"0.

If x is positive, is \(x > 3 \)?

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

\((x - 1)^2 - 2^2 > 0\)

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

The values of x that satisfies the quadratic inequality : \(x < -1\) or \(x > 3\)

Since x is positive , we can confirm that \(x > 3.\)

Statement 1 alone is sufficient.

(2) \((x - 2)^2 > 9\)

\((x - 2)^2 - 3^2 >0\)
\((x-2+3) (x-2-3) >0\)
\((x+1)(x-5)>0\)

The values of x that satisfies the quadratic inequality : \(x < -1 \) or \(x > 5 \)

Since x is positive , we can confirm that \(x > 5.\)

So is x > 3 ? Always YES

Statement 2 alone is sufficient.

Option D is the answer.

Thanks,
Clifin J Francis,
GMAT SME
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1) If (x-1)^2 must be greater than 4 it means that (x-1) must be either greater than 2 OR smaller than -2.
x-1>2 ==> x>3
x-1<-2 ==> x<-1
Because x must be positive then x must be greater than 3. Sufficient.

2) Same reasoning.
x-2>3 ==> x>5
x-2<-3 ==> x<-1
Because x must be positive then x must be greater than 3. Sufficient.

Answer D

Hello,

I'm confused as to how you went from

this:
\((x+1) (x-3)>0\)

to this:
The values of x that satisfies the quadratic inequality : \(x < -1\) or \(x > 3\)

Inequalities is definitely not my strong suit. My approach was basically involved me looking at (x-1)(x-1) > 4 and saying to myself, okay, x-1 must be greater than 2 to satisfy the inequality OR x-1 must be less than -2, which leads to X>3 or X<-1... but apparently from looking at these comments, this is not the way to do it.

Hi Bunuel,

About your approach down below, it doesn't necessarily state that x is an integer, so by using this aproach you could argue that x can equal 2.1. In that case it's not sufficient to answer the question?


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.[/quote]

avigutman GMATCoachBen

I'm not sure if my way of solving this problem is correct but wanted to share it as it took less than 20 seconds (screenshot below). I'd love your input on whether such an approach makes sense for similar questions where the quadratic approach may be too time consuming.

achloes The work you showed works ok for this problem, because it says "x is positive", so the negative case doesn't apply when you take the square root. When you took the square roots, you didn't write out the absolute value signs: |x-1| > 2 and |x-2| > 3

It's good that you are able to work quickly, but as a habit, be careful on that step to avoid the common trap of ignoring the absolute value.


Here's my full solution below, showing all the work of taking the square root on 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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