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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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Intern
Joined: 19 Apr 2018
Posts: 8
Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9  [#permalink]

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06 Nov 2019, 19:54
Bunuel

Hi Bunuel,

I solved the question in the same manner as you did; however, the only difference was I kept the ">" sign for the -1 root as well. So for the first statement the solution I got was x > -1 and x > 3. This is sufficient since the overlap of the two roots is for all the values of X greater than 3. My question is about the statement in bold below. Why does the ">" indicate that the solution lies to the left of the smaller root? Is this a certain theory or concept? I wasn't aware of this so when I solved the inequality I kept the ">" sign for both roots.

(1) (x - 1)^2 > 4 --> (x+1)(x−3)>0(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<−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+1)(x−5)>0(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<−1x<−1 or x>5x>5. Since given that xx is positive then only one range is valid: x>5x>5. Sufficient.

Thank you!
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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 Nov 2019, 23:47
Gmatstudent2018 wrote:
Bunuel

Hi Bunuel,

I solved the question in the same manner as you did; however, the only difference was I kept the ">" sign for the -1 root as well. So for the first statement the solution I got was x > -1 and x > 3. This is sufficient since the overlap of the two roots is for all the values of X greater than 3. My question is about the statement in bold below. Why does the ">" indicate that the solution lies to the left of the smaller root? Is this a certain theory or concept? I wasn't aware of this so when I solved the inequality I kept the ">" sign for both roots.

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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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07 Nov 2019, 07:14
Hi Bunuel,

Apologies for the follow-up question and the back and forth. I read the explanation in the links in your original post, but I'm still not 100% tracking. If the inequality is (x-3) (x+1) > 0, why is that any different than when the equation is = 0? I don't understand the reasoning behind changing the ">" sign for the root -1.

Thank you

Bunuel wrote:
Gmatstudent2018 wrote:
Hi Bunuel,

I solved the question in the same manner as you did; however, the only difference was I kept the ">" sign for the -1 root as well. So for the first statement the solution I got was x > -1 and x > 3. This is sufficient since the overlap of the two roots is for all the values of X greater than 3. My question is about the statement in bold below. Why does the ">" indicate that the solution lies to the left of the smaller root? Is this a certain theory or concept? I wasn't aware of this so when I solved the inequality I kept the ">" sign for both roots.

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Joined: 19 Sep 2019
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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 Nov 2019, 12:48
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:
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.

Hey Bunuel,

can you point out my flaw in logic please?

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

x-1>2 so x>3

or

x-1>-2 so x>-1

Of course, checking the values one sees that it doesn´t work with 1 or 2, but just by taking a look at my results I figured x isn´t necessarily greater than 3.

Did the same for statement to, there I got x>5 and x>-1 and applied the same reasoning.

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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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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, 21:48
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:
(1) $$(x - 2)^2$$ > 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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Re: If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9   [#permalink] 07 Dec 2019, 21:48

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