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# M02-32

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:18
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25% (medium)

Question Stats:

63% (00:40) correct 37% (00:39) wrong based on 235 sessions

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What is the value of $$x^2$$?

(1) $$x = \sqrt{2x - 1}$$

(2) $$x^2 = x^3$$

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Math Expert
Joined: 02 Sep 2009
Posts: 45333

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16 Sep 2014, 00:18
Official Solution:

Statement (1) by itself is sufficient. From Statement (1):
$$x^2 = 2x - 1$$
$$x^2 - 2x + 1 = 0$$
$$(x-1)^2 = 0$$
$$x -1 = 0$$
$$x = 1$$

Thus, $$x^2 = 1$$.

Statement (2) by itself is insufficient. From Statement (2):
$$x^2 - x^3 = 0$$
$$x^2 (1 - x) = 0$$

$$x = 0$$ or $$x =1$$.

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15 Mar 2015, 12:46
I think this question is poor and not helpful.
How can statement (2) be correct if there are two value for x? (x= 0 & 1)
Math Expert
Joined: 02 Sep 2009
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15 Mar 2015, 14:12
KPMM07 wrote:
I think this question is poor and not helpful.
How can statement (2) be correct if there are two value for x? (x= 0 & 1)

Sorry, but it's not clear at all what you mean. There is nothing wrong with the question... Can you please elaborate?
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20 Mar 2015, 07:17
KPMM07 wrote:
I think this question is poor and not helpful.
How can statement (2) be correct if there are two value for x? (x= 0 & 1)

Hi,

Thats exactly the reason why (2) is not correct. We will have two values for x^2.
when x= 0, x^2 = 0 and
when x = 1 , x^2 = 1

Hence (2) is insufficient.

Thanks!
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20 Mar 2015, 07:19
sytabish wrote:
KPMM07 wrote:
I think this question is poor and not helpful.
How can statement (2) be correct if there are two value for x? (x= 0 & 1)

Hi KPMM07 ,

Thats exactly the reason why (2) is not correct. We will have two values for x^2.
when x= 0, x^2 = 0 and
when x = 1 , x^2 = 1

Hence (2) is insufficient.

Thanks!
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27 Apr 2015, 05:37
Bunuel wrote:
Official Solution:

Statement (1) by itself is sufficient. From Statement (1):
$$x^2 = 2x - 1$$
$$x^2 - 2x + 1 = 0$$
$$(x-1)^2 = 0$$
$$x -1 = 0$$
$$x = 1$$

Thus, $$x^2 = 1$$.

Statement (2) by itself is insufficient. From Statement (2):
$$x^2 - x^3 = 0$$
$$x^2 (1 - x) = 0$$

$$x = 0$$ or $$x =1$$.

Hi Bunuel

I have a problem with your analysis of statement (1). x=1 does not satisfy the given equation, x = (2)^0.5 *x -1.

Squaring the given equation will not give 'x^2 = 2x - 1' although it will give you a unique value of x that is real. Please check.

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 45333

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27 Apr 2015, 05:44
Engr2012 wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is sufficient. From Statement (1):
$$x^2 = 2x - 1$$
$$x^2 - 2x + 1 = 0$$
$$(x-1)^2 = 0$$
$$x -1 = 0$$
$$x = 1$$

Thus, $$x^2 = 1$$.

Statement (2) by itself is insufficient. From Statement (2):
$$x^2 - x^3 = 0$$
$$x^2 (1 - x) = 0$$

$$x = 0$$ or $$x =1$$.

Hi Bunuel

I have a problem with your analysis of statement (1). x=1 does not satisfy the given equation, x = (2)^0.5 *x -1.

Squaring the given equation will not give 'x^2 = 2x - 1' although it will give you a unique value of x that is real. Please check.

Thanks

Not sure what exactly are you doing there...

x=1 satisfies x^2 = 2x - 1: 1^1 = 2*1 - 1.
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Joined: 13 Apr 2015
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27 Apr 2015, 06:16
Bunuel wrote:
Engr2012 wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is sufficient. From Statement (1):
$$x^2 = 2x - 1$$
$$x^2 - 2x + 1 = 0$$
$$(x-1)^2 = 0$$
$$x -1 = 0$$
$$x = 1$$

Thus, $$x^2 = 1$$.

Statement (2) by itself is insufficient. From Statement (2):
$$x^2 - x^3 = 0$$
$$x^2 (1 - x) = 0$$

$$x = 0$$ or $$x =1$$.

Hi Bunuel

I have a problem with your analysis of statement (1). x=1 does not satisfy the given equation, x = (2)^0.5 *x -1.

Squaring the given equation will not give 'x^2 = 2x - 1' although it will give you a unique value of x that is real. Please check.

Thanks

Not sure what exactly are you doing there...

x=1 satisfies x^2 = 2x - 1: 1^1 = 2*1 - 1.

How did you get from x = (2)^0.5 *x -1 to x^2 = 2x - 1?
Math Expert
Joined: 02 Sep 2009
Posts: 45333

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27 Apr 2015, 06:20

How did you get from x = (2)^0.5 *x -1 to x^2 = 2x - 1?

Square $$x = \sqrt{2x - 1}$$ to get $$x^2 = 2x - 1$$.
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27 Apr 2015, 06:36
Bunuel wrote:
Engr2012 wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is sufficient. From Statement (1):
$$x^2 = 2x - 1$$
$$x^2 - 2x + 1 = 0$$
$$(x-1)^2 = 0$$
$$x -1 = 0$$
$$x = 1$$

Thus, $$x^2 = 1$$.

Statement (2) by itself is insufficient. From Statement (2):
$$x^2 - x^3 = 0$$
$$x^2 (1 - x) = 0$$

$$x = 0$$ or $$x =1$$.

Hi Bunuel

I have a problem with your analysis of statement (1). x=1 does not satisfy the given equation, x = (2)^0.5 *x -1.

Squaring the given equation will not give 'x^2 = 2x - 1' although it will give you a unique value of x that is real. Please check.

Thanks

Not sure what exactly are you doing there...

x=1 satisfies x^2 = 2x - 1: 1^1 = 2*1 - 1.

Bunuel,

If x^2 =1 , then x = 1 or -1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x -1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x - 1".
Current Student
Joined: 20 Mar 2014
Posts: 2644
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)

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27 Apr 2015, 06:40
Bunuel wrote:

How did you get from x = (2)^0.5 *x -1 to x^2 = 2x - 1?

Square $$x = \sqrt{2x - 1}$$ to get $$x^2 = 2x - 1$$.

Bunuel

The way question is getting displayed, it seems that square root is with 2 only and not with the entire thing (x-1). Please re format the question so that the root is displayed as per your comment.

I think I got my answer to the doubt that I had.
Math Expert
Joined: 02 Sep 2009
Posts: 45333

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27 Apr 2015, 06:41
Engr2012 wrote:
Bunuel,

If x^2 =1 , then x = 1 or -1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x -1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x - 1".

You are NOT correct.

If you square $$x = \sqrt{2x - 1}$$ you get $$x^2 = 2x - 1$$.
Next, if you substitute x=1 into $$x = \sqrt{2x - 1}$$ the equation will hold true.
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Joined: 13 Apr 2015
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27 Apr 2015, 06:44
Bunuel wrote:
Engr2012 wrote:
Bunuel,

If x^2 =1 , then x = 1 or -1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x -1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x - 1".

You are NOT correct.

If you square $$x = \sqrt{2x - 1}$$ you get $$x^2 = 2x - 1$$.
Next, if you substitute x=1 into $$x = \sqrt{2x - 1}$$ the equation will hold true.

I think the confusion is between 2^0.5*x-1 versus (2x-1)^(0.5), in the 2nd case the problem and solution as stated is correct...
Current Student
Joined: 20 Mar 2014
Posts: 2644
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Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
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27 Apr 2015, 07:08
Bunuel wrote:
Engr2012 wrote:
Bunuel,

If x^2 =1 , then x = 1 or -1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x -1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x - 1".

You are NOT correct.

If you square $$x = \sqrt{2x - 1}$$ you get $$x^2 = 2x - 1$$.
Next, if you substitute x=1 into $$x = \sqrt{2x - 1}$$ the equation will hold true.

I think the confusion is between 2^0.5*x-1 versus (2x-1)^(0.5), in the 2nd case the problem and solution as stated is correct...

Yes. Exactly. I was checking for X=1 with the square root with 2 only and not the entire expression. It is not clear from the way the question is getting displayed.
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Joined: 29 Oct 2013
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28 Apr 2015, 11:14
hi Bunuel
can you please solve my query? if x^2= x^3, does not it imply that X is Positive? then that would eliminate 0 as a possibility, making statement 2 sufficient. please tell me where i am going wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 45333

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29 Apr 2015, 02:10
djonga wrote:
hi Bunuel
can you please solve my query? if x^2= x^3, does not it imply that X is Positive? then that would eliminate 0 as a possibility, making statement 2 sufficient. please tell me where i am going wrong.

x^2 = x^3;

x^2 - x^3 = 0;

x^2(1 - x) = 0;

x = 0 or x = 1.
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17 Oct 2015, 14:00
I think this is a high-quality question and I agree with explanation. Why can't I divide both sides of statement 2 by X^2?? This would take me to X=1 and make the statement sufficient.
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17 Oct 2015, 14:45
danjbon wrote:
I think this is a high-quality question and I agree with explanation. Why can't I divide both sides of statement 2 by X^2?? This would take me to X=1 and make the statement sufficient.

You can not divide by x^2 as you don't know whether x=0 or not. Statement 2 is also satisfied by x=0. Be careful with dividing by variables when you either don't know the sign or the values.

Dividing by x^2 implies that you are assuming x can't be =0
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Joined: 05 Oct 2015
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18 Jul 2016, 11:58
Bunuel wrote:
djonga wrote:
hi Bunuel
can you please solve my query? if x^2= x^3, does not it imply that X is Positive? then that would eliminate 0 as a possibility, making statement 2 sufficient. please tell me where i am going wrong.

x^2 = x^3;

x^2 - x^3 = 0;

x^2(1 - x) = 0;

x = 0 or x = 1.

Hi Bunuel,

x^2 = x^3.

why can't we just divide by x^2 to get x = 1?

What am I missing here? I see how both 0 and 1 are solutions but what theorem am I missing? I feel like it is basic and I may just be forgetting.

EDIT: Never mind previous poster answered it.
M02-32   [#permalink] 18 Jul 2016, 11:58

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