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# M02-32

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Joined: 29 May 2017
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30 May 2017, 05:59
I think this is a poor-quality question and I don't agree with the explanation. The question asks the value of x^2. Why do we calculate x and then x^2. If X^2 is the value required, both statements independently be sufficient.

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30 May 2017, 08:06
reachankitarora wrote:
I think this is a poor-quality question and I don't agree with the explanation. The question asks the value of x^2. Why do we calculate x and then x^2. If X^2 is the value required, both statements independently be sufficient.

Please re-read the solution. You just missed the point. The question is flawless.

From (2) we get that x = 0 or x = 1, so x^2 = 0, or x^2 = 1. We have TWO different answers to the question, which means that this statement is NOT sufficient.
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17 Jul 2017, 02:21
I don't agree with the explanation. the answer should be D .If we divide both sides of the 2nd equation with x^2 , we are left with 1=x , thus x^=1

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17 Jul 2017, 02:29
Arjun17 wrote:
I don't agree with the explanation. the answer should be D .If we divide both sides of the 2nd equation with x^2 , we are left with 1=x , thus x^=1

It's always better to read the whole discussion before posting a question. You are making a rookie mistake, which is addressed in a post just above yours.

You cannot divide x^2 = x^3 by x^2 because x^2 could be 0 and we cannot divide by 0. By doing so, you are loosing a root, namely x = 0 (notice that both x = 0 and x = 1 satisfy x^2 = x^3, and this is clearly shown in the original solution).

Takeaway: never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.
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03 Oct 2017, 11:27
Bunuel,

Can you please throw some light on my query.

For statement 2: I made till this point $$x^2$$ (1−x)=0 -- > eq 1

When I segregated $$x^2$$ = 0 and (1-x) = 0. Since $$x^2$$ is always +ve and can't be 0 , so I eliminated this part. Finally, got x = 1.

I have seen such type of questions,even though when $$x^2$$ is 0 some times $$x^2$$ taken +ve and in few solutions it is not( case in our question). Under what cases we need to consider +ve or 0 when we have such eq 1 type.

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03 Oct 2017, 11:30
msk0657 wrote:
Bunuel,

Can you please throw some light on my query.

For statement 2: I made till this point $$x^2$$ (1−x)=0 -- > eq 1

When I segregated $$x^2$$ = 0 and (1-x) = 0. Since $$x^2$$ is always +ve and can't be 0 , so I eliminated this part. Finally, got x = 1.

I have seen such type of questions,even though when $$x^2$$ is 0 some times $$x^2$$ taken +ve and in few solutions it is not( case in our question). Under what cases we need to consider +ve or 0 when we have such eq 1 type.

x^2 could be 0 if x is 0. If we were told that x is not 0, then x^2 will be more than 0.
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31 Oct 2017, 20:14
Quick question, from 1, because we are not given x=/= 0 in the stem, we know we can square the equation because 0 =/= (-1)^(1/2), b/c we cannot take root of -1, is that correct? Is there a general way to check whether we are able to perform operations on variables in equations and inequalities? Thanks in advance.

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31 Oct 2017, 21:10
alekkx wrote:
Quick question, from 1, because we are not given x=/= 0 in the stem, we know we can square the equation because 0 =/= (-1)^(1/2), b/c we cannot take root of -1, is that correct? Is there a general way to check whether we are able to perform operations on variables in equations and inequalities? Thanks in advance.

Check : Manipulating Inequalities (adding, subtracting, squaring etc.).

For more check Ultimate GMAT Quantitative Megathread

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Re: M02-32   [#permalink] 31 Oct 2017, 21:10

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# M02-32

Moderators: chetan2u, Bunuel

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