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M02-32

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Re M02-32 [#permalink]

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New post 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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Re: M02-32 [#permalink]

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New post 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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New post 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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Re: M02-32 [#permalink]

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New post 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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Re: M02-32 [#permalink]

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New post 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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Re: M02-32 [#permalink]

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New post 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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M02-32   [#permalink] 03 Oct 2017, 11:30

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