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M02-32

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Bunuel
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M02-32 [#permalink] New post   16 Sep 2014, 00:18
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Question Stats:

67% (01:00) correct 33% (01:03) wrong based on 877 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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A
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Re M02-32 [#permalink] New post   16 Sep 2014, 00:18
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Official Solution:


What is the value of \(x^2\)?

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

Square:

\(x^2 = 2x - 1\)

\(x^2 - 2x + 1 = 0\)

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

\(x -1 = 0\)

\(x = 1\)

Thus, \(x^2 = 1\). Sufficient/

(2) \(x^2 = x^3\):

\(x^2 - x^3 = 0\)

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

\(x = 0\) or \(x =1\).

Therefore, \(x^2 = 0\) or \(x^2 = 1\). Not sufficient.


Answer: A
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Bunuel
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Re: M02-32 [#permalink] New post   03 Mar 2023, 10:58
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M02-32 [#permalink] New post   12 May 2023, 01:02
I have a doubt regarding the 2nd statement. If you divide both sides by x^2, won't we get 1= x from x^2 = X^3.

Please help clarify my doubt. I chose Option D, since I thought that we can get the value for x from each statement.


Bunuel wrote:
Official Solution:


What is the value of \(x^2\)?


(2) \(x^2 = x^3\):

\(x^2 - x^3 = 0\)

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

\(x = 0\) or \(x =1\).

Therefore, \(x^2 = 0\) or \(x^2 = 1\). Not sufficient.


Answer: A
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Bunuel
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Re: M02-32 [#permalink] New post   12 May 2023, 01:19
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fdfd97 wrote:
I have a doubt regarding the 2nd statement. If you divide both sides by x^2, won't we get 1= x from x^2 = X^3.

Please help clarify my doubt. I chose Option D, since I thought that we can get the value for x from each statement.


Bunuel wrote:
Official Solution:


What is the value of \(x^2\)?


(2) \(x^2 = x^3\):

\(x^2 - x^3 = 0\)

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

\(x = 0\) or \(x =1\).

Therefore, \(x^2 = 0\) or \(x^2 = 1\). Not sufficient.


Answer: A


Note that we cannot divide x^2 = x^3 by x^2 because x^2 (x) can be 0 and division by zero is not allowed. By dividing by x^2, you would be incorrectly assuming that x^2 does not equal zero, potentially excluding a valid solution (observe that x = 0 satisfies the equation). As a rule, never reduce an equation by a variable (or by an expression containing a variable) if you are not certain that the variable (or expression with the variable) does not equal zero. Remember, we cannot divide by zero.
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Re: M02-32 [#permalink] New post   12 May 2023, 02:10
Oh right, that totally slipped out of my mind while solving the question, that's why I got it's value as 1. Thanks.
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Re M02-32 [#permalink] New post   09 Aug 2023, 09:20
I think this is a high-quality question and I agree with explanation.
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Re: M02-32 [#permalink] New post   19 Nov 2023, 20:29
Bunuel wrote:
Official Solution:


What is the value of \(x^2\)?

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

Square:

\(x^2 = 2x - 1\)

\(x^2 - 2x + 1 = 0\)

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

\(x -1 = 0\)

\(x = 1\)

Thus, \(x^2 = 1\). Sufficient/

(2) \(x^2 = x^3\):

\(x^2 - x^3 = 0\)

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

\(x = 0\) or \(x =1\).

Therefore, \(x^2 = 0\) or \(x^2 = 1\). Not sufficient.


Answer: A

Hi Bunuel, I used this method in the first statement after squaring both sides: x^2=2x-1 -> X^2-2x=-1 -> x(x-2)=-1 -> x=(-1) or (1). Why is this incorrect since we've used the same method in statement 2.
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Re: M02-32 [#permalink] New post   19 Nov 2023, 22:14
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ritwick2 wrote:
Bunuel wrote:
Official Solution:


What is the value of \(x^2\)?

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

Square:

\(x^2 = 2x - 1\)

\(x^2 - 2x + 1 = 0\)

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

\(x -1 = 0\)

\(x = 1\)

Thus, \(x^2 = 1\). Sufficient/

(2) \(x^2 = x^3\):

\(x^2 - x^3 = 0\)

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

\(x = 0\) or \(x =1\).

Therefore, \(x^2 = 0\) or \(x^2 = 1\). Not sufficient.


Answer: A

Hi Bunuel, I used this method in the first statement after squaring both sides: x^2=2x-1 -> X^2-2x=-1 -> x(x-2)=-1 -> x=(-1) or (1). Why is this incorrect since we've used the same method in statement 2.


In the second statement, x^2(1 - x) equals 0, so either x^2 or (1 - x) must be 0 due to the zero product property. However, in the first statement, x(x - 2) equals -1, which is different. Here, you cannot conclude that x equals -1 or 1. The rules for solving equations are different when the product is zero compared to when it's a non-zero value.
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Re: M02-32 [#permalink]
19 Nov 2023, 22:14
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