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Re: If (x + 1)^2 – 2x > 2(x + 1) + 2, then x cannot equal [#permalink]
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For me the question is flawed.

Notice how Brent above tested only the first three options

However, if we know that \((x-1)^2 > 4\)

and use D

we do have \(4 > 4\) which is impossible

The same way if you perform a simple substitution we will end up with \(10 > 10\) which is the same result. conceptually.

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Re: If (x + 1)^2 – 2x > 2(x + 1) + 2, then x cannot equal [#permalink]
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Carcass wrote:
For me the question is flawed.

Notice how Brent above tested only the first three options

However, if we know that \((x-1)^2 > 4\)

and use D

we do have \(4 > 4\) which is impossible

The same way if you perform a simple substitution we will end up with \(10 > 10\) which is the same result. conceptually.

Regards


Yes, you're both correct. I just naturally stopped when I found a value that didn't work.

I have a feeling the original question is SUPPOSED to have a "≥" rather than ">"
If it had a "≥" symbol, then the correct answer would be C
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Re: If (x + 1)^2 – 2x > 2(x + 1) + 2, then x cannot equal [#permalink]
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Attachment:
#greprepclub (2).jpg
#greprepclub (2).jpg [ 24.98 KiB | Viewed 2702 times ]


Aside from the typo in the stem: it should be cannot BE equal TO ..............
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Re: If (x + 1)^2 – 2x > 2(x + 1) + 2, then x cannot equal [#permalink]
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Carcass wrote:
Attachment:
#greprepclub (2).jpg


Aside from the typo in the stem: it should be cannot BE equal TO ..............


A good question to ignore :-D
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Re: If (x + 1)^2 – 2x > 2(x + 1) + 2, then x cannot equal [#permalink]
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The take away from the question: test ALWAYS all the answer choices, no matter what.

Even for the most skilled GRE tutor on earth :wink:
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Re: If (x + 1)^2 2x > 2(x + 1) + 2, then x cannot equal [#permalink]
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Re: If (x + 1)^2 2x > 2(x + 1) + 2, then x cannot equal [#permalink]
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