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Bunuel
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Bunuel
If \((x + 1)^2 – 2x > 2(x + 1) + 2\), then x cannot equal which one of the following?


(A) –5
(B) –3
(C) 0
(D) 3
(E) 5

Given,

\((x + 1)^2 – 2x > 2(x + 1) + 2\)

\((x + 1)^2 > 2x + 2(x + 1) + 2\)

\((x + 1)^2 > 2x + 2x + 2 + 2\)

\((x + 1)^2 > 4x + 4\)

Now plug value from options.

x can't be 0.

C is the correct answer.

Please plug D) 3
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Bunuel, there are 2 answer choices. Kindly , review it.

Thanks.
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Bunuel
If \((x + 1)^2 – 2x > 2(x + 1) + 2\), then x cannot equal which one of the following?


(A) –5
(B) –3
(C) 0
(D) 3
(E) 5

Given,

\((x + 1)^2 – 2x > 2(x + 1) + 2\)

\((x + 1)^2 > 2x + 2(x + 1) + 2\)

\((x + 1)^2 > 2x + 2x + 2 + 2\)

\((x + 1)^2 > 4x + 4\)

Now plug value from options.

x can't be 0.

C is the correct answer.

Please plug D) 3

You're right. C is the only correct answer if \((x + 1)^2 – 2x ≥ 2(x + 1) + 2\), but not here.
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LevanKhukhunashvili
Houston I have a problem :)

Lets simplify the left hand:
x^2+1+2x-2x=x^2+1

lets simplify the right hand
2x+4

x^2+1>2x+4
x^2-2x-3>0
(x-3)(x+1)>0

X<-1 and X>3

IMO Both C and D are correct
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Fixed. Thank you.
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Bunuel
If \((x + 1)^2 – 2x > 2(x + 1) + 2\), then x cannot equal which one of the following?


(A) –5
(B) –3
(C) 0
(D) 4
(E) 5

solve \((x + 1)^2 – 2x > 2(x + 1) + 2\)
(x-3)*(x+1)>0
so x cannot be 0 else we get is -ve option
IMO C
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Asked: If \((x + 1)^2 – 2x > 2(x + 1) + 2\), then x cannot equal which one of the following?

x^2 +2x +1 -2x > 2x +2 +2
x^2 +1 -2x -4 >0
x^2 -2x -3 >0
(x -3)(x+1)>0
x>3 or x<-1

x can not be 0
IMO C

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If (x+1)^2 – 2x>2(x+1), then x cannot equal which one of the following?

(A) –5
(B) –3
(C) 0
(D) 4
(E) 5

(x+1)^2 – 2x>2(x+1) ------ > (x-3) (x+1) > 0

This implies that x > 3 or x <-1. 0 is the only option that fails the test.

C is the answer.
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