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655-705 (Hard)|   Algebra|      
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If x^2 does not equal 1, does x^2 + 1= a ? 16 Dec 2010, 14:58
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

57% (02:05) correct 43% (02:05) wrong based on 178 sessions

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If x^2 does not equal 1, does x^2 + 1= a ?

(1) x^4 = a + 1

(2) x < 0
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E
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
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If x^2 does not equal 1, does x^2 + 1= a ? 16 Dec 2010, 15:47
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PASSINGGMAT
If x^2 does not equal 1, does x^2 + 1=a/ ?

1) x^4=a+1

2) x<0
Show more

If x^2 does not equal 1, does x^2 + 1=a?

\(x^2\neq{1}\) means that \(x\neq{1}\) and \(x\neq{-1}\). Question: does \(x^2+1=a\)?

(1) x^4=a+1 --> \(a=x^4-1=(x^2-1)(x^2+1)\) --> if we substitute \(a\) in the above question then the question becomes: does \(x^2+1=(x^2-1)(x^2+1)\) --> reduce by \(x^2+1\) (we can safely do that as \(x^2+1=(non-negative)+(positive)\neq{0}\)) --> does \(1=x^2-1\) --> does \(x^2=2\) --> so finally question becomes: does \(x=\sqrt{2}\) or \(x=-\sqrt{2}\), but we don't know the answer to this question. Not sufficient.

Or: we have \(a=(x^2-1)(x^2+1)\). Now \(a=x^2+1\) to be true then \(x^2-1\) must equal to 1 (as \(x^2+1\neq{0}\)) but we don't know whether \(x^2-1=1\) is true.

(2) x<0 --> clearly insufficient.

(1)+(2) As from (2) \(x<0\) then the question is: does \(x=-\sqrt{2}\), but we don't know the answer to this question. Not sufficient.

Answer: E.

To elaborate if \(x=-\sqrt{2}\) and \(a=3\) (note that these values of \(x\) and \(a\) satisfy stem as well as both statements) then the answer to the question will be YES as \(x^2 + 1=2+1=3=a\) but for other valid values of \(x\) and \(a\) the answer will be No, for example \(x=-2<0\) and \(a=15\) (note that these values of \(x\) and \(a\) also satisfy stem as well as both statements).

As OA is given to be A then I think that there should be \(x^2\neq{2}\) instead of \(x^2\neq{1}\). In this case in the last stage of statement (1) we'll have: does \(x^2=2\) and we can answer No to this question. So statement would be sufficient.

Hope it helps.
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If x^2 does not equal 1, does x^2 + 1= a ? 27 Jul 2019, 21:29
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I propose for the discussion:

If x ^ 2 is not equal to 1, then x is not 1 and x is not -1.
If x is not 1 or -1, then x ^ 2 + 1 cannot be 2, that is, a, cannot be 2, since they ask if: (x ^ 2 + 1) is equal to, a?

1) X ^ 4 = a + 1

If x ^ 2 + 1 = a, we can do x ^ 2 = a - 1. So
x ^ 4 = a ^ 2 - 2a + 1

Then a + 1 = a ^ 2 - 2a + 1
0 = a ^ 2 - 3a
0 = a (a - 3)
Then a = 0, since a = x ^ 2 + 1, a must be greater than zero, a = 0 is not possible.
The other possibility of, a, is 3, that is a = 3, is possible, since, a, is greater than zero.

If a = 3, comply with that, a = x ^ 2 + 1 and that, x, is not 1 and, x, is not -1.
So 1) Sufficient

2) x <0, can be any negative value, including -1, which clearly cannot be assumed by x, Contradiction. INSUFFICIENT.

Answer A

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If x^2 does not equal 1, does x^2 + 1= a ? 27 Jul 2019, 23:52
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Bunuel
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If x^2 does not equal 1, does x^2 + 1=a/ ?

1) x^4=a+1

2) x<0

If x^2 does not equal 1, does x^2 + 1=a?

\(x^2\neq{1}\) means that \(x\neq{1}\) and \(x\neq{-1}\). Question: does \(x^2+1=a\)?

(1) x^4=a+1 --> \(a=x^4-1=(x^2-1)(x^2+1)\) --> if we substitute \(a\) in the above question then the question becomes: does \(x^2+1=(x^2-1)(x^2+1)\) --> reduce by \(x^2+1\) (we can safely do that as \(x^2+1=(non-negative)+(positive)\neq{0}\)) --> does \(1=x^2-1\) --> does \(x^2=2\) --> so finally question becomes: does \(x=\sqrt{2}\) or \(x=-\sqrt{2}\), but we don't know the answer to this question. Not sufficient.

Or: we have \(a=(x^2-1)(x^2+1)\). Now \(a=x^2+1\) to be true then \(x^2-1\) must equal to 1 (as \(x^2+1\neq{0}\)) but we don't know whether \(x^2-1=1\) is true.

(2) x<0 --> clearly insufficient.

(1)+(2) As from (2) \(x<0\) then the question is: does \(x=-\sqrt{2}\), but we don't know the answer to this question. Not sufficient.

Answer: E.

To elaborate if \(x=-\sqrt{2}\) and \(a=3\) (note that these values of \(x\) and \(a\) satisfy stem as well as both statements) then the answer to the question will be YES as \(x^2 + 1=2+1=3=a\) but for other valid values of \(x\) and \(a\) the answer will be No, for example \(x=-2<0\) and \(a=15\) (note that these values of \(x\) and \(a\) also satisfy stem as well as both statements).

As OA is given to be A then I think that there should be \(x^2\neq{2}\) instead of \(x^2\neq{1}\). In this case in the last stage of statement (1) we'll have: does \(x^2=2\) and we can answer No to this question. So statement would be sufficient.

Hope it helps.
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Bunuel
Can you explain this section
(1) x^4=a+1 --> \(a=x^4-1=(x^2-1)(x^2+1)\) --> if we substitute \(a\) in the above question then the question becomes: does \(x^2+1=(x^2-1)(x^2+1)\) --> reduce by \(x^2+1\) (we can safely do that as \(x^2+1=(non-negative)+(positive)\neq{0}\)) --> does \(1=x^2-1\) --> does \(x^2=2\) --> so finally question becomes: does \(x=\sqrt{2}\) or \(x=-\sqrt{2}\), but we don't know the answer to this question. Not sufficient.

not sure how
does \(x^2+1=(x^2-1)(x^2+1)\) --> reduce by \(x^2+1\) (we can safely do that as \(x^2+1=(non-negative)+(positive)\neq{0}\))
equation is derived
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If x^2 does not equal 1, does x^2 + 1= a ? 28 Jul 2019, 03:12
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PASSINGGMAT
If x^2 does not equal 1, does x^2 + 1=a/ ?

1) x^4=a+1

2) x<0

If x^2 does not equal 1, does x^2 + 1=a?

\(x^2\neq{1}\) means that \(x\neq{1}\) and \(x\neq{-1}\). Question: does \(x^2+1=a\)?

(1) x^4=a+1 --> \(a=x^4-1=(x^2-1)(x^2+1)\) --> if we substitute \(a\) in the above question then the question becomes: does \(x^2+1=(x^2-1)(x^2+1)\) --> reduce by \(x^2+1\) (we can safely do that as \(x^2+1=(non-negative)+(positive)\neq{0}\)) --> does \(1=x^2-1\) --> does \(x^2=2\) --> so finally question becomes: does \(x=\sqrt{2}\) or \(x=-\sqrt{2}\), but we don't know the answer to this question. Not sufficient.

Or: we have \(a=(x^2-1)(x^2+1)\). Now \(a=x^2+1\) to be true then \(x^2-1\) must equal to 1 (as \(x^2+1\neq{0}\)) but we don't know whether \(x^2-1=1\) is true.

(2) x<0 --> clearly insufficient.

(1)+(2) As from (2) \(x<0\) then the question is: does \(x=-\sqrt{2}\), but we don't know the answer to this question. Not sufficient.

Answer: E.

To elaborate if \(x=-\sqrt{2}\) and \(a=3\) (note that these values of \(x\) and \(a\) satisfy stem as well as both statements) then the answer to the question will be YES as \(x^2 + 1=2+1=3=a\) but for other valid values of \(x\) and \(a\) the answer will be No, for example \(x=-2<0\) and \(a=15\) (note that these values of \(x\) and \(a\) also satisfy stem as well as both statements).

As OA is given to be A then I think that there should be \(x^2\neq{2}\) instead of \(x^2\neq{1}\). In this case in the last stage of statement (1) we'll have: does \(x^2=2\) and we can answer No to this question. So statement would be sufficient.

Hope it helps.


Bunuel
Can you explain this section
(1) x^4=a+1 --> \(a=x^4-1=(x^2-1)(x^2+1)\) --> if we substitute \(a\) in the above question then the question becomes: does \(x^2+1=(x^2-1)(x^2+1)\) --> reduce by \(x^2+1\) (we can safely do that as \(x^2+1=(non-negative)+(positive)\neq{0}\)) --> does \(1=x^2-1\) --> does \(x^2=2\) --> so finally question becomes: does \(x=\sqrt{2}\) or \(x=-\sqrt{2}\), but we don't know the answer to this question. Not sufficient.

not sure how
does \(x^2+1=(x^2-1)(x^2+1)\) --> reduce by \(x^2+1\) (we can safely do that as \(x^2+1=(non-negative)+(positive)\neq{0}\))
equation is derived
Show more

Divide \(x^2+1=(x^2-1)(x^2+1)\) by x^2 + 1 to get 1 = x^2 - 1.
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If x^2 does not equal 1, does x^2 + 1= a ? 28 Jul 2019, 03:47
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Bunuel
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If x^2 does not equal 1, does x^2 + 1=a/ ?

1) x^4=a+1

2) x<0

If x^2 does not equal 1, does x^2 + 1=a?

\(x^2\neq{1}\) means that \(x\neq{1}\) and \(x\neq{-1}\). Question: does \(x^2+1=a\)?

(1) x^4=a+1 --> \(a=x^4-1=(x^2-1)(x^2+1)\) --> if we substitute \(a\) in the above question then the question becomes: does \(x^2+1=(x^2-1)(x^2+1)\) --> reduce by \(x^2+1\) (we can safely do that as \(x^2+1=(non-negative)+(positive)\neq{0}\)) --> does \(1=x^2-1\) --> does \(x^2=2\) --> so finally question becomes: does \(x=\sqrt{2}\) or \(x=-\sqrt{2}\), but we don't know the answer to this question. Not sufficient.

Or: we have \(a=(x^2-1)(x^2+1)\). Now \(a=x^2+1\) to be true then \(x^2-1\) must equal to 1 (as \(x^2+1\neq{0}\)) but we don't know whether \(x^2-1=1\) is true.

(2) x<0 --> clearly insufficient.

(1)+(2) As from (2) \(x<0\) then the question is: does \(x=-\sqrt{2}\), but we don't know the answer to this question. Not sufficient.

Answer: E.

To elaborate if \(x=-\sqrt{2}\) and \(a=3\) (note that these values of \(x\) and \(a\) satisfy stem as well as both statements) then the answer to the question will be YES as \(x^2 + 1=2+1=3=a\) but for other valid values of \(x\) and \(a\) the answer will be No, for example \(x=-2<0\) and \(a=15\) (note that these values of \(x\) and \(a\) also satisfy stem as well as both statements).

As OA is given to be A then I think that there should be \(x^2\neq{2}\) instead of \(x^2\neq{1}\). In this case in the last stage of statement (1) we'll have: does \(x^2=2\) and we can answer No to this question. So statement would be sufficient.

Hope it helps.
Show more

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If x^2 does not equal 1, does x^2 + 1= a ? Updated on: 28 Jul 2019, 07:21
Originally posted by Shahalikhan on 28 Jul 2019, 07:12.
Last edited by Shahalikhan on 28 Jul 2019, 07:21, edited 1 time in total.
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PASSINGGMAT
If x^2 does not equal 1, does x^2 + 1= a ?

(1) x^4 = a + 1

(2) x < 0
Show more


another simpler aproach

statement 2

take x=2
we have 4+1=a=>5=a
its only true if a =5,all other cases its wrong.so not sufficient



statement 1

x^4=a+1=

from question rearrange as
x^2=a-1=>x^4=(a-1)^2

now we have

a+1=(a-1)^2

thats true only only a=3,0.for all other values its not true.so not sufficient

statement 1 and 2 not sufficient


option e

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If x^2 does not equal 1, does x^2 + 1= a ? 28 Jul 2019, 07:17
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Bunuel
PASSINGGMAT
If x^2 does not equal 1, does x^2 + 1=a/ ?

1) x^4=a+1

2) x<0

If x^2 does not equal 1, does x^2 + 1=a?

\(x^2\neq{1}\) means that \(x\neq{1}\) and \(x\neq{-1}\). Question: does \(x^2+1=a\)?

(1) x^4=a+1 --> \(a=x^4-1=(x^2-1)(x^2+1)\) --> if we substitute \(a\) in the above question then the question becomes: does \(x^2+1=(x^2-1)(x^2+1)\) --> reduce by \(x^2+1\) (we can safely do that as \(x^2+1=(non-negative)+(positive)\neq{0}\)) --> does \(1=x^2-1\) --> does \(x^2=2\) --> so finally question becomes: does \(x=\sqrt{2}\) or \(x=-\sqrt{2}\), but we don't know the answer to this question. Not sufficient.

Or: we have \(a=(x^2-1)(x^2+1)\). Now \(a=x^2+1\) to be true then \(x^2-1\) must equal to 1 (as \(x^2+1\neq{0}\)) but we don't know whether \(x^2-1=1\) is true.

(2) x<0 --> clearly insufficient.

(1)+(2) As from (2) \(x<0\) then the question is: does \(x=-\sqrt{2}\), but we don't know the answer to this question. Not sufficient.

Answer: E.

To elaborate if \(x=-\sqrt{2}\) and \(a=3\) (note that these values of \(x\) and \(a\) satisfy stem as well as both statements) then the answer to the question will be YES as \(x^2 + 1=2+1=3=a\) but for other valid values of \(x\) and \(a\) the answer will be No, for example \(x=-2<0\) and \(a=15\) (note that these values of \(x\) and \(a\) also satisfy stem as well as both statements).

As OA is given to be A then I think that there should be \(x^2\neq{2}\) instead of \(x^2\neq{1}\). In this case in the last stage of statement (1) we'll have: does \(x^2=2\) and we can answer No to this question. So statement would be sufficient.

Hope it helps.

Posted from my mobile device
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BUnuel,

Please why is the answer not c. stat 1 has indicated d possibility of + $ - root 2.

however, Star 2 shows x <0. hence by combining, x should be less than root 2.
from ur explanation....why is the answer not c.
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