Last visit was: 08 Jul 2025, 23:28 It is currently 08 Jul 2025, 23:28
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
WoundedTiger
Joined: 25 Apr 2012
Last visit: 25 Sep 2024
Posts: 523
Own Kudos:
2,479
 [58]
Given Kudos: 740
Location: India
GPA: 3.21
WE:Business Development (Other)
Products:
Posts: 523
Kudos: 2,479
 [58]
4
Kudos
Add Kudos
53
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
smyarga
User avatar
Tutor
Joined: 20 Apr 2012
Last visit: 06 Aug 2020
Posts: 82
Own Kudos:
798
 [23]
Given Kudos: 39
Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE:Education (Education)
Expert
Expert reply
GMAT 2: 730 Q51 V38
Posts: 82
Kudos: 798
 [23]
11
Kudos
Add Kudos
12
Bookmarks
Bookmark this Post
General Discussion
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
thorinoakenshield
Joined: 03 Jan 2015
Last visit: 28 Mar 2015
Posts: 51
Own Kudos:
115
 [2]
Given Kudos: 223
Concentration: Strategy, Marketing
WE:Research (Pharmaceuticals and Biotech)
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
smyarga
For the equation (x^2−x−1)\(^{(x+2)}\)=1 to be true, there are three possibilities:

1. \((x^2-x-1) =1\)
\(x^2-x-2=0\)---> \(x=2\) or \(x=-1\). Both roots are possible. So, we have 2 values.

2. \((x^2-x-1) =-1\) and (x+2) is even
\(x^2-x=0\)--->\(x=1\) or \(x=0\). Check if \(x+2\) is even: if \(x=1\), then \(x+2-odd\); if \(x=0\), then \(x+2-even\).
So only \(x=0\) is possible. +1 value.

3. x+2=0 and \((x^2-x-1)\neq=0\) .
Therefore, x=-2. +1 value

Hence 4 values: -2, -1, 0, 2

D.

Hi,

How did you get (x^2-x-1) to equal "-1"?

TO
User avatar
manpreetsingh86
Joined: 13 Jun 2013
Last visit: 19 Dec 2022
Posts: 219
Own Kudos:
Given Kudos: 14
Posts: 219
Kudos: 1,147
Kudos
Add Kudos
Bookmarks
Bookmark this Post
thorinoakenshield

Hi,

How did you get (x^2-x-1) to equal "-1"?

TO

if (x^2-x-1) is equal to -1, then (x+2) must be even. why ?? because (-1)^2n = 1 here n can take values of 0,1,2,3,,, etc
User avatar
EMPOWERgmatRichC
User avatar
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,788
Own Kudos:
12,485
 [3]
Given Kudos: 450
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Posts: 21,788
Kudos: 12,485
 [3]
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Hi thorinoakenshield,

We're given (X^2 - X - 1)^(X+2) = 1

Given the way that the prompt is written, there are several ways to get a total that = 1:

(1)^(any power) = 1
(-1)^(even power) = 1
(any number)^0 = 1

The 'second option' is arguably the most 'complex' because it depends on 2 things (the base = -1 AND the exponent is EVEN).

IF....
(X^2 - X - 1) = -1
X^2 - X = 0
X(X - 1) = 0

This equation has 2 solutions: 0 and 1. HOWEVER, only one of them fits the given prompt....

IF....
X = 0, we have (-1)^(0+2) = 1, so this is a valid solution.

IF....
X = 1, we have (-1)(1+3) = -1 --> this is NOT a valid solution, since the end result is NOT 1.

GMAT assassins aren't born, they're made,
Rich
User avatar
anik1989
Joined: 06 May 2014
Last visit: 18 Jan 2016
Posts: 10
Own Kudos:
Given Kudos: 7
Posts: 10
Kudos: 11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
EMPOWERgmatRichC
Hi thorinoakenshield,

We're given (X^2 - X - 1)^(X+2) = 1

Given the way that the prompt is written, there are several ways to get a total that = 1:

(1)^(any power) = 1
(-1)^(even power) = 1
(any number)^0 = 1

The 'second option' is arguably the most 'complex' because it depends on 2 things (the base = -1 AND the exponent is EVEN).

IF....
(X^2 - X - 1) = -1
X^2 - X = 0
X(X - 1) = 0

This equation has 2 solutions: 0 and 1. HOWEVER, only one of them fits the given prompt....

IF....
X = 0, we have (-1)^(0+2) = 1, so this is a valid solution.

IF....
X = 1, we have (-1)(1+3) = -1 --> this is NOT a valid solution, since the end result is NOT 1.

GMAT assassins aren't born, they're made,
Rich
what will be the ans? a or d?
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 08 Jul 2025
Posts: 16,101
Own Kudos:
Given Kudos: 475
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,101
Kudos: 74,212
Kudos
Add Kudos
Bookmarks
Bookmark this Post
anik1989
what will be the ans? a or d?

The answer is (D)
There are 4 values x can take: -2, -1, 0, 2 (look at smyarga's solution above)
User avatar
anik1989
Joined: 06 May 2014
Last visit: 18 Jan 2016
Posts: 10
Own Kudos:
Given Kudos: 7
Posts: 10
Kudos: 11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasPrepKarishma
anik1989
what will be the ans? a or d?

The answer is (D)
There are 4 values x can take: -2, -1, 0, 2 (look at smyarga's solution above)
thank you karishma. honestly i dod not get that ans. why power has been ignored in sloution??
avatar
aniketmunshi
Joined: 28 May 2013
Last visit: 30 Nov 2018
Posts: 14
Own Kudos:
Given Kudos: 18
Schools: Mannheim"17
Schools: Mannheim"17
Posts: 14
Kudos: 30
Kudos
Add Kudos
Bookmarks
Bookmark this Post
anik1989
VeritasPrepKarishma
anik1989
what will be the ans? a or d?

The answer is (D)
There are 4 values x can take: -2, -1, 0, 2 (look at smyarga's solution above)
thank you karishma. honestly i dod not get that ans. why power has been ignored in solution??

Hi Anik1989,

No power has been ignored in the solution.

As smyarga mentioned there are three possibilities.

1) Base is 1, then irrespective of the power, answer will be 1. Thus by equating base =1, we get values of x=2, x=-1. Now if we substitute these values in main equation (base^power), equation holds true. Thus we have 2 values of x

2) Base is -1 and power is even number. Then answer will always be 1. Now by equating base to -1, we get x=1,x=0. However, as per our condition if x=1, the power(x+2) becomes odd, thus only x=0 can be taken as a true value. So we have 1 value from this condition.

3) if power is zero(0) then, irrespective of the base, answer will be 1. Thus, by equating power(x+2) to 0, (x+2)=0, x=-2


thus we have totally 4 values(2,-1,0,-2) from all the three above conditions.

Hence answer is D.

Hope it clears your doubt. Power is not ignored, in fact, every possibility is considered!

Cheers
Tx
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 08 Jul 2025
Posts: 16,101
Own Kudos:
74,212
 [1]
Given Kudos: 475
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,101
Kudos: 74,212
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
anik1989
VeritasPrepKarishma
anik1989
what will be the ans? a or d?

The answer is (D)
There are 4 values x can take: -2, -1, 0, 2 (look at smyarga's solution above)
thank you karishma. honestly i dod not get that ans. why power has been ignored in sloution??


Ok, first think about this:
\(a^b = 1\)

In which all cases can this happen?
Case 1: a = 1, b can be anything e.g. \(1^4\), \(1^{-3}\), \(1^0\)
Case 2: a = -1, b must be an even integer e.g. \((-1)^2\), \((-1)^{-4}\)
Case 3: b = 0, a must be a non zero number. e.g. 2^0

So you evaluate each one of these three cases:

Case 1: a = 1, b can be anything
\(x^2−x−1 = 1\)
\(x^2 - x -2 = 0\)
\(x^2 - 2x + x - 2 = 0\)

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

x = -1 or 2

We don't need to worry about b.

Case 2: a = -1, b must be an even integer
\(x^2−x−1 = -1\)
\(x(x - 1) = 0\)

x = 0 or 1

(x+2) must be even integer.
If x = 0, (0+2) is an even integer.
If x = 1, (1+2) is an odd integer so this is not valid.

From this case, we get only one value of x i.e. 0.

Case 3: b = 0, a must be a non zero number.
(x+2) = 0
x = -2
If x = -2, \(x^2−x−1\) is not 0. So this is a valid solution too.

We got four solutions: x = -1, 2, 0, -2
avatar
007mva
Joined: 26 May 2022
Last visit: 01 Aug 2024
Posts: 7
Own Kudos:
Given Kudos: 4
Posts: 7
Kudos: 3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
First of all love this question.

First thing first starting with how to get something to power something 1. If the base is 1 or the power is 0.

So that will give us solutions of x=2, -1 , -2
For the last answer we should be a bit more careful if we make x=0 -1 to the power 2 will again give us x=1.

Thus, x=-2,-1,0,2 4 answers D)

Posted from my mobile device
User avatar
tirthshah2013
Joined: 16 Sep 2022
Last visit: 08 Jul 2025
Posts: 17
Own Kudos:
Given Kudos: 4
Location: India
Concentration: General Management, Technology
Schools: ISB '23 (S)
GPA: 4
Schools: ISB '23 (S)
Posts: 17
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
WoundedTiger
Suppose x is an integer such that (x^2−x−1)^(x+2)=1. How many possible values of x exist?

A. 1
B. 2
C. 3
D. 4
E. 5

Kudos for correct solution

This question is a bit tricky. But if you know all the exponent rules it becomes very easy.

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

1) a^0 = 1
so we can take x+2 = 0, x= -2
this is our first value.

2) 1^a = 1
so here (x^2-x-1) = 1
x^2-x-2 = 0
x^2-2x+x-2=0
x(x-2)+1(x-2)=0
x+1 = 0 ad x -2 = 0
x = -1 and x = 2
these are the other two values. that totals to 3.

3) -1^even = 1
so we can simply take (x^2-x-1) = -1 and x+2 = even.
that gives x^2-x = 0
x(x-1)=0
x = 0 and x = 1. In this case, if we take x = 1, our power will become odd, thus will make our solution wrong. Hence only solution will be x = 0.

So we have 4 unique solutions to this question and they are x=-2,-1,0,2. Hence the answer is D.

Hope it helps.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,360
Own Kudos:
Posts: 37,360
Kudos: 1,010
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Math Expert
102594 posts
PS Forum Moderator
679 posts