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.

x=0 is 1st solution.
x+2=0 implies x=-2 second solution
x^2-x-1=1 implies x^2-x-2=0 implies x=2 and x=-1 another 2 solutions so total 4 solutions.

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

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
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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)
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thank you karishma. honestly i dod not get that ans. why power has been ignored in sloution??

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

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
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