anik1989 wrote:
VeritasPrepKarishma wrote:
anik1989 wrote:
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
_________________
Karishma
Veritas Prep GMAT Instructor
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