[GMAT math practice question]
What is the number of roots of the equation (x^2-5x+5)^{(x^2-5x-6)}=1?
A. 2
B. 3
C. 4
D. 5
E. 6
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What is the number of roots of the equation (x^2-5x+5)^{(x^2
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Max@Math Revolution
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When we have an equation of the form a^b = 1, we need to consider the following three cases:
Case 1: a = 1
Case 2: a = -1 and b is an even number
Case 3: b = 0.
Case 1: x^2-5x+5 = 1
=> x^2-5x+4 = 0
=> (x-1)(x-4) = 0
=> x = 1 or x = 4
Thus, 1 and 4 are roots of the equation.
Case 2: x^2-5x+5 = -1
=> x^2-5x+6 = 0
=> (x-2)(x-3) = 0
=> x = 2 or x = 3
If x = 2, then the exponent b = x^2-5x-6 = 4 - 10 - 6 = -12 is an even number.
If x = 3, then the exponent b = x^2-5x-6 = 9 - 15 - 6 = -12 is an even number.
Thus, 2 and 3 are roots of the equation.
Case 3: x^2-5x-6 = 0
=> (x+1)(x-6) = 0
=> x = -1 or x = 6.
Thus -1 and 6 are roots of the equation.
Hence, 1, 2, 3, 4, 6 and -1 are roots of the equation.
Therefore, E is the answer.
Answer: E
When we have an equation of the form a^b = 1, we need to consider the following three cases:
Case 1: a = 1
Case 2: a = -1 and b is an even number
Case 3: b = 0.
Case 1: x^2-5x+5 = 1
=> x^2-5x+4 = 0
=> (x-1)(x-4) = 0
=> x = 1 or x = 4
Thus, 1 and 4 are roots of the equation.
Case 2: x^2-5x+5 = -1
=> x^2-5x+6 = 0
=> (x-2)(x-3) = 0
=> x = 2 or x = 3
If x = 2, then the exponent b = x^2-5x-6 = 4 - 10 - 6 = -12 is an even number.
If x = 3, then the exponent b = x^2-5x-6 = 9 - 15 - 6 = -12 is an even number.
Thus, 2 and 3 are roots of the equation.
Case 3: x^2-5x-6 = 0
=> (x+1)(x-6) = 0
=> x = -1 or x = 6.
Thus -1 and 6 are roots of the equation.
Hence, 1, 2, 3, 4, 6 and -1 are roots of the equation.
Therefore, E is the answer.
Answer: E
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We have to consider several cases
1) \(x^2-5x+5=1\). In this case, no matter what the result of the power of this expression, the result will be \(1
x=4, x=1\).
2) \(x^2-5x-6=0\). In this case, no matter what is the value of \(x^2-5x+5\), the result will be \(1
x=6, x=-1\)
And it seemed that's all, but...
3) If \(x^2−5x+5=-1\) and \(x^2-5x-6=\) even integer the result will be \(1\)
\(x^2−5x+5=-1\) has the values \(x=3, x=2\) Both of these roots give us even result in \(x^2-5x-6\), so both roots are valid
\(X\) can have \(6\) roots: \(4, 1, 6, -1, 3, 2\).
Therefore, __E__ is the correct answer.
1) \(x^2-5x+5=1\). In this case, no matter what the result of the power of this expression, the result will be \(1
x=4, x=1\).
2) \(x^2-5x-6=0\). In this case, no matter what is the value of \(x^2-5x+5\), the result will be \(1
x=6, x=-1\)
And it seemed that's all, but...
3) If \(x^2−5x+5=-1\) and \(x^2-5x-6=\) even integer the result will be \(1\)
\(x^2−5x+5=-1\) has the values \(x=3, x=2\) Both of these roots give us even result in \(x^2-5x-6\), so both roots are valid
\(X\) can have \(6\) roots: \(4, 1, 6, -1, 3, 2\).
Therefore, __E__ is the correct answer.
