For how many integer values of x does (x² − 5x + 5)ˣ²⁻⁹ˣ⁺²⁰ equal 1?
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Answer & Explanation
Correct answer
D
An expression of the form (base)ⁿ equals 1 in exactly three situations, and you must check all three.
Case 1, base = 1: x² − 5x + 5 = 1 gives x² − 5x + 4 = 0, so (x − 1)(x − 4) = 0, meaning x = 1 or x = 4.
Case 2, exponent = 0 with base ≠ 0: x² − 9x + 20 = 0 gives (x − 4)(x − 5) = 0, meaning x = 4 or x = 5. At x = 5 the base is 5, which is not 0, so x = 5 is valid (x = 4 repeats the solution already found in Case 1).
Case 3, base = −1 with an even exponent: x² − 5x + 5 = −1 gives x² − 5x + 6 = 0, so (x − 2)(x − 3) = 0, meaning x = 2 or x = 3. At x = 2 the exponent is x² − 9x + 20 = 4 − 18 + 20 = 6, which is even, so x = 2 is valid. At x = 3 the exponent is 9 − 27 + 20 = 2, which is even, so x = 3 is valid.
Collecting the distinct solutions across all three cases gives {1, 2, 3, 4, 5}, so there are 5 values.