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, x = 1 or x = 4. Case 2, exponent = 0 with base ≠ 0: x² − 9x + 20 = 0 gives (x − 4)(x − 5) = 0, x = 4 or x = 5; at x = 5 the base is 5, not 0, so x = 5 is valid (x = 4 repeats 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, x = 2 or x = 3; at x = 2 the exponent is 6 (even, valid), at x = 3 the exponent is 2 (even, valid).
Distinct solutions: {1, 2, 3, 4, 5}, so 5 values.
The false summit is choice A, 2: stopping at the base = 1 case alone.