Answer:
x ∈ {2, 3, 4, 5, 6, 7}
Step-by-step explanation:
The equation you have written only has two real solutions. Perhaps you intend ...
The base quadratic equation factors as ...
(x -5)(x -2) +1
so will have the value 1 for x=2 and x=5. The exponent quadratic can have any value and the intended equation will be satisfied.
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The exponent quadratic factors as ...
(x -6)(x -7)
so will be zero for x=6 and x=7. The base quadratic can have any non-zero value and the intended equation will be satisfied.
So, the four solutions found by a typical solver or graphing program are ...
x ∈ {2, 5, 6, 7}
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In general, for values of x between 2.382 and 4.618, the base quadratic will be negative and the left side of the equation will be complex. However, there are 10 values of x where the imaginary part is zero, and two of these where the real part is 1. These values of x are x=3 and x=4, where the equation is (-1)^6 = 1 and (-1)^12=1.
In summary, the solutions are ...
x ∈ {2, 3, 4, 5, 6, 7}
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