Answer: Always true
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Step-by-step explanation:
I'll use x in place of n
Let y = x^2 - 4x + 5
If we complete the square, then,
y = x^2 - 4x + 5
y = (x^2 - 4x) + 5
y = (x^2 - 4x + 4 - 4) + 5
y = (x^2 - 4x + 4) - 4 + 5
y = (x-2)^2 + 1
The quantity (x-2)^2 is never negative as squaring any real number value is never a negative result. Adding on 1 makes the result positive. So y > 0 regardless of whatever x is. Replace x with n, and this shows how n^2 - 4n + 5 is always positive for any integer n.
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You could also use the quadratic formula to find that x^2 - 4x + 5 = 0 has no real solutions, so there are no x intercepts. Either the graph is entirely above the x axis or it is entirely below the x axis.
Plug in any x value you want, say x = 0, and the result is positive. Meaning that whatever x value you plug in will be positive (as the graph can't cross the x axis to go into negative territory)