Question

The number of non-negative values of n for which \(\log_{1/4}(n^2 - 7n + 14)>0\) is ____.

Hint:

Since the base $\tfrac14$ is less than 1, flip the inequality when removing the log: you need $0 \lt n^2-7n+14 \lt 1$. To check whether a quadratic can ever dip below a target value, shift it by that amount and look at the sign of its discriminant -- a negative discriminant with a positive leading coefficient means the expression never changes sign.

Answer 1

Correct Answer: 0

The inequality asks for \(n\) where \(n^2-7n+14\) lies strictly between \(0\) and \(1\), because a base smaller than \(1\) flips the usual direction of a logarithmic inequality. Checking whether the quadratic can ever fall below \(1\) means checking whether \(n^2-7n+13\) (which is \(n^2-7n+14\) shifted down by \(1\)) can ever be negative. Its discriminant is \(49-4(13)=-3\), which is negative, so \(n^2-7n+13\) never touches zero and, since its leading coefficient is positive, stays positive for every real \(n\). That directly means \(n^2-7n+14\) never dips below \(1\), so it can never land in the required window \((0,1)\). With the target range completely unreachable, no value of \(n\), non-negative or otherwise, satisfies the inequality, giving \(\boxed{0}\) as the count.