Consider the quadratic equation x² - 2kx + k² + k - 5 = 0 in x, where k is a real number…

(a) $\tfrac14 D = 5 - k \ge 0$ gives $k \le 5$.

(b) We need $k \le 5$, the vertex $k < 5$, and $f(5) = \left(k - 5\right) \left(k - 4\right) > 0$, which holds for $k < 4$ or $k > 5$. Together: $k < 4$, so $c = 4$.

(c) The parabola opens upward, so 1 lies between the roots exactly when $f(1) = k^{2} - k - 4 < 0$, that is $\frac{1}{2} - \frac{\sqrt{17}}{2} < k < \frac{1}{2} + \frac{\sqrt{17}}{2}$. The length is $\sqrt{17} = 4.12$.

(d) Sum $2k > 0$, product $k^{2} + k - 5 > 0$ (so $k < - \frac{\sqrt{21}}{2} - \frac{1}{2}$ or $k > - \frac{1}{2} + \frac{\sqrt{21}}{2}$, about $-2.79$ and $1.79$), and $k \le 5$. Together $1.79 < k \le 5$: the integers $2, 3, 4, 5$, so 4.

(e) Product $k^{2} + k - 5 < 0$: $-2.79 < k < 1.79$, the integers $-2, -1, 0, 1$, so 4. A negative product forces real roots, so nothing else is needed.