Key Concepts and Formulas

• Discriminant of a Quadratic Equation: For a quadratic equation $ax^2 + bx + c = 0$, the discriminant is given by $D = b^2 - 4ac$. The nature of the roots depends on the discriminant:
  • $D > 0$: Two distinct real roots
  • $D = 0$: Two equal real roots
  • $D < 0$: Two complex conjugate roots
• Location of Roots: If both roots of a quadratic equation $ax^2 + bx + c = 0$ are less than a number $k$, then the following conditions must be satisfied:
  • $D \ge 0$ (Real roots)
  • $a \cdot f(k) > 0$
  • $-\frac{b}{2a} < k$ (Axis of symmetry less than k)

Step-by-Step Solution

Let the given quadratic equation be $f(x) = x^2 - 2kx + k^2 + k - 5 = 0$. Here, $a=1$, $b=-2k$, $c=k^2+k-5$, and we want both roots to be less than $5$, so $k_0=5$.

Step 1: Discriminant Condition ($D \ge 0$)

For the quadratic equation to have real roots, the discriminant must be non-negative. $D = b^2 - 4ac = (-2k)^2 - 4(1)(k^2 + k - 5) = 4k^2 - 4k^2 - 4k + 20 = -4k + 20$ We require $D \ge 0$: $-4k + 20 \ge 0$ $-4k \ge -20$ $k \le 5$ Thus, $k \in (-\infty, 5]$. (Result 1)

Step 2: Function Value Condition ($f(5) > 0$)

Since $a=1 > 0$, the parabola opens upwards. For both roots to be less than 5, the value of the function at $x=5$ must be positive. $f(5) = (5)^2 - 2k(5) + k^2 + k - 5 = 25 - 10k + k^2 + k - 5 = k^2 - 9k + 20$ We require $f(5) > 0$: $k^2 - 9k + 20 > 0$ Factoring the quadratic expression: $(k-4)(k-5) > 0$ The roots of the quadratic equation $k^2 - 9k + 20 = 0$ are $k=4$ and $k=5$. Since the parabola $y = k^2 - 9k + 20$ opens upwards, the inequality $k^2 - 9k + 20 > 0$ holds when $k$ is outside the roots. This means $k < 4$ or $k > 5$, i.e., $k \in (-\infty, 4) \cup (5, \infty)$. (Result 2)

Step 3: Axis of Symmetry Condition ($-\frac{b}{2a} < 5$)

The axis of symmetry is given by $x = -\frac{b}{2a}$. $-\frac{b}{2a} = -\frac{-2k}{2(1)} = \frac{2k}{2} = k$ We require the axis of symmetry to be less than 5: $k < 5$ This means $k \in (-\infty, 5)$. (Result 3)

Step 4: Combining All Conditions

We need to find the intersection of the three intervals:

1. $k \in (-\infty, 5]$
2. $k \in (-\infty, 4) \cup (5, \infty)$
3. $k \in (-\infty, 5)$

The intersection of $(-\infty, 5]$ and $(-\infty, 5)$ is $(-\infty, 5)$. Now, we intersect $(-\infty, 5)$ with $(-\infty, 4) \cup (5, \infty)$. This gives us $(-\infty, 4)$. So, $k \in (-\infty, 4)$.

Common Mistakes & Tips

• Forgetting the Discriminant: Always check the discriminant to ensure real roots.
• Incorrect Inequality Signs: Pay close attention to whether to use strict or non-strict inequalities. In this case, both roots must be less than 5, not less than or equal to 5.
• Interval Notation: Ensure you understand the difference between open and closed intervals and use the correct notation.

Summary

By considering the discriminant, the function value at $x=5$, and the axis of symmetry, we found the intersection of the resulting intervals for $k$. This led us to the conclusion that $k$ must belong to the interval $(-\infty, 4)$. This ensures that both roots of the given quadratic equation are real and strictly less than 5.

The final answer is $\boxed{(-\infty, 4)}$, which corresponds to option (C).