Key Concepts and Formulas

• Quadratic Equation: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Discriminant: The discriminant is $D = b^2 - 4ac$. If $D > 0$, the roots are real and distinct; if $D = 0$, the roots are real and equal; if $D < 0$, the roots are complex.
• Vertex of a Parabola: For a quadratic function $f(x) = ax^2 + bx + c$, the x-coordinate of the vertex is given by $x_v = -\frac{b}{2a}$.
• Roots in an Interval: If $a > 0$ and both roots of $ax^2 + bx + c = 0$ lie in the interval $[p, q]$, then $D > 0$, $af(p) \ge 0$, $af(q) \ge 0$, and $p \le -\frac{b}{2a} \le q$.

Step-by-Step Solution

Step 1: Set up the problem and identify the given information.

We are given the quadratic equation $x^2 - mx + 4 = 0$. We want to find the interval for $m$ such that both roots are real, distinct, and lie in the interval $[1, 5]$. We have $a = 1$, $b = -m$, and $c = 4$.

Step 2: Apply the condition for real and distinct roots.

The discriminant must be greater than zero for real and distinct roots: $D = b^2 - 4ac = (-m)^2 - 4(1)(4) = m^2 - 16 > 0$ This implies $m^2 > 16$, which means $m < -4$ or $m > 4$. Thus, $m \in (-\infty, -4) \cup (4, \infty)$.

Step 3: Apply the condition for the roots to lie in the interval [1, 5].

Since $a=1>0$, we require $f(1) \ge 0$ and $f(5) \ge 0$, where $f(x) = x^2 - mx + 4$.

• $f(1) = (1)^2 - m(1) + 4 = 5 - m \ge 0 \implies m \le 5$
• $f(5) = (5)^2 - m(5) + 4 = 29 - 5m \ge 0 \implies 5m \le 29 \implies m \le \frac{29}{5} = 5.8$

Step 4: Apply the condition for the vertex to lie in the interval [1, 5].

The x-coordinate of the vertex is $x_v = -\frac{b}{2a} = -\frac{-m}{2(1)} = \frac{m}{2}$. We require $1 \le \frac{m}{2} \le 5$. Multiplying by 2, we get $2 \le m \le 10$.

Step 5: Combine all the conditions to find the interval for m.

We have the following conditions:

1. $m \in (-\infty, -4) \cup (4, \infty)$
2. $m \le 5$
3. $m \le 5.8$
4. $2 \le m \le 10$

We need to find the intersection of these intervals. From conditions 2, 3, and 4, we have $2 \le m \le 5$. From condition 1, we have $m \in (-\infty, -4) \cup (4, \infty)$.

Since $2 \le m \le 5$, we consider the intersection with $(-\infty, -4) \cup (4, \infty)$. The intersection of $[2, 5]$ and $(-\infty, -4)$ is empty. The intersection of $[2, 5]$ and $(4, \infty)$ is $(4, 5]$.

Therefore, $m \in (4, 5]$.

Common Mistakes & Tips

• Strict vs. Non-strict Inequalities: Pay attention to whether the interval is open or closed. Since the roots can lie at the endpoints 1 and 5, we use non-strict inequalities ($\ge$ and $\le$).
• Discriminant Condition: Remember that the problem specifies "real and distinct" roots, which means $D > 0$. For real roots (allowing equality), use $D \ge 0$.
• Vertex Condition: The vertex condition is crucial for ensuring that the interval $[1, 5]$ contains both roots, not just one.

Summary

To ensure that both roots of the quadratic equation $x^2 - mx + 4 = 0$ are real, distinct, and lie within the interval $[1, 5]$, we need to satisfy the discriminant condition ($m^2 - 16 > 0$), the function value conditions ($5-m \ge 0$ and $29-5m \ge 0$), and the vertex position condition ($1 \le m/2 \le 5$). Combining these conditions leads to the interval $m \in (4, 5]$.

The final answer is \boxed{(4, 5]}. This corresponds to option (B).