If both the roots of the quadratic equation $$x^2 - mx + 4 = 0$$ are real and distinct and they lie in the interval [1, 5], then $$m$$ lies in the interval:

The discriminant ($$D$$) must be strictly greater than zero:

$$D = b^2 - 4ac > 0$$ $$(-m)^2 - 4(1)(4) > 0$$ $$m^2 - 16 > 0$$ $$m^2 > 16 \implies m \in (-\infty, -4) \cup (4, \infty) \quad \dots \text{(i)}$$

The x-coordinate of the vertex ($$-\frac{b}{2a}$$) must lie within the given interval $$(1, 5)$$ for the roots to be contained within $$[1, 5]$$:

$$1 < \frac{-(-m)}{2(1)} < 5$$ $$1 < \frac{m}{2} < 5$$ $$2 < m < 10 \quad \dots \text{(ii)}$$

Since the coefficient of $$x^2$$ is positive ($$a=1 > 0$$), the parabola opens upwards. For the roots to stay within $$[1, 5]$$, the function values at the boundaries must be greater than or equal to zero: $$f(1) \ge 0$$ and $$f(5) \ge 0$$.

• For $$x = 1$$:$$f(1) = 1^2 - m(1) + 4 \ge 0$$ $$5 - m \ge 0 \implies m \le 5 \quad \dots \text{(iii)}$$
• For $$x = 5$$:$$f(5) = 5^2 - m(5) + 4 \ge 0$$ $$25 - 5m + 4 \ge 0$$ $$29 \ge 5m \implies m \le \frac{29}{5} \quad (m \le 5.8) \quad \dots \text{(iv)}$$

We now find the values of $$m$$ that satisfy all four inequalities:

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

Combining these:

• From (1) and (2), we get $$4 < m < 10$$.
• Applying (3), the range restricts to $$4 < m \le 5$$.
• Condition (4) ($$m \le 5.8$$) is already satisfied by $$m \le 5$$.

The final interval for $$m$$ is:

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