The equation $(p - 1)x^2 + 4x + (p - 5) = 0$, where $p$ is a constant has no real roots - Edexcel - A-Level Maths Pure - Question 6 - 2015 - Paper 1

To find the values of $p$ for which $p^2 - 6p + 1 > 0$, we first determine the roots of the quadratic equation $p^2 - 6p + 1 = 0$.
Using the quadratic formula:

$p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Here, $a = 1$, $b = -6$, and $c = 1$:

$p = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(1)}}{2(1)}$
$= \frac{6 \pm \sqrt{36 - 4}}{2}$
$= \frac{6 \pm \sqrt{32}}{2}$
$= \frac{6 \pm 4\sqrt{2}}{2}$
$= 3 \pm 2\sqrt{2}$

The roots are $p_1 = 3 - 2\sqrt{2}$ and $p_2 = 3 + 2\sqrt{2}$.
Now, we analyze the sign of the quadratic $p^2 - 6p + 1$. Since it opens upwards (as the coefficient of $p^2$ is positive), it will be positive outside the interval between the roots:

Thus, the solution set is:

$p < 3 - 2\sqrt{2} \quad \text{or} \quad p > 3 + 2\sqrt{2}$

Therefore, the set of possible values of $p$ is:
$(-\infty, 3 - 2\sqrt{2}) \cup (3 + 2\sqrt{2}, \infty)$