To solve for the values of $k$ such that the quadratic equation $x^2 + (k-4)x + 1 = 0$ has no real roots, we need to analyze the discriminant of the quadratic equation.

For a general quadratic equation of the form $ax^2 + bx + c = 0$, the discriminant $\Delta$ is given by:

$\Delta = b^2 - 4ac$

In your equation, $a = 1$, $b = (k-4)$, and $c = 1$. Therefore, the discriminant is:

$\Delta = (k-4)^2 - 4(1)(1)$

Simplifying this expression:

$\Delta = (k-4)^2 - 4$

For the quadratic equation to have no real roots, the discriminant must be less than zero:

$(k-4)^2 - 4 < 0$

Now solve this inequality:

$(k-4)^2 < 4$

Taking the square root of both sides:

$|k-4| < 2$

This implies:

$-2 < k-4 < 2$

Adding 4 to all parts of the inequality:

$2 < k < 6$

Thus, the set of possible values of $k$ such that the quadratic equation has no real roots is:

$k \in (2, 6)$

Would you like further details or have any questions?

Related questions:

1. What happens if the discriminant is equal to zero?
2. How would the roots change if $k = 2$?
3. What if $k > 6$, how does the nature of the roots change?
4. Can you give a similar quadratic equation where the roots are imaginary?
5. How does completing the square help in finding the roots?

Tip: Always check the discriminant when analyzing the nature of the roots of quadratic equations!