The problem asks us to find the values of $k$ for which the quadratic equation $x^2 - kx + 3 - k = 0$ has real roots.

1. Problem Description

The problem asks us to find the values of $k$ for which the quadratic equation $x^2 - kx + 3 - k = 0$ has real roots.

2. Solution Steps

For a quadratic equation $ax^2 + bx + c = 0$ to have real roots, the discriminant must be greater than or equal to zero. The discriminant is given by the formula:

$D = b^2 - 4ac$

In our equation, $x^2 - kx + 3 - k = 0$, we have $a = 1$, $b = -k$, and $c = 3 - k$. So, we need to find the values of $k$ for which:

$D = (-k)^2 - 4(1)(3 - k) \ge 0$

$k^2 - 4(3 - k) \ge 0$

$k^2 - 12 + 4k \ge 0$

$k^2 + 4k - 12 \ge 0$

Now we need to factor the quadratic expression $k^2 + 4k - 12$. We are looking for two numbers that multiply to $-12$ and add up to $4$. Those numbers are $6$ and $-2$.

So, we can factor the expression as:

$(k + 6)(k - 2) \ge 0$

Now we analyze the inequality $(k + 6)(k - 2) \ge 0$. The critical points are $k = -6$ and $k = 2$. We test the intervals:

* $k < -6$: Choose $k = -7$. Then $(-7 + 6)(-7 - 2) = (-1)(-9) = 9 > 0$. So, $k < -6$ is part of the solution.

* $-6 < k < 2$: Choose $k = 0$. Then $(0 + 6)(0 - 2) = (6)(-2) = -12 < 0$. So, $-6 < k < 2$ is not part of the solution.

* $k > 2$: Choose $k = 3$. Then $(3 + 6)(3 - 2) = (9)(1) = 9 > 0$. So, $k > 2$ is part of the solution.

Since we have $\ge 0$, we include the critical points $k = -6$ and $k = 2$. Therefore, the solution is $k \le -6$ or $k \ge 2$.

3. Final Answer

$k \le -6$ or $k \ge 2$