Given the inequality $|k - 1|(k - 6) \leq -6$ and the range $-10 \leq k \leq 10$, where $k$ is an integer, we need to determine how many possible integer values of $k$ satisfy this inequality.

Step 1: Consider the inequality

The inequality is $|k - 1|(k - 6) \leq -6$.

Note that the absolute value $|k - 1|$ is always non-negative, so the product $|k - 1|(k - 6)$ is non-negative or zero unless $k - 6$ is negative. The left side must be negative or zero to satisfy $\leq -6$. This implies that the only way for $|k - 1|(k - 6) \leq -6$ is for the left side to be negative, and since $|k - 1| \geq 0$, we must have $|k - 1|(k - 6) < 0$.

However, for the left side to equal $-6$ exactly, the product would need to match that value, but it cannot since $|k - 1|(k - 6)$ can't be negative by definition unless the absolute value function is multiplied by a negative term.

Step 2: Break down the inequality

Consider the two cases for the absolute value $|k - 1|$:

1. Case 1: $k - 1 \geq 0$ (i.e., $k \geq 1$) \ Then $|k - 1| = k - 1$. The inequality becomes: $(k - 1)(k - 6) \leq -6$

2. Case 2: $k - 1 < 0$ (i.e., $k < 1$) \ Then $|k - 1| = -(k - 1)$. The inequality becomes: $-(k - 1)(k - 6) \leq -6 \implies (k - 1)(k - 6) \geq 6$

Step 3: Analyze the inequality in both cases

Case 1: $k \geq 1$

The inequality is $(k - 1)(k - 6) \leq -6$.

Expanding it:

$k^2 - 7k + 6 \leq -6$ $k^2 - 7k + 12 \leq 0$

To solve $k^2 - 7k + 12 \leq 0$, we find the roots of the quadratic equation $k^2 - 7k + 12 = 0$ using the quadratic formula:

$k = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(12)}}{2(1)} = \frac{7 \pm \sqrt{49 - 48}}{2} = \frac{7 \pm 1}{2}$

So, the roots are $k = 4$ and $k = 3$. The inequality $k^2 - 7k + 12 \leq 0$ holds when $3 \leq k \leq 4$.

Case 2: $k < 1$

The inequality is $(k - 1)(k - 6) \geq 6$.

Expanding it:

$k^2 - 7k + 6 \geq 6$ $k^2 - 7k \geq 0$

This inequality holds when $k \leq 0$ or $k \geq 7$.

Step 4: Combine results with the given range $-10 \leq k \leq 10$

From Case 1:

• $3 \leq k \leq 4$ gives $k = 3, 4$.

From Case 2:

• $k \leq 0$ gives $k = -10, -9, \ldots, 0$.

Thus, the possible values for $k$ that satisfy the inequality are: $k = -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 3, 4$

Final Count

There are $11 + 2 = 13$ possible integers for $k$ that satisfy the inequality.

Would you like further details on any step, or do you have any additional questions?

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Related Questions:

1. How do we solve inequalities involving absolute values and quadratic expressions?
2. How would the solution change if the inequality was $|k-1|(k-6) \geq -6$?
3. What happens if the range of $k$ was different?
4. Can you solve similar inequalities without the integer constraint on $k$?
5. How do absolute value inequalities differ when the constant on the right side is positive?

Tip: Always consider splitting the inequality into cases when dealing with absolute values, as it often simplifies the process of solving the inequality.