Given that  −10≤k≤10
and  k
is an integer. How many possible integers  k
so that the following inequality holds? |k-1|(k-6)\le -6

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.

Learn how to solve absolute value inequalities involving quadratic expressions with a step-by-step explanation. This problem examines the conditions under which |k-1|(k-6) <= -6 holds true, covering key algebraic concepts like inequalities, absolute values, and quadratic equations. Suitable for high school students.

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