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Let's break down and solve the two parts of this problem.

### Part (a) \( 6 \leq x^2 + x < 12 + 2x \)

We are given a compound inequality:

\[
6 \leq x^2 + x < 12 + 2x
\]

**Step 1: Break this into two inequalities.**

The first inequality is:

\[
6 \leq x^2 + x
\]

The second inequality is:

\[
x^2 + x < 12 + 2x
\]

#### Solving the first inequality: \( 6 \leq x^2 + x \)

Rearrange the inequality:

\[
x^2 + x - 6 \geq 0
\]

Now, solve \( x^2 + x - 6 = 0 \) by factoring:

\[
(x - 2)(x + 3) = 0
\]

The roots are \( x = 2 \) and \( x = -3 \).

The inequality \( x^2 + x - 6 \geq 0 \) holds in the intervals \( (-\infty, -3] \cup [2, \infty) \) because the quadratic opens upwards.

So, the solution to this inequality is:

\[
x \in (-\infty, -3] \cup [2, \infty)
\]

#### Solving the second inequality: \( x^2 + x < 12 + 2x \)

Rearrange the inequality:

\[
x^2 + x - 12 - 2x < 0
\]

Simplify:

\[
x^2 - x - 12 < 0
\]

Now, solve \( x^2 - x - 12 = 0 \) by factoring:

\[
(x - 4)(x + 3) = 0
\]

The roots are \( x = 4 \) and \( x = -3 \).

The inequality \( x^2 - x - 12 < 0 \) holds between the roots, so the solution is:

\[
x \in (-3, 4)
\]

#### Combine the results:

From the first inequality, we have \( x \in (-\infty, -3] \cup [2, \infty) \), and from the second inequality, we have \( x \in (-3, 4) \). The solution is the intersection of these sets:

\[
x \in [-3, -3] \cup [2, 4) = \{-3\} \cup [2, 4)
\]

Thus, the solution for part (a) is:

\[
x \in \{-3\} \cup [2, 4)
\]

### Part (b) \( |4x - 1| \geq x - 6 \)

This is an absolute value inequality. We need to consider two cases based on the definition of absolute value.

#### Case 1: \( 4x - 1 \geq x - 6 \)

Rearrange the inequality:

\[
4x - x \geq -6 + 1
\]

\[
3x \geq -5
\]

\[
x \geq -\frac{5}{3}
\]

#### Case 2: \( -(4x - 1) \geq x - 6 \)

Simplify:

\[
-4x + 1 \geq x - 6
\]

Rearrange the inequality:

\[
-4x - x \geq -6 - 1
\]

\[
-5x \geq -7
\]

\[
x \leq \frac{7}{5}
\]

#### Combine the results:

From Case 1, we have \( x \geq -\frac{5}{3} \), and from Case 2, we have \( x \leq \frac{7}{5} \).

Thus, the solution is:

\[
x \in \left[-\frac{5}{3}, \frac{7}{5}\right]
\]

### Final Answers:

1. For part (a): \( x \in \{-3\} \cup [2, 4) \)
2. For part (b): \( x \in \left[-\frac{5}{3}, \frac{7}{5}\right] \)

---

Do you want further clarification on any of the steps or details?

Here are five related questions to expand your understanding:
1. How do we determine the sign of a quadratic expression to solve inequalities?
2. What is the significance of the intersection when solving compound inequalities?
3. How can you use the graph of a quadratic function to solve inequalities?
4. What changes when solving absolute value inequalities compared to regular inequalities?
5. How do you test the intervals when solving inequalities involving quadratic expressions?

**Tip**: Always check for extraneous solutions when dealing with absolute value inequalities by substituting the solutions back into the original inequality.

Solve the inequalities: (a) 6 ≤ x² + x < 12 + 2x and (b) |4x - 1| ≥ x - 6

Learn how to solve the compound inequality 6 ≤ x² + x < 12 + 2x and the absolute value inequality |4x - 1| ≥ x - 6. This detailed solution covers factoring quadratics, sign analysis, and absolute value properties. Suitable for students in Grades 9-11.

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