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Let's go through the quadratic inequalities one by one and solve them to identify the neutral numbers and solution sets. After solving, we can determine the solution set intervals and graph them accordingly.

### 1. \( x^2 \leq 18 - 7x \)

We first rearrange the inequality to bring all terms to one side:

\[
x^2 + 7x - 18 \leq 0
\]

Now solve the quadratic equation \( x^2 + 7x - 18 = 0 \) using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 7 \), and \( c = -18 \).

\[
x = \frac{-7 \pm \sqrt{7^2 - 4(1)(-18)}}{2(1)} = \frac{-7 \pm \sqrt{49 + 72}}{2} = \frac{-7 \pm \sqrt{121}}{2}
\]

\[
x = \frac{-7 \pm 11}{2}
\]

Thus, \( x = 2 \) or \( x = -9 \).

The quadratic factors as \( (x - 2)(x + 9) \leq 0 \). Using the sign chart method, the solution set is:

\[
[-9, 2]
\]

### 2. \( x^2 - 14 < 9x \)

Rearrange the inequality:

\[
x^2 - 9x - 14 < 0
\]

Solve the quadratic equation \( x^2 - 9x - 14 = 0 \) using the quadratic formula:

\[
x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(-14)}}{2(1)} = \frac{9 \pm \sqrt{81 + 56}}{2} = \frac{9 \pm \sqrt{137}}{2}
\]

This gives two roots, approximately:

\[
x = \frac{9 \pm 11.7}{2} \implies x \approx 10.35 \quad \text{or} \quad x \approx -1.35
\]

The quadratic factors as \( (x + 1.35)(x - 10.35) < 0 \). The solution set is:

\[
(-1.35, 10.35)
\]

### 3. \( 3x^2 - 10x - 8 \leq 0 \)

Solve the quadratic inequality. First, solve the quadratic equation:

\[
3x^2 - 10x - 8 = 0
\]

Using the quadratic formula:

\[
x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(-8)}}{2(3)} = \frac{10 \pm \sqrt{100 + 96}}{6} = \frac{10 \pm \sqrt{196}}{6}
\]

\[
x = \frac{10 \pm 14}{6}
\]

Thus, \( x = 4 \) or \( x = -\frac{4}{3} \).

The quadratic factors as \( (x - 4)(3x + 4) \leq 0 \). The solution set is:

\[
\left[-\frac{4}{3}, 4\right]
\]

### 4. \( x^2 + 6x + 10 > 0 \)

This quadratic does not factor easily, and solving \( x^2 + 6x + 10 = 0 \) gives imaginary roots, meaning the quadratic is always positive. Therefore, the solution set is:

\[
(-\infty, \infty)
\]

### 5. \( 6x^2 \geq 1 - x \)

Rearrange the inequality:

\[
6x^2 + x - 1 \geq 0
\]

Solve the quadratic equation:

\[
6x^2 + x - 1 = 0
\]

Using the quadratic formula:

\[
x = \frac{-1 \pm \sqrt{1^2 - 4(6)(-1)}}{2(6)} = \frac{-1 \pm \sqrt{1 + 24}}{12} = \frac{-1 \pm \sqrt{25}}{12}
\]

\[
x = \frac{-1 \pm 5}{12}
\]

Thus, \( x = \frac{4}{12} = \frac{1}{3} \) or \( x = \frac{-6}{12} = -\frac{1}{2} \).

The quadratic factors as \( (x - \frac{1}{3})(x + \frac{1}{2}) \geq 0 \). The solution set is:

\[
(-\infty, -\frac{1}{2}] \cup \left[\frac{1}{3}, \infty\right)
\]

---

Would you like to go over the details for any specific step or would you like help graphing these solutions?

Here are 5 related questions you could explore:
1. How do you graph a quadratic inequality on a number line?
2. What is the significance of neutral numbers in inequalities?
3. How do you solve a quadratic inequality using the sign chart method?
4. Can all quadratic inequalities be solved using factoring?
5. How do imaginary roots affect the solution set of a quadratic inequality?

**Tip:** When solving quadratic inequalities, always check the endpoints of the intervals to determine if they should be included in the solution set.

Identify the neutral numbers and solution set of the following Quadratic Inequality and draw its graph.

This problem focuses on solving quadratic inequalities, identifying neutral numbers, and determining solution sets. Learn how to solve inequalities like x² ≤ 18 - 7x and draw graphs to represent the solutions. Suitable for Grades 9-11.

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