https://file.aiquickdraw.com/mathbot/file/1727028524886z84g03jf.jpg

Let's solve the quadratic inequalities given in the image.

### 1. \( x^2 \leq 18 - 7x \)
Rewrite this inequality as a standard quadratic inequality:
\[
x^2 + 7x - 18 \leq 0
\]
Now, solve the equation \(x^2 + 7x - 18 = 0\) by factoring:
\[
(x + 9)(x - 2) = 0
\]
So, \(x = -9\) and \(x = 2\) are the roots. These are the *neutral numbers* or critical points. Now, test intervals around these roots to find the solution set.

- Test \(x < -9\) (e.g., \(x = -10\)): \( (-10)^2 + 7(-10) - 18 = 100 - 70 - 18 > 0 \), so not included.
- Test \(-9 < x < 2\) (e.g., \(x = 0\)): \( (0)^2 + 7(0) - 18 = -18 < 0 \), so included.
- Test \(x > 2\) (e.g., \(x = 3\)): \( (3)^2 + 7(3) - 18 = 9 + 21 - 18 = 12 > 0 \), so not included.

The solution set is \([-9, 2]\), and the *neutral numbers* are \(x = -9\) and \(x = 2\).

### 2. \( x^2 - 14 < 9x \)
Rewrite this inequality:
\[
x^2 - 9x - 14 < 0
\]
Solve \(x^2 - 9x - 14 = 0\) using the quadratic formula:
\[
x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(-14)}}{2(1)} = \frac{9 \pm \sqrt{81 + 56}}{2} = \frac{9 \pm \sqrt{137}}{2}
\]
The roots are irrational, so the solution set will lie between these roots. Approximating the roots as \(x_1 \approx -1.34\) and \(x_2 \approx 10.34\), the solution set is approximately \((-1.34, 10.34)\).

### 3. \( 3x^2 - 10x - 8 \leq 0 \)
Solve \(3x^2 - 10x - 8 = 0\) using the quadratic formula:
\[
x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(-8)}}{2(3)} = \frac{10 \pm \sqrt{100 + 96}}{6} = \frac{10 \pm \sqrt{196}}{6} = \frac{10 \pm 14}{6}
\]
So, the roots are:
\[
x = \frac{10 + 14}{6} = 4 \quad \text{and} \quad x = \frac{10 - 14}{6} = -\frac{2}{3}
\]
The solution set is \([- \frac{2}{3}, 4]\), and the *neutral numbers* are \(x = -\frac{2}{3}\) and \(x = 4\).

### 4. \( x^2 + 6x + 10 > 0 \)
First, solve \(x^2 + 6x + 10 = 0\) using the quadratic formula:
\[
x = \frac{-6 \pm \sqrt{(6)^2 - 4(1)(10)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 40}}{2} = \frac{-6 \pm \sqrt{-4}}{2} = \frac{-6 \pm 2i}{2} = -3 \pm i
\]
This equation has no real roots. Since \(x^2 + 6x + 10 > 0\) is always positive (the discriminant is negative, so the parabola lies above the x-axis), the solution set is \(\mathbb{R}\) (all real numbers).

### 5. \( 6x^2 \geq 1 - x \)
Rewrite the inequality:
\[
6x^2 + x - 1 \geq 0
\]
Solve \(6x^2 + x - 1 = 0\) using the quadratic formula:
\[
x = \frac{-1 \pm \sqrt{(1)^2 - 4(6)(-1)}}{2(6)} = \frac{-1 \pm \sqrt{1 + 24}}{12} = \frac{-1 \pm \sqrt{25}}{12} = \frac{-1 \pm 5}{12}
\]
The roots are:
\[
x = \frac{-1 + 5}{12} = \frac{4}{12} = \frac{1}{3} \quad \text{and} \quad x = \frac{-1 - 5}{12} = \frac{-6}{12} = -\frac{1}{2}
\]
The solution set is \( (-\infty, -\frac{1}{2}] \cup [\frac{1}{3}, \infty) \), and the *neutral numbers* are \(x = -\frac{1}{2}\) and \(x = \frac{1}{3}\).

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Let me know if you would like more details or have any questions!

Here are 5 related questions:
1. How do you graph the solution set of a quadratic inequality?
2. What is the significance of the discriminant in solving quadratic inequalities?
3. Can a quadratic inequality have no real solution?
4. How does factoring help solve quadratic inequalities?
5. How are quadratic inequalities applied in real-world problems?

**Tip**: When solving quadratic inequalities, always check the signs of intervals around the roots to correctly determine the solution set.

Identify the neutral numbers and solution set of the following quadratic inequalities: 1. x² ≤ 18 - 7x, 2. x² - 14 < 9x, 3. 3x² - 10x - 8 ≤ 0, 4. x² + 6x + 10 > 0, 5. 6x² ≥ 1 - x.

This problem covers solving quadratic inequalities such as x² ≤ 18 - 7x and identifying neutral numbers using factoring and the quadratic formula. Learn how to find the solution sets and graph each inequality, suitable for high school students.

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