https://file.aiquickdraw.com/mathbot/file/1725566348152jjaeyknm.jpg

The image contains three inequalities that need to be solved. Let's go through them one by one:

### 1. \((x^2 - 10x + 9)(4x + 1)^2 > 0\)

Here, we need to solve the inequality:
\[
(x^2 - 10x + 9)(4x + 1)^2 > 0
\]

1. **Factorize** the quadratic expression \(x^2 - 10x + 9\):
   \[
   x^2 - 10x + 9 = (x - 9)(x - 1)
   \]
   So the inequality becomes:
   \[
   (x - 9)(x - 1)(4x + 1)^2 > 0
   \]

2. The term \((4x + 1)^2\) is always non-negative (since a square is always \(\geq 0\)). It will equal 0 only when \(x = -\frac{1}{4}\). So, we just need to find where \((x - 9)(x - 1)\) is positive.

3. Solve \((x - 9)(x - 1) > 0\) using a sign chart:
   - The roots are \(x = 9\) and \(x = 1\), which divide the number line into intervals: \((-\infty, 1)\), \((1, 9)\), and \((9, \infty)\).
   - Test points from each interval:
     - For \(x < 1\), both factors are negative, so the product is positive.
     - For \(1 < x < 9\), one factor is negative and one is positive, so the product is negative.
     - For \(x > 9\), both factors are positive, so the product is positive.

4. Since \((4x + 1)^2 > 0\) for all \(x \neq -\frac{1}{4}\), the solution to the inequality is:
   \[
   (-\infty, 1) \cup (9, \infty)
   \]
   excluding \(x = -\frac{1}{4}\).

---

### 2. \((x^2 - 10x + 9)(4x + 1)^2 \leq 0\)

This inequality is similar to the first, except we now have \(\leq 0\).

1. Factorize the quadratic again: \((x - 9)(x - 1)(4x + 1)^2 \leq 0\).

2. As \((4x + 1)^2\) is non-negative and only zero when \(x = -\frac{1}{4}\), we focus on \((x - 9)(x - 1)\):
   - \((x - 9)(x - 1) = 0\) when \(x = 1\) or \(x = 9\).
   - The quadratic product \((x - 9)(x - 1)\) is negative between \(1\) and \(9\).

3. The solution will include the points where the product equals zero (\(x = 1\) or \(x = 9\)) and the interval where the product is negative. Therefore, the solution is:
   \[
   [1, 9]
   \]

---

### 3. \((x^2 - 6x + 8)(x^2 - 4) \leq 0\)

Here, we need to solve:
\[
(x^2 - 6x + 8)(x^2 - 4) \leq 0
\]

1. Factorize each quadratic:
   - \(x^2 - 6x + 8 = (x - 4)(x - 2)\)
   - \(x^2 - 4 = (x - 2)(x + 2)\)

   The inequality becomes:
   \[
   (x - 4)(x - 2)(x + 2)(x - 2) \leq 0
   \]
   or, simplified:
   \[
   (x - 4)(x - 2)^2(x + 2) \leq 0
   \]

2. The term \((x - 2)^2\) is always non-negative and equals zero when \(x = 2\).

3. Analyze the sign of the product \((x - 4)(x + 2)\) using a sign chart:
   - The roots are \(x = -2\) and \(x = 4\), dividing the number line into intervals: \((-\infty, -2)\), \((-2, 4)\), and \((4, \infty)\).
   - Test points in each interval:
     - For \(x < -2\), both factors are negative, so the product is positive.
     - For \(-2 < x < 4\), one factor is negative and the other positive, so the product is negative.
     - For \(x > 4\), both factors are positive, so the product is positive.

4. The solution is where the product is zero or negative. The solution is:
   \[
   [-2, 4]
   \]

---

### Summary of Solutions:
1. \((- \infty, 1) \cup (9, \infty)\), excluding \(x = -\frac{1}{4}\).
2. \([1, 9]\)
3. \([-2, 4]\)

Would you like more details on any specific steps or clarifications?

### Related Questions:
1. How do we factor quadratic expressions efficiently?
2. What is the importance of a sign chart in solving inequalities?
3. Why does \((4x + 1)^2 \geq 0\) for all real \(x\)?
4. How do we solve inequalities involving higher powers or more complex expressions?
5. What are the key differences between strict (\(>\)) and non-strict (\(\geq\)) inequalities?

**Tip:** Always test values between roots to understand how the sign of an expression changes across different intervals.

Solve the following inequalities:
1) (x^2 - 10x + 9)(4x + 1)^2 > 0
2) (x^2 - 10x + 9)(4x + 1)^2 ≤ 0
3) (x^2 - 6x + 8)(x^2 - 4) ≤ 0

Learn how to solve quadratic inequalities like (x^2 - 10x + 9)(4x + 1)^2 > 0 and others using factorization and the sign chart method. This detailed solution breaks down each step, covers key concepts, and provides clear explanations. Suitable for students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation