To solve the inequality \(|x^2 - 6x + 9| < 25\), we need to consider the absolute value function. The absolute value inequality means that the expression inside the absolute value must be between -25 and 25.

### Step-by-Step Solution

1. **Understand the Absolute Value Inequality**:

   \[
   |x^2 - 6x + 9| < 25
   \]

   This implies:

   \[
   -25 < x^2 - 6x + 9 < 25
   \]

2. **Split into Two Inequalities**:

   We can split this compound inequality into two separate inequalities:

   \[
   x^2 - 6x + 9 > -25
   \]
   \[
   x^2 - 6x + 9 < 25
   \]

3. **Solve Each Inequality Separately**:

   - **First Inequality**:

     \[
     x^2 - 6x + 9 > -25
     \]

     Add 25 to both sides:

     \[
     x^2 - 6x + 34 > 0
     \]

     This is a quadratic inequality. Let's solve it by finding the roots of the corresponding equation:

     \[
     x^2 - 6x + 34 = 0
     \]

     The discriminant (\(\Delta\)) of this quadratic equation is:

     \[
     \Delta = (-6)^2 - 4 \cdot 1 \cdot 34 = 36 - 136 = -100
     \]

     Since the discriminant is negative (\(\Delta < 0\)), the quadratic has no real roots, and \(x^2 - 6x + 34\) is always positive. Thus, the inequality \(x^2 - 6x + 34 > 0\) is true for all real \(x\).

   - **Second Inequality**:

     \[
     x^2 - 6x + 9 < 25
     \]

     Subtract 25 from both sides:

     \[
     x^2 - 6x - 16 < 0
     \]

     Solve for the roots of the corresponding equation:

     \[
     x^2 - 6x - 16 = 0
     \]

     Find the roots using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6\), and \(c = -16\):

     \[
     x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-16)}}{2(1)}
     \]

     Simplify the equation:

     \[
     x = \frac{6 \pm \sqrt{36 + 64}}{2}
     \]

     \[
     x = \frac{6 \pm \sqrt{100}}{2}
     \]

     \[
     x = \frac{6 \pm 10}{2}
     \]

     The roots are:

     \[
     x_1 = \frac{6 + 10}{2} = 8, \quad x_2 = \frac{6 - 10}{2} = -2
     \]

     So, the roots are \(x = 8\) and \(x = -2\).

4. **Determine the Intervals**:

   To solve the inequality \(x^2 - 6x - 16 < 0\), we consider the intervals defined by the roots \(x = -2\) and \(x = 8\):

   - Test the intervals \(x < -2\), \(-2 < x < 8\), and \(x > 8\):

     - For \(x < -2\), pick \(x = -3\): \(x^2 - 6x - 16 = 9 + 18 - 16 = 11\) (positive)
     - For \(-2 < x < 8\), pick \(x = 0\): \(x^2 - 6x - 16 = 0 - 0 - 16 = -16\) (negative)
     - For \(x > 8\), pick \(x = 9\): \(x^2 - 6x - 16 = 81 - 54 - 16 = 11\) (positive)

     The quadratic \(x^2 - 6x - 16 < 0\) is negative in the interval \((-2, 8)\).

5. **Conclusion**:

   The solution to the inequality \(|x^2 - 6x + 9| < 25\) is:

   \[
   x \in (-2, 8)
   \]

Thus, the solution set is \(x \in (-2, 8)\).

### Would you like further details or have any questions?

Here are 5 related questions to consider:

1. What if the inequality were \(|x^2 - 6x + 9| \leq 25\)?
2. How do we solve an absolute value inequality like \(|x^2 - 4x + 4| > 10\)?
3. How do we determine the sign of a quadratic expression over different intervals?
4. Can absolute value inequalities have solutions that are non-continuous intervals?
5. How does the discriminant of a quadratic equation affect the inequality solutions?

**Tip**: When solving inequalities, always check the sign of the expression over the intervals defined by critical points.

Learn how to solve the absolute value inequality |x^2 - 6x + 9| < 25 with a detailed step-by-step solution. This problem covers key algebraic concepts, including absolute value inequalities and quadratic equations. Suitable for students in Grades 10-12.

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