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.