Let's solve the inequality \(|x-1|^2 + 2|x-1| \leq 2\).

### Step 1: Set \( y = |x - 1| \)
This substitution simplifies the inequality:
\[
y^2 + 2y \leq 2
\]

### Step 2: Rearrange the inequality
\[
y^2 + 2y - 2 \leq 0
\]

### Step 3: Solve the quadratic inequality
To solve \( y^2 + 2y - 2 \leq 0 \), we first find the roots of the quadratic equation \( y^2 + 2y - 2 = 0 \) by using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 2 \), and \( c = -2 \):
\[
y = \frac{-2 \pm \sqrt{2^2 - 4(1)(-2)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 8}}{2} = \frac{-2 \pm \sqrt{12}}{2} = \frac{-2 \pm 2\sqrt{3}}{2} = -1 \pm \sqrt{3}
\]
So the roots are:
\[
y = -1 + \sqrt{3} \quad \text{and} \quad y = -1 - \sqrt{3}
\]
Since \( y = |x - 1| \geq 0 \), we discard \( y = -1 - \sqrt{3} \) (which is negative) and focus on \( y = -1 + \sqrt{3} \).

### Step 4: Determine the range of \( y \)
The quadratic inequality \( y^2 + 2y - 2 \leq 0 \) is satisfied when \( y \) is between the roots:
\[
-1 + \sqrt{3} \leq y \leq 1 - \sqrt{3}
\]

But \( y = |x - 1| \geq 0 \), and \( -1 + \sqrt{3} \approx 0.732 \), so:
\[
0 \leq y \leq \sqrt{3} - 1
\]

### Step 5: Convert back to \( x \)
Since \( y = |x - 1| \), we have:
\[
0 \leq |x - 1| \leq \sqrt{3} - 1
\]

This implies:
\[
-\sqrt{3} + 1 \leq x - 1 \leq \sqrt{3} - 1
\]
Adding 1 to all parts of the inequality:
\[
2 - \sqrt{3} \leq x \leq \sqrt{3}
\]

### Final Answer
The solution set for the inequality is:
\[
x \in \left[2 - \sqrt{3}, \sqrt{3}\right]
\]

Would you like more details or have any questions?

Here are some related questions for further exploration:
1. What happens if the inequality sign is reversed?
2. How would the solution change if the constant term on the right-hand side is altered?
3. What if the absolute value expression is replaced by another function?
4. How does this problem relate to the concept of distance on a number line?
5. Can you solve similar inequalities with higher powers?

**Tip:** When dealing with inequalities involving absolute values, try substituting the absolute value with a new variable to simplify the problem.

Learn how to solve the inequality |x-1|^2 + 2|x-1| <= 2 with detailed steps. This problem involves quadratic inequalities and absolute value functions, providing insights into algebraic manipulations and root finding. Suitable for Grades 10-12 students and those preparing for college entrance exams.

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