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.