To solve the inequality $|x-1|^2 + 2|x-1| \leq 2$, let's use a substitution to simplify it. Let $y = |x-1|$. The inequality then becomes:

$y^2 + 2y \leq 2$

1. Rearrange the inequality:

   $y^2 + 2y - 2 \leq 0$

2. Solve the quadratic inequality $y^2 + 2y - 2 = 0$ to find the boundary points. We can use the quadratic formula:

   $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 2$, and $c = -2$:

   $y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1}$ $y = \frac{-2 \pm \sqrt{4 + 8}}{2}$ $y = \frac{-2 \pm \sqrt{12}}{2}$ $y = \frac{-2 \pm 2\sqrt{3}}{2}$ $y = -1 \pm \sqrt{3}$

   Therefore, the roots are $y = -1 + \sqrt{3}$ and $y = -1 - \sqrt{3}$. Since $y = |x-1| \geq 0$, we only consider the positive root:

   $y = -1 + \sqrt{3}$

3. Determine the intervals where the inequality holds:

   Since $y = |x-1|$ is non-negative, the relevant root is $y = -1 + \sqrt{3}$. We test the inequality $y^2 + 2y - 2 \leq 0$ within the interval $[0, -1 + \sqrt{3}]$:

   • When $y = 0$: $0^2 + 2(0) - 2 = -2$ (which satisfies the inequality).
   • When $y = -1 + \sqrt{3}$: $(-1 + \sqrt{3})^2 + 2(-1 + \sqrt{3}) - 2 = 0$ (the boundary point).

   Thus, the inequality holds for $0 \leq y \leq -1 + \sqrt{3}$.

4. Translate back to $x$ using $y = |x-1|$:

   $0 \leq |x-1| \leq -1 + \sqrt{3}$

   This implies:

   $-1 + \sqrt{3} \geq x - 1 \geq -(-1 + \sqrt{3})$

   $1 - \sqrt{3} \leq x - 1 \leq \sqrt{3} - 1$

   $1 + 1 - \sqrt{3} \leq x \leq 1 + \sqrt{3} - 1$

   $2 - \sqrt{3} \leq x \leq \sqrt{3}$

Therefore, the solution set for the inequality $|x-1|^2 + 2|x-1| \leq 2$ is:

$x \in [2 - \sqrt{3}, \sqrt{3}]$

Would you like more details or have any questions?

Related Questions:

1. How would the solution change if the inequality were strict, $|x-1|^2 + 2|x-1| < 2$?
2. What if the quadratic term had a different sign, such as $|x-1|^2 - 2|x-1| \leq 2$?
3. How can we graphically represent the inequality and its solution set?
4. How do absolute values affect the domain and range of functions?
5. How would the inequality change if it involved a cubic term instead?

Tip:

When dealing with inequalities involving absolute values, substituting a variable for the absolute value term can simplify the problem significantly, allowing for easier algebraic manipulation.