Question

Find the sum of all real solutions of
$|x + 1|^2 - 4|x + 1| + 3 = 0$

• A. -2
• B. 0
• C. 2
• D. 4

Hint:

Equations of the form $a|f(x)|^2 + b|f(x)| + c = 0$ can always be simplified by substituting $y=|f(x)|$. After finding the possible values for $y$, remember to check if they are non-negative, since the absolute value cannot be negative. Then solve for $x$ for each valid $y$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an equation involving the absolute value of a linear term. The equation is quadratic in form with respect to the absolute value term. We need to find all real solutions and then compute their sum. 
Step 2: Key Formula or Approach: 
1. Recognize that the equation is a quadratic in the variable $|x-1|$. 
2. Substitute a temporary variable, say $y = |x-1|$, to solve the quadratic equation for $y$. 
3. Once the values for $y$ are found, substitute back to get $|x-1| = y$. 
4. Solve the absolute value equations. Remember that $|A| = B$ (for $B>0$) implies $A=B$ or $A=-B$. 
Step 3: Detailed Explanation: 
Assuming the corrected equation: 
\[ |x - 1|^2 - 4|x - 1| + 3 = 0 \] Let $y = |x - 1|$. The equation becomes: 
\[ y^2 - 4y + 3 = 0 \] This is a simple quadratic equation. We can factor it: 
\[ (y - 1)(y - 3) = 0 \] This gives two possible values for y: $y = 1$ or $y = 3$. 
Now, substitute back $y = |x - 1|$: 
Case 1: $|x - 1| = 1$ 
This gives two possibilities: 
- $x - 1 = 1 \implies x = 2$ 
- $x - 1 = -1 \implies x = 0$ 
Case 2: $|x - 1| = 3$ 
This gives two possibilities: 
- $x - 1 = 3 \implies x = 4$ 
- $x - 1 = -3 \implies x = -2$ 
The set of all real solutions is $\{-2, 0, 2, 4\}$. 
The sum of all these solutions is: 
\[ \text{Sum} = (-2) + 0 + 2 + 4 = 4 \] Step 4: Final Answer: 
The sum of all real solutions is 4.