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: Rewrite the given equation

The given equation is:

|x − 1|² − 4|x − 1| + 3 = 0

Since |x − 1|² = (x − 1)², the equation becomes:

(x − 1)² − 4|x − 1| + 3 = 0

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Step 2: Use interval-wise analysis

The absolute value expression changes form depending on the value of x. So we consider two cases.

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Case 1: x ≥ 1

For x ≥ 1,
|x − 1| = x − 1

Substitute in the equation:

(x − 1)² − 4(x − 1) + 3 = 0

Simplifying:

x² − 6x + 8 = 0

(x − 2)(x − 4) = 0

x = 2, 4 (both satisfy x ≥ 1)

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Case 2: x < 1

For x < 1,
|x − 1| = 1 − x

Substitute in the equation:

(x − 1)² − 4(1 − x) + 3 = 0

Simplifying:

x² − 2x + 1 − 4 + 4x + 3 = 0

x² + 2x = 0

x(x + 2) = 0

x = 0, −2 (both satisfy x < 1)

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Step 3: List all real solutions

The real solutions are:
−2, 0, 2, 4

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Step 4: Calculate the sum

Sum = (−2) + 0 + 2 + 4

Sum = 4

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Final Answer:

The sum of all real solutions is
4