Question

Sum of the roots of the equation \( |x-3|^2 + |x-3| - 2 = 0 \) is equal to

• A. \(2\)
• B. \(4\)
• C. \(6\)
• D. \(16\)
• E. \(-2\)

Hint:

Always replace expressions inside modulus with a new variable and remember the condition that modulus is always non-negative.

Answer 1

The Correct Option is C

Concept: Whenever an equation involves modulus (absolute value), we simplify by introducing a substitution: \[ |x-3| = t \quad \text{where } t \ge 0 \] This converts the equation into a standard quadratic equation in \(t\), which is easier to solve. After solving for \(t\), we back-substitute to find \(x\).

Step 1: Substitute \( |x-3| = t \)
Given equation: \[ |x-3|^2 + |x-3| - 2 = 0 \] Substitute: \[ t^2 + t - 2 = 0 \]

Step 2: Solve the quadratic equation
\[ t^2 + t - 2 = 0 \] Factor: \[ (t+2)(t-1) = 0 \] Thus, \[ t = -2 \quad \text{or} \quad t = 1 \]

Step 3: Apply constraint \( t \ge 0 \)
Since \( t = |x-3| \ge 0 \), we discard: \[ t = -2 \] So valid value: \[ t = 1 \]

Step 4: Back-substitute
\[ |x-3| = 1 \] This gives two cases: Case 1: \[ x - 3 = 1 \Rightarrow x = 4 \] Case 2: \[ x - 3 = -1 \Rightarrow x = 2 \]

Step 5: Find sum of roots
\[ \text{Roots are } x = 4 \text{ and } x = 2 \] \[ \text{Sum} = 4 + 2 = 6 \] Final Answer: \[ \boxed{6} \]