Question

Difficulty: MediumLinear Equations in One and Two Variables

If xx is a real number that satisfies the equation ∣2x−1∣=3x+11|2x - 1| = 3x + 11, what is the value of x2+2xx^2 + 2x?

  1. 00Answer
  2. B
    33
  3. C
    88
  4. D
    3535
  5. E
    120120

Answer

The value of x2+2xx^2 + 2x is 00.
Solving the equation ∣2x−1∣=3x+11|2x - 1| = 3x + 11 produces two algebraic candidates: x=−12x = -12 and x=−2x = -2. Substituting x=−12x = -12 into the right side yields 3(−12)+11=−253(-12) + 11 = -25. Since absolute value expressions cannot be negative, x=−12x = -12 is an extraneous solution. Substituting x=−2x = -2 yields ∣−5∣=5|-5| = 5, which is valid. Evaluating x2+2xx^2 + 2x at x=−2x = -2 gives (−2)2+2(−2)=0(-2)^2 + 2(-2) = 0.

Step-by-Step Solution

1
Set up the two linear cases for the absolute value equation ∣2x−1∣=3x+11|2x - 1| = 3x + 11.
Case 1: 2x−1=3x+112x - 1 = 3x + 11; Case 2: 2x−1=−(3x+11)2x - 1 = -(3x + 11).
By definition, ∣u∣=c|u| = c implies u=cu = c or u=−cu = -c (provided c≥0c \geq 0).
2
Solve each linear equation for candidate values of xx.
From Case 1: x=−12x = -12. From Case 2: 2x−1=−3x−11  ⟹  5x=−10  ⟹  x=−22x - 1 = -3x - 11 \implies 5x = -10 \implies x = -2.
Isolate xx algebraically in both equations.
3
Check candidate solutions in the original equation to eliminate extraneous roots.
For x=−12x = -12: ∣2(−12)−1∣=∣−25∣=25|2(-12) - 1| = |-25| = 25, but 3(−12)+11=−25≠253(-12) + 11 = -25 \neq 25 (extraneous). For x=−2x = -2: ∣2(−2)−1∣=∣−5∣=5|2(-2) - 1| = |-5| = 5, and 3(−2)+11=53(-2) + 11 = 5 (valid solution).
An absolute value cannot equal a negative number; substituting back is mandatory.
4
Evaluate the target expression x2+2xx^2 + 2x using the valid root x=−2x = -2.
(−2)2+2(−2)=4−4=0(-2)^2 + 2(-2) = 4 - 4 = 0.
Substitute the verified real solution into the given expression.

Key Concept

Solving Absolute Value Linear Equations and Checking for Extraneous Solutions
Estimated Time:1m 30s
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