Solving a System of Absolute Value Equations: When Does |x+2|=0?

Question

{x+2=x2x+2=0 \begin{cases} |x+2|=|x-2| \\ \lvert x+2\rvert=0 \end{cases}

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Solve the equation x+2=0 |x+2| = 0
    This implies x+2=0 x+2 = 0 , hence x=2 x = -2 .
  • Step 2: Solve the equation x+2=x2 |x+2| = |x-2| .
    This leads to two cases:
    - Case 1: x+2=x2 x+2 = x-2 results in 2=2 2 = -2 , which is false.
    - Case 2: x+2=(x2) x+2 = -(x-2) simplifies to 2x=0 2x = 0 , hence x=0 x = 0 .
  • Step 3: Check for intersection of solutions.
    We have solutions x=2 x = -2 from the first equation and x=0 x = 0 from the absolute value equation.
  • Step 4: Verify if any common x x satisfies both equations.

Substituting x=2 x = -2 into x+2=x2 |x+2| = |x-2| , we find that it does not satisfy x2=4=4 |x-2| = |-4| = 4 . Therefore, x=2 x = -2 is not a solution to the system.

Substituting x=0 x = 0 into both equations does not satisfy x+2=0 |x+2| = 0 . Therefore, there is no overlap of solutions.

The solution to the system is No solution.

Answer

No solution