sqrt(-x + 8) = x - 2What value of x is the solution to the equation above?

1. APPLY CONSTRAINTS to establish valid solution domain

• Given: \(\sqrt{-\mathrm{x} + 8} = \mathrm{x} - 2\)
• Domain requirements:
  • Square root requires: \(-\mathrm{x} + 8 \geq 0\), so \(\mathrm{x} \leq 8\)
  • Equation validity requires: \(\mathrm{x} - 2 \geq 0\), so \(\mathrm{x} \geq 2\)
  • Combined domain: \(2 \leq \mathrm{x} \leq 8\)

2. SIMPLIFY by eliminating the radical

• Square both sides: \([\sqrt{-\mathrm{x} + 8}]^2 = (\mathrm{x} - 2)^2\)
• This gives us: \(-\mathrm{x} + 8 = \mathrm{x}^2 - 4\mathrm{x} + 4\)

3. SIMPLIFY to solve the resulting quadratic

• Rearrange: \(0 = \mathrm{x}^2 - 4\mathrm{x} + \mathrm{x} + 4 - 8\)
• Combine like terms: \(0 = \mathrm{x}^2 - 3\mathrm{x} - 4\)
• Factor: \(0 = (\mathrm{x} - 4)(\mathrm{x} + 1)\)
• Potential solutions: \(\mathrm{x} = 4\) or \(\mathrm{x} = -1\)

4. APPLY CONSTRAINTS to verify solutions

• Check \(\mathrm{x} = 4\) in original equation:
  • Left side: \(\sqrt{-(4) + 8} = \sqrt{4} = 2\)
  • Right side: \(4 - 2 = 2\)
  • Since \(2 = 2\), \(\mathrm{x} = 4\) is valid ✓
• Check \(\mathrm{x} = -1\) in original equation:
  • Left side: \(\sqrt{-(-1) + 8} = \sqrt{9} = 3\)
  • Right side: \(-1 - 2 = -3\)
  • Since \(3 \neq -3\), \(\mathrm{x} = -1\) is extraneous ✗

Answer: 4

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak APPLY CONSTRAINTS skill: Students solve the quadratic correctly but fail to check their solutions in the original equation.

After factoring to get \(\mathrm{x} = 4\) and \(\mathrm{x} = -1\), they assume both solutions are valid without verification. Since \(\mathrm{x} = -1\) appears to satisfy the domain restriction \(\mathrm{x} \leq 8\), they might incorrectly conclude there are two solutions or select the wrong answer if they only check one solution superficially.

This may lead them to select Choice A) -1 or cause confusion about which answer to choose.

Second Most Common Error:

Inadequate APPLY CONSTRAINTS reasoning: Students forget to establish domain restrictions before solving, leading them to accept any algebraic solution without considering the constraints of radical equations.

Without recognizing that \(\mathrm{x} - 2 \geq 0\) is required for the original equation to be meaningful, they may work with solutions outside the valid domain and get confused when checking doesn't work properly.

This leads to confusion and guessing among the given choices.

The Bottom Line:

Radical equations require careful attention to both domain restrictions and solution verification. The algebraic manipulation creates the potential for extraneous solutions, making the checking step absolutely essential rather than optional.