\(\sqrt{(\mathrm{x} + 2)(\mathrm{x} - 5)(\mathrm{x} + 9)} = 0\)What is a positive solution to the given equation?

1. TRANSLATE the equation notation

• Given: \(\sqrt{\mathrm{x + 2}}(\mathrm{x - 5})(\mathrm{x + 9}) = 0\)
• This means: \(\sqrt{\mathrm{x + 2}} \cdot (\mathrm{x - 5}) \cdot (\mathrm{x + 9}) = 0\)
• The square root applies only to the first factor \((\mathrm{x + 2})\)

2. INFER the domain restriction

• For \(\sqrt{\mathrm{x + 2}}\) to be defined in real numbers: \(\mathrm{x + 2} \geq 0\)
• This gives us: \(\mathrm{x} \geq -2\)
• This is our domain - any solution must satisfy this condition

3. APPLY the zero product property

• For the product \(\sqrt{\mathrm{x + 2}} \cdot (\mathrm{x - 5}) \cdot (\mathrm{x + 9}) = 0\), at least one factor must be zero
• Setting each factor to zero:
  • \(\sqrt{\mathrm{x + 2}} = 0\) → \(\mathrm{x + 2} = 0\) → \(\mathrm{x} = -2\)
  • \(\mathrm{x - 5} = 0\) → \(\mathrm{x} = 5\)
  • \(\mathrm{x + 9} = 0\) → \(\mathrm{x} = -9\)

4. APPLY CONSTRAINTS to validate solutions

• Check which solutions satisfy \(\mathrm{x} \geq -2\):
  • \(\mathrm{x} = -2\): Valid ✓ (since \(-2 \geq -2\))
  • \(\mathrm{x} = 5\): Valid ✓ (since \(5 \geq -2\))
  • \(\mathrm{x} = -9\): Invalid ✗ (since \(-9 \lt -2\))

5. INFER the final answer

• Valid solutions: \(\mathrm{x} = -2\) and \(\mathrm{x} = 5\)
• The problem asks for the positive solution
• Therefore: \(\mathrm{x} = 5\)

Answer: C (5)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Misinterpreting the radical notation as \(\sqrt{(\mathrm{x + 2})(\mathrm{x - 5})(\mathrm{x + 9})} = 0\) instead of \(\sqrt{\mathrm{x + 2}} \cdot (\mathrm{x - 5}) \cdot (\mathrm{x + 9}) = 0\)

Students might think the entire product is under the square root, leading them to find a different domain condition. This creates confusion about which solutions are valid and may cause them to get stuck and randomly select an answer.

Second Most Common Error:

Poor APPLY CONSTRAINTS reasoning: Finding all three algebraic solutions (\(\mathrm{x} = -9, -2, 5\)) but failing to check the domain restriction \(\mathrm{x} \geq -2\)

Students solve the zero product property correctly but forget that \(\mathrm{x} = -9\) violates the domain. They might see that both -2 and 5 work algebraically and choose the wrong positive value, or include -9 as a valid solution, leading to confusion and guessing.

The Bottom Line:

This problem tests whether students can correctly interpret radical notation and consistently apply domain restrictions throughout the solution process.