If x gt 0, which of the following is a solution to the equation sqrt(x^2 + 27) = 2x?

1. TRANSLATE the problem information

• Given equation: \(\sqrt{\mathrm{x}^2 + 27} = 2\mathrm{x}\)
• Constraint: \(\mathrm{x} \gt 0\)
• Need to find which answer choice satisfies both conditions

2. INFER the solution strategy

• The square root on the left side prevents direct algebraic manipulation
• Key insight: Square both sides to eliminate the radical
• This transforms the equation into a more manageable form

3. SIMPLIFY by squaring both sides

• \((\sqrt{\mathrm{x}^2 + 27})^2 = (2\mathrm{x})^2\)
• Left side: \(\mathrm{x}^2 + 27\)
• Right side: \(4\mathrm{x}^2\)
• New equation: \(\mathrm{x}^2 + 27 = 4\mathrm{x}^2\)

4. SIMPLIFY to isolate the x² term

• Subtract x² from both sides: \(27 = 4\mathrm{x}^2 - \mathrm{x}^2\)
• Combine like terms: \(27 = 3\mathrm{x}^2\)
• Divide by 3: \(\mathrm{x}^2 = 9\)

5. SIMPLIFY to find x values

• Take the square root: \(\mathrm{x} = \pm\sqrt{9} = \pm 3\)
• This gives us two potential solutions: \(\mathrm{x} = 3\) and \(\mathrm{x} = -3\)

6. APPLY CONSTRAINTS to select the final answer

• The problem states \(\mathrm{x} \gt 0\)
• Since \(-3 \lt 0\), we reject \(\mathrm{x} = -3\)
• Therefore: \(\mathrm{x} = 3\)

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skills: Students don't recognize that squaring both sides is the correct approach to eliminate the radical. They might attempt to isolate the square root differently or avoid the squaring method entirely due to fear of creating extraneous solutions.

Without this key strategic insight, students get stuck on the radical equation and cannot make progress algebraically. This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly square both sides but make algebraic errors in the subsequent steps, such as incorrectly combining \(\mathrm{x}^2 + 27 = 4\mathrm{x}^2\) or making arithmetic mistakes when solving \(3\mathrm{x}^2 = 27\).

These calculation errors lead to wrong values for x, potentially causing them to select Choice A (2) or Choice D (9) based on their incorrect work.

The Bottom Line:

This problem tests whether students can strategically handle radical equations through the squaring method while maintaining accuracy through multi-step algebraic manipulation. The key breakthrough is recognizing that squaring both sides transforms a difficult radical equation into a straightforward quadratic relationship.