How many real solutions does the equation sqrt(x + 9) = x - 3 have?Exactly oneExactly twoInfinitely manyZero

1. APPLY CONSTRAINTS to determine the valid domain

• The square root \(\sqrt{x + 9}\) requires: \(x + 9 \geq 0\), so \(x \geq -9\)
• Since square roots are always non-negative, we need: \(x - 3 \geq 0\), so \(x \geq 3\)
• Combined domain restriction: \(x \geq 3\)

2. INFER the solving strategy

• To eliminate the square root, square both sides of the equation
• This will create a quadratic equation we can solve algebraically

3. SIMPLIFY by squaring both sides

• Square both sides: \((\sqrt{x + 9})^2 = (x - 3)^2\)
• This gives us: \(x + 9 = x^2 - 6x + 9\)

4. SIMPLIFY the algebraic equation

• Subtract 9 from both sides: \(x = x^2 - 6x\)
• Rearrange to standard form: \(x^2 - 7x = 0\)
• Factor: \(x(x - 7) = 0\)
• Solutions: \(x = 0\) or \(x = 7\)

5. APPLY CONSTRAINTS to check validity

• Test \(x = 0\): Does not satisfy \(x \geq 3\), so reject this solution
• Test \(x = 7\): Satisfies \(x \geq 3\), so this could be valid

6. Verify the solution in the original equation

• For \(x = 7\): \(\sqrt{7 + 9} = \sqrt{16} = 4\), and \(7 - 3 = 4\) ✓
• The solution checks out

Answer: A (Exactly one)

Why Students Usually Falter on This Problem

Most Common Error Path:

Poor APPLY CONSTRAINTS reasoning: Students find both algebraic solutions \(x = 0\) and \(x = 7\) but fail to check them against the domain restriction \(x \geq 3\). They see two solutions from the factored form and assume both are valid.

This may lead them to select Choice B (Exactly two)

Second Most Common Error:

Weak INFER skill: Students don't recognize that squaring both sides can introduce extraneous solutions, so they skip the verification step entirely. Even if they apply domain constraints correctly, they might miss that verification in the original equation is crucial.

This causes confusion about which solutions are actually valid and may lead to guessing.

The Bottom Line:

Radical equations require careful attention to domain restrictions AND verification of solutions because algebraic manipulation can introduce invalid answers. The key insight is that algebraic solutions must always be tested against both domain constraints and the original equation.