Question:x + 2 = sqrt(3x + 10)What is the solution set of the equation above?

1. INFER the solution strategy

• We have \(\mathrm{x + 2 = \sqrt{3x + 10}}\)
• The square root makes this challenging to solve directly
• Key insight: Square both sides to eliminate the square root
• Important: Squaring can create extraneous solutions, so we must check all answers

2. SIMPLIFY by squaring both sides

\(\mathrm{(x + 2)^2 = (\sqrt{3x + 10})^2}\)

\(\mathrm{(x + 2)^2 = 3x + 10}\)

3. SIMPLIFY by expanding and collecting terms

Expand left side: \(\mathrm{x^2 + 4x + 4 = 3x + 10}\)

Move all terms to one side: \(\mathrm{x^2 + 4x + 4 - 3x - 10 = 0}\)

Combine like terms: \(\mathrm{x^2 + x - 6 = 0}\)

4. SIMPLIFY by factoring the quadratic

• Look for two numbers that multiply to -6 and add to 1
• Those numbers are 3 and -2: \(\mathrm{(x + 3)(x - 2) = 0}\)
• This gives us \(\mathrm{x = -3}\) or \(\mathrm{x = 2}\)

5. APPLY CONSTRAINTS by checking solutions

• Must verify both solutions work in the original equation

For x = -3:

• Left side: \(\mathrm{(-3) + 2 = -1}\)
• Right side: \(\mathrm{\sqrt{3(-3) + 10} = \sqrt{1} = 1}\)
• Since \(\mathrm{-1 \neq 1}\), this solution is extraneous

For x = 2:

• Left side: \(\mathrm{2 + 2 = 4}\)
• Right side: \(\mathrm{\sqrt{3(2) + 10} = \sqrt{16} = 4}\)
• Since \(\mathrm{4 = 4}\), this solution is valid

Answer: B. \(\mathrm{\{2\}}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak APPLY CONSTRAINTS execution: Students solve the quadratic correctly to get \(\mathrm{x = -3}\) and \(\mathrm{x = 2}\), but fail to check these solutions in the original equation. They assume both solutions are valid since they came from proper algebraic manipulation.

Without checking, they conclude the solution set is \(\mathrm{\{-3, 2\}}\) and select Choice C (\(\mathrm{\{-3, 2\}}\)).

Second Most Common Error:

Incomplete INFER reasoning: Students recognize they need to square both sides but don't understand why checking for extraneous solutions is necessary. They may check \(\mathrm{x = -3}\), see it doesn't work, but think they made an error somewhere and include it anyway, or they get confused about which solutions to keep.

This leads to uncertainty about whether the answer is \(\mathrm{\{-3\}}\), \(\mathrm{\{2\}}\), or \(\mathrm{\{-3, 2\}}\), causing them to guess or abandon systematic solution.

The Bottom Line:

The key challenge is understanding that squaring both sides of an equation can introduce solutions that don't satisfy the original equation. Students must develop the habit of verification, especially with radical equations.