Question:sqrt(2x - 5) = -3How many distinct real solutions does the given equation have?

1. INFER the fundamental constraint

• The equation is \(\sqrt{2\mathrm{x} - 5} = -3\)
• Key insight: The principal square root \(\sqrt{\mathrm{expression}}\) always gives non-negative results
• Since -3 is negative, this equation appears impossible from the start

2. SIMPLIFY algebraically to verify our reasoning

• Square both sides to eliminate the square root:

\((\sqrt{2\mathrm{x} - 5})^2 = (-3)^2\)

• This gives us: \(2\mathrm{x} - 5 = 9\)
• Solve for x: \(2\mathrm{x} = 14\), so \(\mathrm{x} = 7\)

3. INFER that we must check our potential solution

• Substitute x = 7 back into the original equation:

\(\sqrt{2(7) - 5} = \sqrt{9} = 3\)

• But we need this to equal -3, not 3
• Since \(3 \neq -3\), our potential solution doesn't work

4. INFER the final conclusion

• The algebraic manipulation gave us a candidate solution, but it failed verification
• This confirms our initial insight: the equation has no real solutions

Answer: (A) Zero

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students square both sides, find x = 7, and immediately select an answer suggesting "exactly one solution" without checking their work in the original equation.

They think: "I solved 2x - 5 = 9 and got x = 7, so there's one solution." They don't realize that squaring both sides can introduce extraneous solutions that don't satisfy the original equation.

This may lead them to select Choice (B) (Exactly one).

Second Most Common Error:

Missing conceptual knowledge: Students don't remember that the principal square root function only outputs non-negative values, so they don't immediately recognize the equation as impossible.

Without this foundational understanding, they rely entirely on algebraic manipulation and miss the conceptual shortcut that makes the problem much simpler.

This leads to confusion and potentially guessing between multiple answer choices.

The Bottom Line:

This problem tests whether students understand both the definition of the principal square root function AND the importance of checking solutions when working with radical equations. The algebraic work is straightforward, but the conceptual understanding determines whether students recognize the trap.