sqrt(2x - 6) = -3 How many distinct real solutions does the given equation have?...

1. TRANSLATE the problem information

• Given equation: \(\sqrt{2x - 6} = -3\)
• Question asks: How many distinct real solutions?

2. INFER the fundamental constraint

• Recall that the principal square root function √(...) always produces non-negative outputs
• This means \(\sqrt{2x - 6} \geq 0\) for any real value where the expression is defined
• The right side of our equation is -3, which is negative

3. APPLY CONSTRAINTS to determine solution existence

• Since \(\sqrt{2x - 6} \geq 0\) but we need it to equal -3
• This creates an impossible situation: we need something non-negative to equal something negative
• Therefore, no real solutions exist

Answer: (A) Zero

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students immediately try to solve algebraically by squaring both sides without first considering whether the equation makes sense.

They might square both sides:

\((\sqrt{2x - 6})^2 = (-3)^2\)

This gives:

\(2x - 6 = 9\)

Solving:

\(2x = 15\), so \(x = 7.5\)

Then they might check:

\(\sqrt{2(7.5) - 6} = \sqrt{9} = 3 \neq -3\)

This approach leads to confusion because they find a value that makes the expression under the radical positive, but don't realize that squaring both sides introduced an extraneous solution. This may lead them to select Choice (B) (Exactly one) thinking they found a solution, or causes confusion and guessing.

The Bottom Line:

The key insight is recognizing fundamental constraints before attempting algebraic manipulation. Square root functions have restricted ranges, and some equations are impossible by definition rather than requiring algebraic solution.