Which of the following is an equivalent form of \((2\mathrm{x} - 1.5)^2 + (0.5\mathrm{x}^2 - 2.5)\)?

1. INFER the solution strategy

• This problem asks for an equivalent form, which means we need to expand and simplify
• We have a binomial square plus another polynomial expression
• Strategy: Expand the square first, then combine all like terms

2. SIMPLIFY the binomial square \((2\mathrm{x} - 1.5)²\)

• Use the formula \((\mathrm{a} - \mathrm{b})² = \mathrm{a}² - 2\mathrm{ab} + \mathrm{b}²\)
• \((2\mathrm{x} - 1.5)² = (2\mathrm{x})² - 2(2\mathrm{x})(1.5) + (1.5)²\)
• \(= 4\mathrm{x}² - 6\mathrm{x} + 2.25\)

3. SIMPLIFY by adding the second polynomial

• Now we have: \(4\mathrm{x}² - 6\mathrm{x} + 2.25 + 0.5\mathrm{x}² - 2.5\)
• Rewrite to group like terms: \(4\mathrm{x}² + 0.5\mathrm{x}² - 6\mathrm{x} + 2.25 - 2.5\)

4. SIMPLIFY by combining like terms

• x² terms: \(4\mathrm{x}² + 0.5\mathrm{x}² = 4.5\mathrm{x}²\)
• x terms: \(-6\mathrm{x}\) (only one x term)
• Constants: \(2.25 - 2.5 = -0.25\)

Answer: C. \(4.5\mathrm{x}² - 6\mathrm{x} - 0.25\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students correctly expand \((2\mathrm{x} - 1.5)²\) to get \(4\mathrm{x}² - 6\mathrm{x} + 2.25\), but forget to include the \(0.5\mathrm{x}²\) term when combining like terms.

They focus only on the expanded binomial and the constant -2.5, getting:
\(4\mathrm{x}² - 6\mathrm{x} + 2.25 - 2.5 = 4\mathrm{x}² - 6\mathrm{x} - 0.25\)

This leads them to select Choice A (\(4\mathrm{x}² - 6\mathrm{x} - 0.25\))

Second Most Common Error:

Poor SIMPLIFY execution with decimal arithmetic: Students make sign errors when combining the constant terms, especially getting confused with 2.25 - 2.5.

Some students might calculate \(2.25 - 2.5 = +0.25\) instead of \(-0.25\), or make other decimal calculation mistakes.

This may lead them to select Choice D (\(4.5\mathrm{x}² - 6\mathrm{x} + 0.25\)) if they get the x² terms right but mess up the constant.

The Bottom Line:

This problem tests careful algebraic manipulation with decimals. Success requires systematically expanding the binomial, then methodically combining ALL like terms without losing track of any pieces of the expression.