Consider the quadratic function y = 2x^2 - 12x + 7. The graph intersects the x-axis at two points A...

1. TRANSLATE the problem information

• Given: \(\mathrm{y = 2x^2 - 12x + 7}\) intersects x-axis at points A and B
• Find: x-coordinate of midpoint of segment AB
• What this means: If \(\mathrm{A = (r, 0)}\) and \(\mathrm{B = (s, 0)}\), we need \(\mathrm{\frac{r + s}{2}}\)

2. INFER the most efficient approach

• Key insight: For any parabola, the midpoint between x-intercepts lies on the axis of symmetry
• We can find this directly without solving for the actual intercepts
• Two methods available: axis of symmetry formula or sum of roots

3. SIMPLIFY using the axis of symmetry method

• For \(\mathrm{y = ax^2 + bx + c}\), axis of symmetry: \(\mathrm{x = \frac{-b}{2a}}\)
• Here: \(\mathrm{a = 2, b = -12}\)
• \(\mathrm{x = \frac{-(-12)}{2 \cdot 2}}\)
  \(\mathrm{= \frac{12}{4}}\)
  \(\mathrm{= 3}\)

4. VERIFY using sum of roots method

• For \(\mathrm{2x^2 - 12x + 7 = 0}\), sum of roots \(\mathrm{= \frac{-b}{a}}\)
  \(\mathrm{= \frac{-(-12)}{2}}\)
  \(\mathrm{= 6}\)
• Midpoint x-coordinate \(\mathrm{= \frac{sum\:of\:roots}{2}}\)
  \(\mathrm{= \frac{6}{2}}\)
  \(\mathrm{= 3}\) ✓

Answer: B (3)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the connection between axis of symmetry and midpoint of x-intercepts

Many students think they must solve the complete quadratic equation using the quadratic formula: \(\mathrm{x = \frac{12 \pm \sqrt{144 - 56}}{4}}\). While this approach works, it's unnecessarily complex and increases chances of arithmetic errors. Students who take this route often make calculation mistakes or get discouraged by the messy square root \(\mathrm{\sqrt{88}}\).

This leads to confusion and abandoning the systematic approach for guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Making sign errors when applying formulas

Students correctly identify the need to use \(\mathrm{x = \frac{-b}{2a}}\) but make errors like:

• Using \(\mathrm{x = \frac{-(-12)}{2 \cdot 2} = \frac{-12}{4} = -3}\) (forgetting the negative of negative b)
• Computing \(\mathrm{x = \frac{-(-12)}{2 \cdot 2} = \frac{12}{8} = 1.5}\) (arithmetic error in denominator)

This may lead them to select Choice A (0) if they get confused with signs.

The Bottom Line:

This problem tests whether students can recognize elegant shortcuts in quadratic problems. The key insight is that parabolas are symmetric, so you don't always need to find exact intercepts to answer questions about their midpoint.