3t^2 - 8t + 2 = 0What are the solutions to the given equation?(8 ± sqrt(64 - 24))/6(8 ± sqrt(9...

1. TRANSLATE the equation into standard form

• Given: \(3\mathrm{t}^2 - 8\mathrm{t} + 2 = 0\)
• This is already in standard form \(\mathrm{ax}^2 + \mathrm{bx} + \mathrm{c} = 0\) where:
  • \(\mathrm{a} = 3\) (coefficient of \(\mathrm{t}^2\))
  • \(\mathrm{b} = -8\) (coefficient of \(\mathrm{t}\))
  • \(\mathrm{c} = 2\) (constant term)

2. INFER that the quadratic formula is needed

• Since this quadratic doesn't factor easily, we need the quadratic formula
• Formula: \(\mathrm{t} = \frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2 - 4\mathrm{ac}}}{2\mathrm{a}}\)

3. SIMPLIFY by substituting our values

• \(\mathrm{t} = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(2)}}{2(3)}\)
• The key step: \(-(-8) = +8\) in the numerator
• \(\mathrm{t} = \frac{8 \pm \sqrt{64 - 24}}{6}\)
• \(\mathrm{t} = \frac{8 \pm \sqrt{40}}{6}\)

Answer: A. \(\frac{8 \pm \sqrt{64 - 24}}{6}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Sign confusion when substituting \(\mathrm{b} = -8\) into the quadratic formula

Students often forget that the quadratic formula has "-b" in the numerator, so when \(\mathrm{b} = -8\), they need \(-(-8) = +8\). Instead, they might write:

\(\mathrm{t} = \frac{-8 \pm \sqrt{64 - 24}}{6}\)

This doesn't match any of the given choices directly and leads to confusion and guessing.

Second Most Common Error:

Poor coefficient identification in TRANSLATE: Confusing the value of \(\mathrm{b}^2\) in the discriminant

Students might think \(\mathrm{b} = 8\) instead of \(\mathrm{b} = -8\), leading to the same \(\mathrm{b}^2 = 64\) but missing the sign error in the numerator. Or they might calculate \(\mathrm{b}^2\) incorrectly as 9 instead of 64.

This may lead them to select Choice B (\(\frac{8 \pm \sqrt{9 - 24}}{6}\)) or get confused by the negative discriminant.

The Bottom Line:

The trickiest part is handling the negative coefficient correctly throughout the quadratic formula, especially remembering that \(-\mathrm{b}\) becomes \(+8\) when \(\mathrm{b} = -8\).