Question:3x^2 + bx + 8 = 0For the quadratic equation shown, b is a constant. If one of the solutions...

1. TRANSLATE the problem information

• Given information:
  • Quadratic equation: \(3\mathrm{x}^2 + \mathrm{bx} + 8 = 0\)
  • One solution is \(\mathrm{x} = \frac{4}{3}\)
  • Need to find the value of b
• What this tells us: If \(\mathrm{x} = \frac{4}{3}\) is truly a solution, then substituting this value into the equation must make the left side equal zero.

2. INFER the solution approach

• Since we know one solution works, we can substitute it directly into the equation
• This will give us an equation with only one unknown (b) that we can solve

3. SIMPLIFY through substitution

• Substitute \(\mathrm{x} = \frac{4}{3}\) into \(3\mathrm{x}^2 + \mathrm{bx} + 8 = 0\):
  \(3\left(\frac{4}{3}\right)^2 + \mathrm{b}\left(\frac{4}{3}\right) + 8 = 0\)

• Calculate \(\left(\frac{4}{3}\right)^2\):
  \(\left(\frac{4}{3}\right)^2 = \frac{16}{9}\)

• Substitute this result:
  \(3\left(\frac{16}{9}\right) + \mathrm{b}\left(\frac{4}{3}\right) + 8 = 0\)

4. SIMPLIFY the equation

• Multiply:
  \(3\left(\frac{16}{9}\right) = \frac{48}{9} = \frac{16}{3}\)
• Rewrite:
  \(\frac{16}{3} + \frac{4\mathrm{b}}{3} + 8 = 0\)

• Convert 8 to thirds:
  \(8 = \frac{24}{3}\)
• Combine:
  \(\frac{16}{3} + \frac{4\mathrm{b}}{3} + \frac{24}{3} = 0\)
• Simplify:
  \(\frac{16 + 4\mathrm{b} + 24}{3} = 0\)
• Therefore:
  \(40 + 4\mathrm{b} = 0\)

5. SIMPLIFY to find b

• Solve:
  \(4\mathrm{b} = -40\)
• Therefore:
  \(\mathrm{b} = -10\)

Answer: B) -10

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may not recognize that "\(\mathrm{x} = \frac{4}{3}\) is a solution" means they can substitute this value directly into the equation to make it equal zero. Instead, they might try to factor the quadratic first or use the quadratic formula, making the problem unnecessarily complex.

This leads to confusion and abandoning the systematic approach, resulting in guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when working with fractions, particularly when squaring 4/3 or combining fractions with different denominators. Common mistakes include:

• Calculating \(\left(\frac{4}{3}\right)^2\) as \(\frac{4}{9}\) instead of \(\frac{16}{9}\)
• Incorrectly combining \(\frac{16}{3} + \frac{24}{3}\)
• Sign errors when solving \(40 + 4\mathrm{b} = 0\)

This may lead them to select Choice A (-12), Choice C (-8), or Choice E (10) depending on the specific calculation error.

The Bottom Line:

This problem tests whether students understand the fundamental definition of what it means for a value to be a solution to an equation, combined with careful fraction arithmetic. The key insight is recognizing that you can work backwards from the given solution.