\(\mathrm{x(x + 1) - 56 = 4x(x - 7)}\)What is the sum of the solutions to the given equation?

1. SIMPLIFY the equation by expanding both sides

• Given equation: \(\mathrm{x(x + 1) - 56 = 4x(x - 7)}\)
• Left side: \(\mathrm{x(x + 1) - 56 = x^2 + x - 56}\)
• Right side: \(\mathrm{4x(x - 7) = 4x^2 - 28x}\)
• Equation becomes: \(\mathrm{x^2 + x - 56 = 4x^2 - 28x}\)

2. SIMPLIFY to standard quadratic form

• Move all terms to one side:
  \(\mathrm{x^2 + x - 56 - 4x^2 + 28x = 0}\)
• Combine like terms: \(\mathrm{-3x^2 + 29x - 56 = 0}\)
• Multiply by -1: \(\mathrm{3x^2 - 29x + 56 = 0}\)

3. INFER the solution approach

• We have a quadratic equation in standard form \(\mathrm{ax^2 + bx + c = 0}\)
• Use the quadratic formula with \(\mathrm{a = 3, b = -29, c = 56}\)
• The question asks for the sum of solutions, so we'll need both roots

4. SIMPLIFY using the quadratic formula

• \(\mathrm{x = \frac{29 \pm \sqrt{(-29)^2 - 4(3)(56)}}{2(3)}}\)
• Calculate discriminant:
  \(\mathrm{(-29)^2 - 4(3)(56) = 841 - 672 = 169}\)
• \(\mathrm{x = \frac{29 \pm \sqrt{169}}{6} = \frac{29 \pm 13}{6}}\)
• Two solutions:
  \(\mathrm{x_1 = \frac{29 + 13}{6} = \frac{42}{6} = 7}\)
  and
  \(\mathrm{x_2 = \frac{29 - 13}{6} = \frac{16}{6} = \frac{8}{3}}\)

5. SIMPLIFY to find the sum

• Sum = \(\mathrm{7 + \frac{8}{3} = \frac{21}{3} + \frac{8}{3} = \frac{29}{3}}\)

Answer: 29/3 (or 9.666, 9.667)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skills: Sign errors when expanding or combining like terms

Students often make mistakes like:

• Expanding \(\mathrm{4x(x - 7)}\) as \(\mathrm{4x^2 + 28x}\) instead of \(\mathrm{4x^2 - 28x}\)
• Getting the wrong signs when moving terms: writing \(\mathrm{3x^2 + 29x + 56 = 0}\) instead of \(\mathrm{3x^2 - 29x + 56 = 0}\)
• Arithmetic errors in the quadratic formula calculations

These algebraic mistakes compound through the solution, leading to completely wrong final answers and causing confusion when trying to match answer choices.

Second Most Common Error:

Weak INFER reasoning: Finding only one solution or forgetting to add solutions

Students might correctly solve the quadratic but:

• Only report one of the two x-values as their final answer
• Calculate both solutions correctly but forget that the question asks for their sum
• Get confused about what the question is actually asking for

This leads to selecting one of the individual solution values instead of their sum.

The Bottom Line:

This problem requires sustained algebraic accuracy through multiple steps, combined with careful reading of what the question asks for. The expansion and rearrangement phases are particularly error-prone, and students must maintain precision while keeping track of the ultimate goal.