\(2\mathrm{x}(\mathrm{x} + 3) - 20 = 5\mathrm{x}(\mathrm{x} - 2)\)What is the sum of the solutions to the given equation?Express your...

1. SIMPLIFY the equation by expanding both sides

• Left side: \(\mathrm{2x(x + 3) - 20 = 2x^2 + 6x - 20}\)
• Right side: \(\mathrm{5x(x - 2) = 5x^2 - 10x}\)

Your equation becomes: \(\mathrm{2x^2 + 6x - 20 = 5x^2 - 10x}\)

2. SIMPLIFY by collecting all terms on one side

Move everything to the left side:

\(\mathrm{2x^2 + 6x - 20 - 5x^2 + 10x = 0}\)

Combine like terms: \(\mathrm{-3x^2 + 16x - 20 = 0}\)

Multiply by -1 to make the leading coefficient positive: \(\mathrm{3x^2 - 16x + 20 = 0}\)

3. INFER the solution method needed

Since this is a quadratic equation in standard form (\(\mathrm{ax^2 + bx + c = 0}\)), you need the quadratic formula with \(\mathrm{a = 3, b = -16, c = 20}\).

4. SIMPLIFY using the quadratic formula

\(\mathrm{x = \frac{16 ± \sqrt{(-16)^2 - 4(3)(20)}}{2(3)}}\)

\(\mathrm{x = \frac{16 ± \sqrt{256 - 240}}{6}}\)

\(\mathrm{x = \frac{16 ± \sqrt{16}}{6}}\)

\(\mathrm{x = \frac{16 ± 4}{6}}\)

This gives you two solutions:

• \(\mathrm{x = \frac{16 + 4}{6} = \frac{20}{6} = \frac{10}{3}}\)
• \(\mathrm{x = \frac{16 - 4}{6} = \frac{12}{6} = 2}\)

5. SIMPLIFY to find the sum

Sum = \(\mathrm{\frac{10}{3} + 2 = \frac{10}{3} + \frac{6}{3} = \frac{16}{3}}\)

Answer: \(\mathrm{\frac{16}{3}}\) (or 5.333 as a decimal)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make errors when expanding the polynomials or combining like terms. For example, they might get \(\mathrm{2x^2 + 6x - 20 = 5x^2 - 10x}\) but then incorrectly combine terms as \(\mathrm{2x^2 - 5x^2 + 6x + 10x - 20 = 0}\), leading to \(\mathrm{-3x^2 + 16x - 20 = 0}\) becoming something like \(\mathrm{-3x^2 + 4x - 20 = 0}\). This produces entirely different solutions and an incorrect sum.

Second Most Common Error:

Incomplete INFER reasoning: Students solve for one solution correctly but forget they need both solutions from the quadratic formula. They might find \(\mathrm{x = \frac{10}{3}}\) and submit that as their final answer, missing that the problem asks for the sum of solutions. This leads to answering \(\mathrm{\frac{10}{3}}\) instead of \(\mathrm{\frac{16}{3}}\).

The Bottom Line:

This problem tests sustained algebraic manipulation accuracy through multiple steps. One small error in expanding, collecting terms, or applying the quadratic formula cascades into a completely wrong final answer.