Question:8x^2 + 5 = 6x + 5Which of the following is a non-zero solution to the equation above?

1. TRANSLATE the problem information

• Given equation: \(8x^2 + 5 = 6x + 5\)
• Find: The non-zero solution from the answer choices

2. SIMPLIFY by eliminating the constant terms

• Subtract 5 from both sides:

\(8x^2 + 5 - 5 = 6x + 5 - 5\)

\(8x^2 = 6x\)

3. SIMPLIFY to standard quadratic form

• Move all terms to one side:

\(8x^2 - 6x = 0\)

4. INFER the factoring strategy

• Notice both terms contain x, so we can factor out the greatest common factor
• Factor out 2x: \(2x(4x - 3) = 0\)

5. INFER the solution method

• Apply Zero Product Property: if \(2x(4x - 3) = 0\), then either \(2x = 0\) or \(4x - 3 = 0\)

6. SIMPLIFY to find both solutions

• From \(2x = 0\): \(x = 0\)
• From \(4x - 3 = 0\): \(4x = 3\), so \(x = \frac{3}{4}\)

7. APPLY CONSTRAINTS to select the required solution

• We found \(x = 0\) and \(x = \frac{3}{4}\)
• The question asks for the non-zero solution
• Therefore: \(x = \frac{3}{4}\)

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make algebraic errors when rearranging the equation or factoring. They might incorrectly subtract terms, forget to move all terms to one side, or factor incorrectly (such as factoring out just x instead of 2x). This leads to wrong solutions that don't match any answer choice, causing confusion and guessing.

Second Most Common Error:

Poor APPLY CONSTRAINTS reasoning: Students correctly find both solutions \(x = 0\) and \(x = \frac{3}{4}\), but misread the question and select \(x = 0\) as their answer. Since \(x = 0\) isn't among the choices, this leads them to second-guess their work and potentially select a wrong answer or abandon their systematic approach.

The Bottom Line:

This problem tests whether students can systematically work through a quadratic equation that doesn't initially appear in standard form, while carefully reading what the question actually asks for. The key insight is recognizing that finding all solutions is just the first step - you must then apply the constraint about which solution is wanted.