2x^2 - 4x = t In the equation above, t is a constant. If the equation has no real solutions,...

1. TRANSLATE the problem requirements

• Given: \(\mathrm{2x^2 - 4x = t}\) where \(\mathrm{t}\) is a constant
• Find: Value of \(\mathrm{t}\) for which the equation has no real solutions
• Key insight: "No real solutions" relates to the discriminant of a quadratic equation

2. INFER the solution approach

• To analyze solutions, I need the equation in standard form \(\mathrm{ax^2 + bx + c = 0}\)
• The discriminant \(\mathrm{b^2 - 4ac}\) determines the number of real solutions
• For no real solutions, the discriminant must be negative

3. SIMPLIFY to standard form

• Starting with: \(\mathrm{2x^2 - 4x = t}\)
• Subtract \(\mathrm{t}\) from both sides: \(\mathrm{2x^2 - 4x - t = 0}\)
• Now in standard form with \(\mathrm{a = 2, b = -4, c = -t}\)

4. SIMPLIFY the discriminant condition

• Discriminant \(\mathrm{= b^2 - 4ac = (-4)^2 - 4(2)(-t)}\)
• \(\mathrm{= 16 - 4(2)(-t)}\)
• \(\mathrm{= 16 + 8t}\)
• For no real solutions: \(\mathrm{16 + 8t \lt 0}\)

5. SIMPLIFY the inequality

• \(\mathrm{16 + 8t \lt 0}\)
• \(\mathrm{8t \lt -16}\)
• \(\mathrm{t \lt -2}\)

6. APPLY CONSTRAINTS to select the answer

• Need \(\mathrm{t \lt -2}\) from the given choices
• A. -3: Since \(\mathrm{-3 \lt -2}\) ✓
• B. -1: Since \(\mathrm{-1 \gt -2}\) ✗
• C. 1: Since \(\mathrm{1 \gt -2}\) ✗
• D. 3: Since \(\mathrm{3 \gt -2}\) ✗

Answer: A. -3

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't connect "no real solutions" to the discriminant being negative. Instead, they might try to solve the equation directly or substitute answer choices without understanding the underlying relationship.

This leads to confusion about what the problem is actually asking, causing them to get stuck and guess randomly.

Second Most Common Error:

Poor TRANSLATE reasoning: Students incorrectly identify the coefficient \(\mathrm{c}\) as \(\mathrm{+t}\) instead of \(\mathrm{-t}\) when converting \(\mathrm{2x^2 - 4x - t = 0}\) to standard form. This leads to discriminant \(\mathrm{= 16 - 8t}\) instead of \(\mathrm{16 + 8t}\).

With the wrong discriminant, they get \(\mathrm{16 - 8t \lt 0}\), which gives \(\mathrm{t \gt 2}\). This might lead them to select Choice D (3) as it's the only option greater than 2.

The Bottom Line:

This problem requires students to bridge abstract quadratic theory (discriminant) with concrete problem-solving. The key insight is recognizing that "no real solutions" isn't about solving for x, but about finding when the discriminant is negative.