Digital SAT Math
SAT Math: 4x² + bx + 49 = 0 Has Exactly One Real Solution, Find Positive Value of b
• Direct Answer / Key Takeaway
Delta = b^2 - 4(4)(49) = 0 => b^2 = 784 => b = 28.
Exact Question Text
For what positive value of b does the equation 4x^2 + bx + 49 = 0 have exactly one real solution?
1. Step-by-Step Formal Solution
- Condition for 1 solution: b^2 - 4ac = 0.
- b^2 - 4(4)(49) = 0 => b^2 - 784 = 0.
- b = sqrt(784) = 28.
2. Fast Desmos / Calculator Shortcut
Type b^2 - 4(4)(49) = 0 into Desmos to get b = 28.
3. Distractor Trap Analysis
- Trap A (Sign Flip): Forgetting to reverse inequality or dropping negative sign.
- Trap B (Partial Evaluation): Solving for intermediate variable and stopping early.
- Trap C (Extreme Assumption): Over-inferring beyond textual evidence.
- Option Correct: Accurately satisfies all algebraic and logical constraints.
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