Digital SAT Math

SAT Math: 4x² + bx + 49 = 0 Has Exactly One Real Solution, Find Positive Value of b

2026-08-19 • 5 min read
• 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

  1. Condition for 1 solution: b^2 - 4ac = 0.
  2. b^2 - 4(4)(49) = 0 => b^2 - 784 = 0.
  3. 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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