SAT Math Practice Question #924 (Hard (800)) | Test Citadel
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SAT Math Difficulty: Hard (800)

Digital SAT Math Practice Question #924

Test your problem-solving accuracy on this official-caliber item. Select an answer choice to check your reasoning instantly.

In the \( xy \)-plane, the system of equations below has exactly one real solution \( (x, y) \): \[ y = 3x^2 + (-6)x + 2 \] \[ y = 6x - k \] where \( k \) is a constant. What is the value of \( k \)?
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Socratic AI Engine Step-by-Step Derivation
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Tactical Insight (Hint 1)

Equate the two expressions for \( y \) to form a single quadratic equation \( Ax^2 + Bx + C = 0 \). Remember that having exactly one solution means the discriminant \( B^2 - 4AC \) must equal 0.

Elimination Framework (Hint 2)

Desmos Shortcut: Enter the parabola in line 1: \( y = 3x^2 + (-6)x + 2 \). Enter line 2: \( y = 6x - k \) with a slider for \( k \). Move the slider until the line is perfectly tangent to the parabola at a single point.

Masterclass Solution & Distractor Trap Analysis

To find when a quadratic and a line intersect at exactly one real point, set the two equations equal: \[ 3x^2 + (-6)x + 2 = 6x - k \] \[ 3x^2 + (-12)x + (2 + k) = 0 \] A quadratic equation has exactly one real solution if and only if its di...

Distractor Analysis: Trap choice eliminates careless test-takers who confuse roots with coordinates...

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