Digital SAT Hard Math Practice Flashcards

If x and y are positive real numbers such that log₂(x) + log₄(y) = 5 and log₄(x) + log₂(y) = 7, what is the value of log₂(xy)?

Answer: 8

Let a = log₂(x) and b = log₂(y). Since log₄(t) = log₂(t)/2, the equations become a + b/2 = 5 and a/2 + b = 7. Adding both equations: (3/2)(a + b) = 12, so a + b = 8. Since log₂(xy) = log₂(x) + log₂(y) = a + b, the answer is 8.

A monic cubic polynomial p(x) has roots at x = 1, x = −1, and x = k (where k ≠ ±1). If p(2) = 12, what is the value of k?

Answer: −2

Since p(x) is monic with roots 1, −1, and k: p(x) = (x − 1)(x + 1)(x − k) = (x² − 1)(x − k). Substituting x = 2: (4 − 1)(2 − k) = 3(2 − k) = 12, so 2 − k = 4 and k = −2.

The quadratic function f(x) = ax² + bx + c has its vertex at (3, −7) and passes through the point (1, 5). What is the value of a + b + c?

Answer: 5

Key insight: a + b + c = f(1) for any quadratic. Using vertex form f(x) = a(x − 3)² − 7, substitute the point (1, 5): 5 = a(−2)² − 7 = 4a − 7, so a = 3. Then f(1) = 3(1 − 3)² − 7 = 12 − 7 = 5.

For what nonzero value of a does the system below have infinitely many solutions? 6x − 4y = 10 9x − 6y = a

Answer: 15

For infinitely many solutions the equations must be scalar multiples of each other. The ratio of the x-coefficients is 9/6 = 3/2, and −6/−4 = 3/2 confirms the same ratio on y. Multiplying the right side of the first equation by 3/2: (3/2)(10) = 15. Therefore a = 15.

A downward-opening parabola has x-intercepts at x = −5 and x = 1 and a maximum value of 18. What is f(0)?

Answer: 10

The axis of symmetry is x = (−5 + 1)/2 = −2. Write f(x) = a(x + 5)(x − 1). At the vertex x = −2: f(−2) = a(3)(−3) = −9a = 18, so a = −2. Then f(0) = −2(0 + 5)(0 − 1) = −2(5)(−1) = 10.

The equation x² + y² − 6x + 8y = k defines a circle C₁ for k > −25. A second circle C₂ is centered at the origin and internally tangent to C₁ when k = 75. What is the radius of C₂?

Answer: 5

Complete the square: (x − 3)² + (y + 4)² = k + 25. When k = 75, C₁ has center (3, −4) and radius √100 = 10. The distance from the origin to (3, −4) is √(9 + 16) = 5. Since the origin lies inside C₁, circle C₂ is inside C₁. Internal tangency requires r₁ − r₂ = d, so 10 − r₂ = 5, giving r₂ = 5.