Digital SAT Hard Math Flashcards

How many distinct points of intersection exist between the circle x² + y² = 25 and the parabola y = x² − 5?

Answer: 3

Substitute y = x² − 5 into x² + y² = 25: x² + (x² − 5)² = 25 → x² + x⁴ − 10x² + 25 = 25 → x⁴ − 9x² = 0 → x²(x² − 9) = 0. So x = 0, x = 3, or x = −3, giving three points: (0, −5), (3, 4), and (−3, 4). All three satisfy both equations, so there are exactly 3 intersection points.

If f(x) = (3x − 1)/(x + 2), what is f⁻¹(5)?

Answer: −11/2

To find f⁻¹(5), solve f(x) = 5: (3x − 1)/(x + 2) = 5 → 3x − 1 = 5(x + 2) → 3x − 1 = 5x + 10 → −11 = 2x → x = −11/2. You can verify: f(−11/2) = (3(−11/2) − 1)/((−11/2) + 2) = (−33/2 − 2/2)/(−11/2 + 4/2) = (−35/2)/(−7/2) = 5 ✓.

A jar contains 5 red, 3 blue, and 2 green marbles. Two marbles are drawn at random without replacement. Given that at least one marble is red, what is the probability that both marbles are red?

Answer: 2/7

Use conditional probability: P(both red | at least one red) = P(both red) / P(at least one red). P(both red) = C(5,2)/C(10,2) = 10/45 = 2/9. P(no red) = C(5,2)/C(10,2) = 10/45 = 2/9, so P(at least one red) = 1 − 2/9 = 7/9. Therefore P(both red | at least one red) = (2/9) ÷ (7/9) = 2/7.

The cubic polynomial p(x) = x³ + ax² + bx + c has roots at x = 1, x = −2, and x = 3. What is the value of a + b + c?

Answer: −1

By Vieta's formulas: a = −(1 + (−2) + 3) = −2; b = (1)(−2) + (1)(3) + (−2)(3) = −2 + 3 − 6 = −5; c = −(1)(−2)(3) = 6. So a + b + c = −2 − 5 + 6 = −1. Elegant shortcut: since x = 1 is a root, p(1) = 1 + a + b + c = 0, so a + b + c = −1 directly.

For x > 0, the equation log₃(x + 4) + log₃(x) = 2 has one valid solution. What is it?

Answer: −2 + √13

Combine the logarithms: log₃(x(x + 4)) = 2 → x(x + 4) = 3² = 9 → x² + 4x − 9 = 0. By the quadratic formula: x = (−4 ± √(16 + 36))/2 = −2 ± √13. Since x > 0 is required (logarithm domain), we discard x = −2 − √13 (negative). The valid solution is x = −2 + √13 ≈ 1.61.

In the xy-plane, the circle defined by x² + y² − 8x + 6y − 11 = 0 is tangent to the line y = x + k for exactly two values of k. What is the sum of those two values of k?

Answer: −14

Complete the square: (x − 4)² + (y + 3)² = 11 + 16 + 9 = 36, so center (4, −3) and radius 6. Rewrite y = x + k as x − y + k = 0. The tangency condition requires the distance from the center to the line to equal the radius: |4 − (−3) + k|/√2 = 6 → |7 + k| = 6√2 → k = −7 + 6√2 or k = −7 − 6√2. The sum is (−7 + 6√2) + (−7 − 6√2) = −14.