Digital SAT Math Algebra Practice Flashcards

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

Answer: 3

Write the function in vertex form: f(x) = a(x − 3)² − 7. Substitute the point (1, 5): 5 = a(1 − 3)² − 7 → 5 = 4a − 7 → 4a = 12 → a = 3.

If (x − p)(x − q) = x² − 6x + k and q = p + 2, what is the value of k?

Answer: 8

Expanding gives p + q = 6 (sum of roots) and pq = k (product of roots). Substituting q = p + 2: p + (p + 2) = 6 → 2p = 4 → p = 2 and q = 4. Therefore k = pq = 2 × 4 = 8.

The system 4x − 6y = 10 and mx − 9y = n has infinitely many solutions. What is the value of m + n?

Answer: 21

For infinitely many solutions the equations must be proportional. The y-coefficient ratio is −9/−6 = 3/2. Applying the same ratio: m = 4 × (3/2) = 6 and n = 10 × (3/2) = 15. So m + n = 6 + 15 = 21.

If f(x) = 2x + 1 and g(f(x)) = 4x² + 4x − 4, which of the following defines g(x)?

Answer: x² − 5

Let u = f(x) = 2x + 1, so x = (u − 1)/2. Substitute: 4((u−1)/2)² + 4((u−1)/2) − 4 = (u−1)² + 2(u−1) − 4 = u² − 2u + 1 + 2u − 2 − 4 = u² − 5. Therefore g(x) = x² − 5.

The equation |2x − 7| = x + 1 has two solutions. What is the product of those two solutions?

Answer: 16

Case 1: 2x − 7 = x + 1 → x = 8. Check: |16 − 7| = 9 = 8 + 1 ✓. Case 2: −(2x − 7) = x + 1 → −2x + 7 = x + 1 → 3x = 6 → x = 2. Check: |4 − 7| = 3 = 2 + 1 ✓. Both solutions are valid. Product = 8 × 2 = 16.

If r and s are the two roots of 3x² − 12x + 7 = 0, what is the value of (r − s)²?

Answer: 20/3

By Vieta's formulas: r + s = 12/3 = 4 and rs = 7/3. Apply the identity (r − s)² = (r + s)² − 4rs = 16 − 4(7/3) = 16 − 28/3 = 48/3 − 28/3 = 20/3.