Nonlinear Equations and Systems Flashcards

The system of equations y = x² − 4x + k and y = 2x − 1 has exactly one solution. What is the value of k?

Answer: 8

Substituting the linear equation into the quadratic: x² − 4x + k = 2x − 1, which simplifies to x² − 6x + (k + 1) = 0. For exactly one solution, the discriminant must equal zero: (−6)² − 4(1)(k + 1) = 0 → 36 − 4k − 4 = 0 → 4k = 32 → k = 8.

The system x² + y² = 25 and y = x + 1 has two solutions, (x₁, y₁) and (x₂, y₂). What is the value of x₁ + x₂ + y₁ + y₂?

Answer: 0

Substituting y = x + 1 into x² + y² = 25 gives x² + (x+1)² = 25 → 2x² + 2x − 24 = 0 → x² + x − 12 = 0. By Vieta's formulas, x₁ + x₂ = −1. Since y = x + 1, we have y₁ + y₂ = (x₁ + 1) + (x₂ + 1) = −1 + 2 = 1. Therefore x₁ + x₂ + y₁ + y₂ = −1 + 1 = 0.

How many real solutions does √(3x + 4) = x − 2 have?

Answer: 1

Squaring both sides: 3x + 4 = x² − 4x + 4 → x² − 7x = 0 → x(x − 7) = 0, giving x = 0 or x = 7. Checking x = 0: √4 = 2, but x − 2 = −2. Since the right side must be non-negative for a square root equation, x = 0 is extraneous. Checking x = 7: √25 = 5 and 7 − 2 = 5 ✓. Only one real solution exists.

The equation 2x² − kx + 8 = 0 has two real solutions, both of which are positive. Which of the following must be true about k?

Answer: k ≥ 8

By Vieta's formulas, the sum of the roots = k/2 and the product = 4. For both roots to be positive, their sum must be positive: k/2 > 0 → k > 0. For two real solutions, the discriminant must be non-negative: k² − 64 ≥ 0 → k ≥ 8 or k ≤ −8. Combining with k > 0, we get k ≥ 8. The product being 4 > 0 is automatically satisfied.

The two roots of x² + px + q = 0 satisfy r + s = 5 and r² + s² = 13. What is the value of q?

Answer: 6

By Vieta's formulas, r + s = −p = 5 and rs = q. Using the identity r² + s² = (r + s)² − 2rs: 13 = 25 − 2q → 2q = 12 → q = 6. Note that p = −5, and the equation is x² − 5x + 6 = 0, which factors as (x − 2)(x − 3) = 0, confirming r = 2, s = 3.

How many real solutions does the system x² + y² = 4 and y = x² − 2 have?

Answer: 3

Since y = x² − 2, we have x² = y + 2. Substituting into the circle equation: (y + 2) + y² = 4 → y² + y − 2 = 0 → (y + 2)(y − 1) = 0, so y = −2 or y = 1. For y = −2: x² = 0 → x = 0, giving one point (0, −2). For y = 1: x² = 3 → x = ±√3, giving two points (√3, 1) and (−√3, 1). The total is 3 real solutions — a result that surprises students expecting a symmetric even count.