Nonlinear Equations and Systems Flashcards

The line y = 3x + k is tangent to the parabola y = x² − 2x + 5. What is the value of k?

Answer: -5/4

Tangency means exactly one intersection, so the discriminant of the resulting quadratic must equal zero. Setting the equations equal: x² − 2x + 5 = 3x + k simplifies to x² − 5x + (5 − k) = 0. Applying the discriminant condition b² − 4ac = 0: (−5)² − 4(1)(5 − k) = 0 → 25 − 20 + 4k = 0 → 4k = −5 → k = −5/4. The other choices result from common sign errors or failing to distribute the −4 correctly.

How many ordered pairs (x, y) satisfy both y = x² − 4 and y = |x| + 2?

Answer: 2

Split into cases based on the absolute value. For x ≥ 0: x² − 4 = x + 2 → x² − x − 6 = 0 → (x − 3)(x + 2) = 0, giving x = 3 (valid) and x = −2 (rejected, outside this case). For x < 0: x² − 4 = −x + 2 → x² + x − 6 = 0 → (x + 3)(x − 2) = 0, giving x = −3 (valid) and x = 2 (rejected, outside this case). This yields exactly two ordered pairs: (3, 5) and (−3, 5). Choosing 4 is the most common error — students who keep both roots from each case without checking domain validity.

The system of equations x² + y² = 50 and y = x + 4 has two solutions. What is the product of the x-coordinates of these two solutions?

Answer: -17

Substituting y = x + 4 into the circle equation: x² + (x + 4)² = 50 → x² + x² + 8x + 16 = 50 → 2x² + 8x − 34 = 0 → x² + 4x − 17 = 0. Rather than solving this quadratic (which yields irrational roots), apply Vieta's formulas: for x² + 4x − 17 = 0, the product of the roots equals c/a = −17/1 = −17. Choosing −4 is the mistake of computing the sum of roots instead of the product.

For all real values of k, the equation x² − (k + 3)x + 3k = 0 always has exactly one root whose value does not depend on k. What is the other root, expressed in terms of k?

Answer: k

Regroup by factoring out terms with k: x² − kx − 3x + 3k = 0 → x(x − 3) − k(x − 3) = 0 → (x − k)(x − 3) = 0. The two roots are x = 3 (constant, independent of k) and x = k (varies with k). The root that is always fixed is 3, making k the other root. Students who apply Vieta's sum-of-roots formula (sum = k + 3) without recognizing the factored form often select k + 3, confusing the sum with an individual root.

In which of the following systems do the graphs of the two equations share no points in common?

Answer: y = x² + 4x + 5 and y = 2x + 1

Set each pair equal and evaluate the discriminant of the resulting quadratic. (A): x² + 4x + 5 = 2x + 1 → x² + 2x + 4 = 0; discriminant = 4 − 16 = −12 < 0 → no real intersections. (B): x² − 4x + 5 = 2x − 4 → x² − 6x + 9 = 0 = (x−3)²; discriminant = 0 → one (tangent) intersection. (C): x² + 2x + 2 = x + 4 → x² + x − 2 = 0 → (x+2)(x−1)=0; two intersections. (D): x² − 2x + 1 = −x + 3 → x² − x − 2 = 0 → (x−2)(x+1)=0; two intersections. Only system (A) has a negative discriminant.

What is the sum of all real solutions to x⁴ − 10x² + 9 = 0?

Answer: 0

Substitute u = x² to convert to a quadratic: u² − 10u + 9 = 0 → (u − 1)(u − 9) = 0, giving u = 1 or u = 9. Substituting back: x² = 1 → x = ±1, and x² = 9 → x = ±3. All four values are real. Their sum is (1) + (−1) + (3) + (−3) = 0. A common trap is choosing 4 (adding only the positive roots 1 + 3) or 10 (adding the u-values 1 + 9 rather than the x-values).