SAT Topic 7

Nonlinear systems

Solve when one equation is a quadratic (or other curve) and the other a line.

Concept

The standard move: substitution. Plug one equation into the other, simplify to a single-variable equation (usually a quadratic), and solve.

Line + parabola — at most 2 intersection points (the resulting quadratic has at most 2 real roots).
Tangent — exactly 1 intersection ⇔ the resulting quadratic has discriminant 0.
No intersection — discriminant < 0 (the line and parabola never meet).

The number of solutions to a nonlinear system is the geometric question: how many points of intersection do the two graphs share?

Solution

Set equal. x² − 1 = x + 1

Rearrange. x² − x − 2 = 0

Factor. (x − 2)(x + 1) = 0 ⇒ x = 2 or x = −1.

Find y. At x = 2: y = 3. At x = −1: y = 0.

Two solutions: (2, 3) and (−1, 0).

Solution

Set equal. x² = kx − 1 ⇒ x² − kx + 1 = 0

Tangent. Exactly one solution ⇒ discriminant = 0: k² − 4 = 0

Solve. k = ±2.

Two tangent lines: y = 2x − 1 and y = −2x − 1.

8 questions on intersections of lines and parabolas, tangent conditions, and counting solutions.

Practice test · 8 questions Question 1 of 8 · Score 0