SAT: Nonlinear Systems

Replace one variable using the linear equation and substitute into the nonlinear equation to produce a single equation in one variable, typically a quadratic.

For $ax^2 + bx + c = 0$, the discriminant $D = b^2 - 4ac$ determines the number of real solutions: $D > 0$ gives two, $D = 0$ gives one, $D < 0$ gives none.

Setting a linear equation equal to a quadratic produces a quadratic equation whose solutions are the $x$-coordinates of intersection points. A tangent line gives $D = 0$.

For $ax^2 + bx + c = 0$ with roots $r_1$ and $r_2$: sum $= -\frac{b}{a}$ and product $= \frac{c}{a}$. These allow answering SAT questions without fully solving.

Substituting a linear equation into $x^2 + y^2 = r^2$ yields a quadratic in one variable. The discriminant determines 0, 1, or 2 intersection points.

A line is tangent to a parabola when the resulting quadratic has exactly one solution, meaning $D = 0$. Solving $D = 0$ for the parameter finds the tangency condition.

The SAT often asks 'how many solutions' a system has. After substitution, use the discriminant: $D > 0$ yields 2, $D = 0$ yields 1, $D < 0$ yields 0.

A parabola $y = (x - h)^2 + k$ has its minimum (or maximum) at $y = k$. A horizontal line $y = c$ intersects at one point when $c = k$ and at two points when $c > k$ (for upward-opening).

Setting $y = \frac{1}{x}$ equal to a linear equation $y = mx + b$ produces $\frac{1}{x} = mx + b$, which simplifies to $mx^2 + bx - 1 = 0$, a quadratic in $x$.

The graph of $y = |x|$ is a V-shape. A horizontal line $y = c$ ($c > 0$) intersects it at two points ($x = \pm c$); $y = 0$ touches at one point; $y < 0$ gives no intersection.