Nonlinear equations and systems of equations

Nonlinear systems on the SAT typically combine a linear equation ($y = mx + b$) and a quadratic equation ($y = ax^2 + bx + c$), or involve circles and higher-degree polynomials. Questions usually ask for the number of solutions, specific points of intersection $(x, y)$, or the value of a constant that yields a certain number of solutions.

Key Solving Strategies

• Substitution Method: The most reliable method for nonlinear systems. Isolate one variable in the linear equation (usually $y$) and substitute that expression into the nonlinear equation to solve for $x$.

• Elimination Method: Useful when both equations share identical squared terms (e.g., $x^2 + y^2 = 25$ and $x^2 - y = 5$).

• Discriminant Analysis: When a linear-quadratic system simplifies to a single quadratic equation $Ax^2 + Bx + C = 0$, the discriminant ($B^2 - 4AC$) determines the number of real intersection points:

  • $B^2 - 4AC > 0$: Two distinct real solutions (line intersects the curve twice).

  • $B^2 - 4AC = 0$: Exactly one real solution (line is tangent to the curve).

  • $B^2 - 4AC < 0$: No real solutions (line does not touch the curve).

Systems Involving Radical & Rational Equations

• Extranous Solutions: When squaring both sides of a radical equation, always test your candidate solutions back in the original equation to eliminate false solutions.

• Rational Equations: Multiply every term by the least common denominator (LCD) to clear fractions, then check that solutions do not make any denominator zero.

1. Use the Discriminant to Find Constants for Solution Counts. When an SAT question asks for a value of $k$ that makes a system have exactly one solution, set the two equations equal, rearrange to $Ax^2 + Bx + C = 0$, and solve $B^2 - 4AC = 0$. 

   Example: The system of equations below has exactly one real solution. What is the value of $c$?

   $$y = x^2 + 4x + c$$
   $$y = 2x + 1$$

   • Set the equations equal to each other:

     $$x^2 + 4x + c = 2x + 1$$

   • Rearrange into standard quadratic form ($Ax^2 + Bx + C = 0$):

     $$x^2 + 2x + (c - 1) = 0$$

   • Identify coefficients: $A = 1$, $B = 2$, $C = c - 1$. Set the discriminant to zero:

     $$B^2 - 4AC = (2)^2 - 4(1)(c - 1) = 0$$
     $$4 - 4c + 4 = 0 \implies 8 - 4c = 0 \implies c = 2$$

2. Watch Out for Extraneous Solutions in Radical Equations. Squaring both sides can introduce valid algebraic values that fail in the original radical context. 

   Example: Solve for $x$: $\sqrt{2x + 6} = x - 1$.

   • Square both sides:

     $$2x + 6 = (x - 1)^2 \implies 2x + 6 = x^2 - 2x + 1$$

   • Rearrange into standard form:

     $$x^2 - 4x - 5 = 0 \implies (x - 5)(x + 1) = 0$$

   • Candidate solutions: $x = 5$ or $x = -1$.

   • Check candidates in the original equation:

     • Test $x = 5$: $\sqrt{2(5) + 6} = 5 - 1 \implies \sqrt{16} = 4$ (True).

     • Test $x = -1$: $\sqrt{2(-1) + 6} = -1 - 1 \implies \sqrt{4} = -2$ (False, principal square root is positive).

   • The only valid solution is $x = 5$.

3. Solve Systems by Substituting $y$ Directly. When asked for the sum or product of $x$-values for a system, set $y = y$ immediately to create a single quadratic. 

   Example: How many real solutions $(x, y)$ does the system have?

   $$y = 3x^2 - x + 5$$
   $$y = 2x + 1$$

   • Set $y$ expressions equal:

     $$3x^2 - x + 5 = 2x + 1 \implies 3x^2 - 3x + 4 = 0$$

   • Evaluate the discriminant ($B^2 - 4AC$):

     $$(-3)^2 - 4(3)(4) = 9 - 48 = -39$$

   • Since $-39 < 0$, the system has 0 real solutions.