Nonlinear Equations and Systems (incl. quadratics)

Nonlinear equations are equations where the variable is squared, under a root, in a denominator, or otherwise not just multiplied by a number. On the SAT this mostly means quadratic equations (with an $x^2$ term), radical equations (with $\sqrt{\ }$), and nonlinear systems, where a line meets a curve. You will be asked for solutions, for the number of solutions, or for a constant that makes something true.

• A quadratic equation can have zero, one, or two real solutions. Graphically, these are the places where the parabola crosses the $x$-axis.
• If a product equals zero, at least one factor equals zero. This is why factoring works: $(x - a)(x - b) = 0$ means $x = a$ or $x = b$.
• The discriminant $b^2 - 4ac$ tells you how many real solutions there are without solving.
• Squaring both sides of an equation can create extraneous solutions: values that solve the squared equation but not the original. Always check them.
• Solving a system of a line and a parabola means setting the two expressions equal. The number of intersection points equals the number of solutions.

For $ax^2 + bx + c = 0$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• Vertex of $y = ax^2 + bx + c$: $x = -\frac{b}{2a}$; substitute back for the $y$-value.
• Sum of solutions $= -\frac{b}{a}$; product of solutions $= \frac{c}{a}$.
• Perfect squares: $(x + k)^2 = x^2 + 2kx + k^2$.

Solve $x^2 - x - 12 = 0$.

1. Look for two numbers that multiply to $-12$ and add to $-1$: these are $-4$ and $3$.
2. Factor: $(x - 4)(x + 3) = 0$.
3. Set each factor to zero: $x = 4$ or $x = -3$.

Check one: $16 - 4 - 12 = 0$. Correct.

For what positive value of $k$ does $x^2 - kx + 16 = 0$ have exactly one real solution?

1. Exactly one solution means the discriminant is zero: $(-k)^2 - 4(1)(16) = 0$.
2. So $k^2 = 64$ and $k = \pm 8$.
3. The question asks for the positive value: $k = 8$.

Check: $x^2 - 8x + 16 = (x - 4)^2$, which is zero only at $x = 4$. One solution, as required.

• Sign errors when reading solutions from factors. $(x + 5)$ gives $x = -5$, not $5$.
• Forgetting the negative root. $x^2 = 25$ has two solutions, $5$ and $-5$, unless the question restricts $x$.
• Not checking radical equations. After squaring $\sqrt{x + 7} = x - 5$, you get $x = 2$ and $x = 9$, but only $9$ works.
• Answering the wrong quantity. If the question asks for $y$, or for $k$, or for the positive solution, re-read before you bubble.
• Mixing up $b$ in the discriminant. In $2x^2 - 4x + 5$, $b = -4$, so $b^2 = 16$ (positive), and the discriminant is $16 - 40$, not $16 + 40$.