How to solve SAT nonlinear equations and systems

This skill covers equations that are not linear, such as quadratics, radical equations, absolute value equations, and equations with the variable in a denominator. It also covers systems where at least one equation is nonlinear, usually a line and a parabola. On the Digital SAT you solve these equations, count their solutions, and find where graphs intersect. Advanced Math is about 35% of the Math section, or 13 to 15 of the 44 questions, and has only three skills, so expect several questions from this one.

What the questions look like

Solve a quadratic. You find the solutions of an equation like $x^2 - 11x + 18 = 0$, or a related value such as the positive solution, the sum of the solutions, or the product.

Count solutions. You decide how many real solutions an equation has, or find the constant that makes a system have exactly one solution. These usually come down to the discriminant.

Radical, rational, and absolute value equations. You solve something like $\sqrt{x + 7} = x - 5$ or $|2x - 3| = 9$. Squaring both sides or clearing a denominator can introduce answers that don't actually work, so checking matters here.

Systems with a curve. A line and a parabola, or a line and a circle, are given. You find an intersection point or the number of intersection points.

Step-by-step method

1. Set the equation equal to zero. Move every term to one side so you have $ax^2 + bx + c = 0$.
2. Factor if it is quick. If not, use the quadratic formula:
   $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
3. Use the discriminant to count solutions. For $ax^2 + bx + c = 0$:
   • $b^2 - 4ac > 0$ means two real solutions.
   • $b^2 - 4ac = 0$ means exactly one real solution.
   • $b^2 - 4ac < 0$ means no real solutions.
4. Use sum and product shortcuts. The solutions add to $-\frac{b}{a}$ and multiply to $\frac{c}{a}$. These save time when a question asks for the sum or product directly.
5. Check for extraneous solutions. After squaring both sides or multiplying by an expression containing $x$, substitute every candidate back into the original equation.
6. For systems, substitute. Replace $y$ in the nonlinear equation with the linear expression. You get a single equation in $x$, usually a quadratic, and the method above applies.

Worked example

Example 1. What is the solution set of $\sqrt{x + 7} = x - 5$?

Square both sides:

$$x + 7 = (x - 5)^2 = x^2 - 10x + 25$$

Set it equal to zero and factor:

$$x^2 - 11x + 18 = 0$$
$$(x - 9)(x - 2) = 0$$

The candidates are $x = 9$ and $x = 2$. Check both in the original equation.

• $x = 9$: the left side is $\sqrt{16} = 4$ and the right side is $9 - 5 = 4$. It works.
• $x = 2$: the left side is $\sqrt{9} = 3$ and the right side is $2 - 5 = -3$. It fails, because a square root is never negative.

The solution set is $\{9\}$. A choice listing both 9 and 2 is the trap.

Example 2. In the system below, $c$ is a constant.

$$y = x^2 - 6x + 11$$
$$y = 2x + c$$

For what value of $c$ does the system have exactly one real solution?

Set the expressions for $y$ equal:

$$x^2 - 6x + 11 = 2x + c$$
$$x^2 - 8x + (11 - c) = 0$$

Exactly one solution means the discriminant is zero. Here $a = 1$, $b = -8$, and the constant term is $11 - c$:

$$(-8)^2 - 4(1)(11 - c) = 0$$
$$64 - 44 + 4c = 0$$
$$c = -5$$

Check: with $c = -5$, the equation becomes $x^2 - 8x + 16 = 0$, or $(x - 4)^2 = 0$, so $x = 4$. The line gives $y = 2(4) - 5 = 3$, and the parabola gives $16 - 24 + 11 = 3$. The line touches the parabola at the single point $(4, 3)$.

Common traps

• Dividing by the variable. In $x^2 = 5x$, dividing by $x$ gives only $x = 5$ and loses $x = 0$. Instead, write $x^2 - 5x = 0$ and factor: $x(x - 5) = 0$.
• Keeping extraneous solutions. Squaring both sides can create answers that fail the original equation. Always plug candidates back in after squaring or clearing denominators.
• Using the wrong $b$ or $c$ in the discriminant. Move everything to one side first. In Example 2, the constant is $11 - c$, not 11, and $b$ is $-8$, not $-6$.
• Getting the sign of the sum wrong. The sum of the solutions is $-\frac{b}{a}$. For $2x^2 - 10x + 3 = 0$, the sum is $5$, not $-5$.
• Solving only one case of an absolute value. $|2x - 3| = 9$ means $2x - 3 = 9$ or $2x - 3 = -9$, so $x = 6$ or $x = -3$. An absolute value set equal to a negative number has no solution.

Desmos tip

Graph the left side and the right side as two separate functions, such as $y = \sqrt{x + 7}$ and $y = x - 5$. Their intersection points are the real solutions. Desmos never shows extraneous roots, because it graphs the original equation, so in Example 1 you see only one intersection at $x = 9$. For "exactly one solution" questions, type the line with its constant, use the slider, and watch for the moment the line just touches the curve.