SAT Quadratic Equations and Nonlinear Systems

Quadratic equations and linear-quadratic systems fall under the Advanced Math domain of the digital SAT Math section, tested through the skill Nonlinear equations in one variable and systems of equations in two variables. The domain measures your understanding of absolute value, quadratic, exponential, polynomial, rational, radical, and other nonlinear equations. It also measures your ability to connect different representations of a nonlinear relationship between two variables.

This guide walks through factoring, completing the square, and the quadratic formula. It then covers vertex form and graph transformations, discriminant reasoning, and systems pairing a line with a parabola, solved both algebraically and on a graph. Each worked example below is original, built to show every step, not copied from a released question.

Where Quadratics and Nonlinear Systems Fit in Advanced Math

Advanced Math carries the same question share as Algebra on the digital SAT Math section: about 35% of Math questions, 13 to 15 questions across both modules combined. The domain groups three official skills: equivalent expressions, nonlinear functions, and nonlinear equations in one variable and systems of equations in two variables. The third skill covers quadratics and linear-quadratic systems, the subject of this guide.

The digital SAT Math section runs across two modules. For the full layout of question counts and timing, see our SAT Math section guide. The other two Advanced Math skills, equivalent expressions and nonlinear functions, appear in our SAT Polynomials and Equivalent Expressions guide.

College Board does not publish a percentage split among the three skills inside Advanced Math. Quadratics and nonlinear systems therefore share the domain's 13 to 15 question count with equivalent expressions and nonlinear functions questions, per the math specifications documentation. The table below sets out the four Math domains and their share of the test.

Advanced Math splits into three official skills, and this guide focuses on one of them. Knowing which skill a question belongs to helps you pick the right method fast. The table below maps each skill to its coverage here.

Factoring, Completing the Square, and the Quadratic Formula

Three methods solve a quadratic equation in standard form, ax² + bx + c = 0. Pick the method fitting the numbers in front of you. Factoring works fastest when the numbers are clean, completing the square exposes the vertex, and the quadratic formula solves any quadratic without exception.

Factoring

Factoring rewrites the quadratic as a product of two binomials, then sets each factor to zero. Look for two numbers multiplying to a times c and adding to b. Special patterns also cut the work, such as a difference of squares, a² - b² = (a - b)(a + b).

Here is an original example. Solve x² - 5x + 6 = 0. Two numbers multiply to 6 and add to -5: -2 and -3. Factor the equation: (x - 2)(x - 3) = 0. Setting each factor to zero gives x = 2 or x = 3.

Here is a second original example, using the difference-of-squares pattern. Solve x² - 16 = 0. Rewrite it as (x - 4)(x + 4) = 0. The two solutions are x = 4 and x = -4.

Completing the Square

Completing the square turns the left side of the equation into a perfect square trinomial, then solves by taking a square root. The steps below start from standard form, ax² + bx + c = 0. Run them in order, and note the first step matters whenever a is not 1.

1. Divide every term by a when a is not 1, so the x² term has a coefficient of 1.
2. Move the constant term to the right side of the equation.
3. Take half of the b coefficient, square it, and add the result to both sides.
4. Factor the left side into a perfect square binomial.
5. Take the square root of both sides, keeping the plus-or-minus sign.
6. Solve the two resulting linear equations for x.

Here is an original example. Solve x² + 6x - 7 = 0. The coefficient of x² is already 1, so move the constant: x² + 6x = 7. Half of 6 is 3, and 3 squared is 9, so add 9 to both sides: x² + 6x + 9 = 16. The left side factors to (x + 3)² = 16, and taking the square root gives x + 3 = ±4, so x = 1 or x = -7.

Here is a second original example, with a leading coefficient other than 1. Solve 2x² + 8x - 10 = 0. Divide every term by 2 first: x² + 4x - 5 = 0. Move the constant and add 4 to both sides: x² + 4x + 4 = 9, so (x + 2)² = 9. Taking the square root gives x + 2 = ±3, so x = 1 or x = -5.

The Quadratic Formula

Every quadratic in standard form solves with the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a). Use it whenever factoring fails to produce clean integers. It handles any leading coefficient with no dividing step first.

Here is an original example. Solve 2x² + 3x - 5 = 0. Here a = 2, b = 3, and c = -5. The value under the square root is 3² - 4(2)(-5) = 9 + 40 = 49, and √49 = 7. So x = (-3 ± 7) / 4, giving x = 1 or x = -2.5.

Vertex Form and Graph Transformations

A quadratic equation graphs as a parabola, a U-shaped curve with one highest or lowest point, the vertex. The curve is symmetric about a vertical line through the vertex, the axis of symmetry. Reading the vertex straight off the equation saves time on any question asking for a maximum, a minimum, or the axis of symmetry.

Vertex Form and the Vertex

Vertex form writes a quadratic as y = a(x - h)² + k, with the vertex sitting at the point (h, k). A positive value of a opens the parabola upward, and the vertex marks the minimum value. A negative value of a opens the parabola downward, and the vertex marks the maximum value.

Here is an original example. The equation y = 2(x - 3)² + 4 has a vertex at (3, 4). Since a = 2 is positive, the parabola opens upward. The minimum value of y is 4.

Converting from standard form to vertex form uses the completing-the-square steps above. Here is an original example. Rewrite y = x² - 4x + 1 in vertex form. Completing the square on x² - 4x gives (x - 2)² - 4, so y = (x - 2)² - 4 + 1 = (x - 2)² - 3. The vertex sits at (2, -3).

Graph Transformations

The values of a, h, and k in vertex form move and reshape the parent graph y = x² without changing its basic U shape. Each value acts on its own, so a question changing one of them changes one feature of the graph. The table below lists each change and its effect.

The Discriminant and Number of Solutions

The discriminant, b² - 4ac, sits under the square root in the quadratic formula. Its sign tells you how many real solutions a quadratic equation has before you solve it. Questions asking for a solution count, or for the value of a constant making the count match a condition, are answered with the discriminant alone. The table below pairs each sign with its solution count.

Zero Discriminant: A Repeated Root

Here is an original example. For x² + 4x + 4 = 0, the discriminant is 4² - 4(1)(4) = 16 - 16 = 0, so the equation has one repeated solution. Factoring confirms it: (x + 2)² = 0, giving x = -2. On a graph, this parabola touches the x-axis at exactly one point.

Negative Discriminant: No Real Solution

Here is an original example. For x² + x + 1 = 0, the discriminant is 1² - 4(1)(1) = 1 - 4 = -3, a negative value, so the equation has no real solution. On a graph, this parabola never crosses the x-axis. A question asking only for the number of real solutions is finished at this point, with no further work.

Solving Linear-Quadratic Systems

A linear-quadratic system pairs a straight line with a parabola, and its solutions are the points where the two graphs cross. Such a system has two solutions, one solution, or no solution. Substitution turns the pair into a single quadratic equation, and the sign of the discriminant of the result gives the count.

Solving Algebraically by Substitution

Substitution reduces the system to one quadratic equation in one variable. Run the steps below in order, and finish by pairing each x-value with its y-value. Stopping before the last step is the most common way to lose the point.

1. Solve the linear equation for one variable, usually y.
2. Substitute the result into the quadratic equation, so one variable remains.
3. Rearrange the equation into standard form, ax² + bx + c = 0.
4. Solve the quadratic by factoring, completing the square, or the quadratic formula.
5. Back-substitute each solution into the linear equation to find the paired coordinate.
6. Write each solution as an ordered pair.

Here is an original example. Solve the system y = x + 1 and y = x² - 5. Set the two expressions for y equal: x + 1 = x² - 5. Rearranged, this reads x² - x - 6 = 0, which factors to (x - 3)(x + 2) = 0, giving x = 3 or x = -2. Substituting back into y = x + 1 gives y = 4 when x = 3, and y = -1 when x = -2, so the system has two solutions: (3, 4) and (-2, -1).

Reading Solutions From a Graph

A line crossing a parabola twice gives two solutions, touching it at one point gives one solution, and missing it entirely gives no solution. The discriminant of the quadratic equation formed after substitution predicts the count before you graph anything. Graphing both equations is a fast check on your algebra when the numbers stay small.

Here is an original example, continuing the system above. After substitution, the system reduces to x² - x - 6 = 0. Its discriminant is (-1)² - 4(1)(-6) = 1 + 24 = 25, a positive value, confirming two solutions. The line and the parabola cross at two points on a graph, matching the count.

Practice Quadratics and Nonlinear Systems With Us

We built free drill sets for quadratic functions and equations, and for nonlinear systems, at three difficulty levels. Every set gives instant feedback with a full explanation for each question. No signup is required.

• Quadratic functions and equations: easy set, medium set, hard set.
• Nonlinear systems: easy set, medium set, hard set.

Start at the easy set, then move up once you solve every question without checking your steps. For full-length timed practice covering Advanced Math alongside the other three Math domains, visit our SAT practice page. Working the drills and a full test together turns factoring, the quadratic formula, and system-solving into steps you run without thinking twice.

Sources

• College Board: The Math Section Overview, SAT Suite
• College Board: Advanced Math, SAT Suite
• College Board: Math Specifications, SAT Suite