Advanced Math · SAT Math

Nonlinear Systems on the Digital SAT

Nonlinear systems on the Digital SAT pair a quadratic with a linear equation and ask where they intersect. The trick is recognizing disguised quadratics (equations with fractions, roots, or higher powers that simplify to quadratics) and using substitution to solve. This lesson assumes you've worked through quadratics already. 6 practice problems with full solutions.
By the Prepiii Editorial TeamUpdated 2026-05-25~10 min read

What counts as a nonlinear equation

A nonlinear equation is any equation where the variable appears:

  • with a power higher than 1 (e.g., x², x³)
  • inside a root (√x, ∛x)
  • in a denominator (e.g., 1/x)
  • inside an absolute-value sign or exponent

On the Digital SAT, by far the most common nonlinear type is the quadratic — directly or in disguise. Master quadratics (see the quadratics lesson) and you'll handle 90% of nonlinear questions.

Disguised quadratics: clear the disguise first

The SAT loves to give equations that look nonlinear but simplify to quadratics after one or two algebraic moves. The recipe: get rid of the disguise, then it's just a quadratic.

Fractions: multiply both sides by the denominator. 1/(x+5)² = 4 → 1 = 4(x+5)² → quadratic.

Square roots: square both sides. √(3x+34) = (x-2) → 3x + 34 = (x-2)² → quadratic.

Higher powers: if you have x⁴ terms, substitute u = x² to reduce to a quadratic in u.

Always bring everything to one side (=0) before solving. If you don't, the quadratic formula and factoring won't apply.

Line-meets-parabola systems

The most common nonlinear-system pattern: one quadratic equation and one linear equation. The solution(s) are the point(s) where the parabola and the line intersect.

The method: substitution. The linear equation is usually easy to solve for one variable. Plug that expression into the quadratic and solve the resulting single-variable quadratic.

Example. Solve y = x² and y = 2x + 3.

Set the two expressions for y equal: x² = 2x + 3.

Move everything to one side: x² - 2x - 3 = 0.

Factor: (x - 3)(x + 1) = 0 → x = 3 or x = -1. Two solutions = two intersection points.

How many intersections? Use the discriminant

On "how many solutions does this system have?" questions, you don't need to actually solve. After substituting and bringing everything to one side, you have a quadratic ax² + bx + c = 0. The discriminant b² - 4ac tells you the answer:

b² - 4ac > 0 → 2 intersections (line crosses parabola in two places)

b² - 4ac = 0 → 1 intersection (line is tangent to the parabola)

b² - 4ac < 0 → 0 intersections (line and parabola don't meet)

The SAT often gives a system with an unknown k and asks for the value of k that makes the system have exactly one solution. That's a discriminant-equals-zero question.

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Solving with Desmos

Desmos turns nonlinear systems into one-step problems. Three techniques:

1. Graph both equations, click the intersection

Type both equations as written. Desmos shows the curves and marks intersection points automatically. Click each one for exact (x, y) coordinates.

2. Find k that gives exactly one intersection — use a slider

For "find k such that y = x² and y = kx + 1 have exactly one solution": graph both with a slider for k, drag until the line just touches the parabola. Read k.

3. Solve disguised quadratics directly

For "solve 1/(x+5)² = 4", just type the equation. Desmos returns the solutions without manual clearing of the fraction.

For the full set of Desmos techniques across the entire test, see our Desmos & Test Tools guides.

Common mistakes

Forgetting to move everything to one side before factoring

You can't factor x² + 5x = 6 directly. First rewrite as x² + 5x - 6 = 0, THEN factor. This applies to all disguised quadratics: clear the disguise, move everything to one side, then solve.

Forgetting the ± when taking square roots

If x² = 16, x can be 4 OR -4. On nonlinear systems, both solutions are usually intersection points (two-solution case). Forgetting the negative root drops half the answer.

Squaring without checking for extraneous solutions

When you square both sides of an equation involving square roots, you can introduce false solutions. Always plug your answers back into the ORIGINAL equation to verify.

Confusing 'no real solutions' with 'no solutions'

On the SAT, 'no real solutions' (b² - 4ac < 0) means the line and parabola don't intersect on the real plane. There are still complex solutions — but the SAT only cares about real ones.

Practice problems

6 problems adapted from College Board released questions and internal Prepiii sets. Click each one to reveal the solution.

1

What is the positive solution to -4x² - 7x = -36?

  1. −4
  2. −9/4
  3. 9/4
  4. 4

Click to reveal solution →

Answer: (C) 9/4

Move everything to one side: -4x² - 7x + 36 = 0 → 4x² + 7x - 36 = 0 (multiplied by −1).

Factor: (4x - 9)(x + 4) = 0 → x = 9/4 or x = -4. The positive solution is 9/4.

2

Which is a solution to the equation 2x² - 2 = 2x + 3?

  1. 2
  2. 1 − √11
  3. (1 + √11)/2
  4. (2 + √44)/4

Click to reveal solution →

Answer: (C) (1 + √11)/2

Standard form: 2x² - 2x - 5 = 0, so a = 2, b = -2, c = -5.

Quadratic formula: x = (2 ± √(4 + 40))/4 = (2 ± √44)/4 = (1 ± √11)/2.

3

A system consists of y = x² - 4 and y = 5. How many real solutions does this system have?

  1. 0
  2. 1
  3. 2
  4. Infinitely many

Click to reveal solution →

Answer: (C) 2

Substitute: x² - 4 = 5 → x² = 9 → x = ±3. Two real solutions: (3, 5) and (-3, 5).

4

For what value of k does the system y = x² and y = 2x + k have exactly one solution?

  1. −1
  2. 0
  3. 1
  4. 2

Click to reveal solution →

Answer: (A) −1

Substitute: x² = 2x + k → x² - 2x - k = 0.

For exactly one solution, discriminant = 0: (-2)² - 4(1)(-k) = 0 → 4 + 4k = 0 → k = -1.

5

What is a solution to 2x² - 4 = x²?

  1. 1
  2. 2
  3. 3
  4. 4

Click to reveal solution →

Answer: (B) 2

Subtract x²: x² - 4 = 0 → x² = 4 → x = ±2.

Of the choices, 2 is a solution.

6

If √(x + 7) = x - 5 and x > 5, what is the value of x?

Click to reveal solution →

Answer: x = 9

Square both sides: x + 7 = (x - 5)² → x + 7 = x² - 10x + 25.

Move everything to one side: x² - 11x + 18 = 0 → (x - 9)(x - 2) = 0 → x = 9 or x = 2.

Since x > 5, only x = 9 works. x = 2 would give √9 = -3, which is false (extraneous solution from squaring).

Frequently asked questions

What's a nonlinear equation?

+
Any equation where the variable appears with a power higher than 1, inside a root, in a denominator, in an absolute value, or in an exponent. The most common type on the SAT is the quadratic — directly or in disguise.

What's a disguised quadratic?

+
An equation that doesn't look quadratic at first but simplifies to one after algebraic manipulation. Examples: equations with fractions (multiply through by the denominator), with square roots (square both sides), or with x⁴ terms (substitute u = x²). Always bring everything to one side after clearing the disguise.

How do I solve a line-meets-parabola system?

+
Substitution. Solve the linear equation for one variable, then substitute that expression into the quadratic. You'll get a single-variable quadratic — solve it using factoring or the quadratic formula. Each solution corresponds to an intersection point.

How do I tell how many intersections a line and parabola have?

+
After substituting, you'll have a quadratic ax² + bx + c = 0. The discriminant b² − 4ac tells you the answer: positive = 2 intersections, zero = 1 intersection (tangent), negative = 0 intersections (no real solutions).

Why do I have to check for extraneous solutions when I square both sides?

+
Because squaring can introduce false solutions. For example, √(x + 7) = x − 5 has the algebraic solutions x = 9 and x = 2, but x = 2 gives √9 = -3, which is impossible. Always plug solutions back into the ORIGINAL equation when you've squared anywhere.

Keep going

Quadratics

Foundation — read this first

Systems of equations

The linear-only sibling

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