SAT Nonlinear Equations and Systems: worked methods

Nonlinear questions become more manageable when you identify the structure before calculating. Is the prompt asking for roots, an intersection, a parameter, or an expression involving the solutions? The requested quantity determines how much algebra you need.

Match the structure to a method

Structure	Useful first move	Final check
A product equals zero	Set each factor equal to zero	Include every valid root
One variable is isolated in a system	Substitute into the other equation	Verify both coordinates
Number of quadratic roots	Use the discriminant	Confirm the equation remains quadratic
A denominator or radical	Record restrictions before transforming	Test candidates in the original equation

Graphing can help locate intersections and check a result. For an exact value or a parameter condition, use the equations to verify what the graph suggests. A graphing window can hide an intersection or make two close roots look like one.

Worked example: substitution and two intersections

Every example in this guide is an original teaching exercise, not a released SAT question.

Solve y = x² − 5 and x + y = 1. Substitute to obtain x + x² − 5 = 1, so x² + x − 6 = 0. Factoring gives (x + 3)(x − 2) = 0, hence x = −3 or x = 2.

For x = −3, y = 4; for x = 2, y = −1. Both pairs satisfy the original equations: −3 + 4 = 1 and 2 + (−1) = 1. The solutions are (−3, 4) and (2, −1). If asked for the sum of the x-coordinates, report −1 rather than either coordinate alone.

Worked example: a squared variable does not create a free sign choice

For y² = 4x + 1 and y = x + 1, substitution gives (x + 1)² = 4x + 1. Thus x² − 2x = 0 and x(x − 2) = 0. The coordinate pairs are (0, 1) and (2, 3). They satisfy both original equations.

Do not attach an arbitrary negative y-coordinate after solving for x: the equation y = x + 1 fixes y. If you take a square root during another method, retain both branches where appropriate and then check the second equation.

Worked example: squaring can introduce an extraneous root

Solve √(x + 3) = x − 1. The right side must be nonnegative, so x ≥ 1. Squaring gives x + 3 = x² − 2x + 1, or x² − 3x − 2 = 0. Its roots are (3 ± √17)/2.

Only (3 + √17)/2 is at least 1, so it is the valid solution. The other root is negative and cannot satisfy the original equation's nonnegative right side. The restriction eliminates it; substitution into the original equation confirms the remaining root. A transformed equation supplies candidates, not automatic solutions.

Worked example: a cancelled factor still restricts the domain

Solve (x − 2)/(x² − 4) = 1. The original denominator excludes x = 2 and x = −2. Factoring and cancelling for permitted x gives 1/(x + 2) = 1, so x = −1. Checking the original gives (−3)/(−3) = 1.

Multiplying through without restrictions produces x − 2 = x² − 4, or (x − 2)(x + 1) = 0. The candidate x = 2 must still be rejected because the original expression is undefined there.

Worked example: a hidden quadratic

For x⁴ − 5x² + 4 = 0, let u = x². Then u² − 5u + 4 = (u − 1)(u − 4) = 0. Thus x² = 1 or x² = 4, giving x = −2, −1, 1, 2. Returning from u to x is essential: 1 and 4 are not the final roots.

Parameters and the number of real solutions

For x² − 6x + k = 0, the discriminant is 36 − 4k. There is one distinct real root when k = 9, two when k < 9, and none when k > 9. This classification works because the coefficient of x² is fixed and nonzero.

If a parameter controls the leading coefficient, inspect the case where that coefficient is zero separately; the equation may become linear. For a contextual problem, also apply conditions such as positive lengths or whole-number counts.

Review errors by method selection, sign, omitted root, domain restriction, and wrong requested quantity. For more on root relationships, read Vieta and the zero-product rule. For linear systems, use the linear systems guide.

Official sources

• College Board: Advanced Math
• College Board: Math question types