Subtopics

Equivalent expressions

Factoring polynomials, expanding, combining like terms, and working with rational expressions.

Nonlinear equations in one variable

Solving quadratic, polynomial, and rational equations. Using the discriminant to classify solutions.

Systems with a linear and nonlinear equation

Solving systems where one equation is linear and one is quadratic or another curve.

Nonlinear functions

Interpreting quadratic, exponential, and radical functions from equations, tables, and graphs.

Function notation and transformations

Evaluating f(x), g(f(x)), and identifying the effect of a, h, k on f(x) = a·f(x−h)+k.

Key formulas

x = (−b ± √(b² − 4ac)) / 2a (quadratic formula)

y = a(x − h)² + k (vertex form)

D = b² − 4ac (discriminant)

a² − b² = (a+b)(a−b) (difference of squares)

(a ± b)² = a² ± 2ab + b² (perfect square)

f(t) = a · bᵗ (exponential growth/decay)

See full formula sheet →

Test-taking tips

Factor first: most quadratics on the SAT factor cleanly without needing the formula.
The discriminant tells you how many real solutions exist before you solve.
Vertex form is fastest for finding the maximum or minimum value.
For exponential functions, b > 1 means growth; 0 < b < 1 means decay.

Fully worked Advanced Math example

Worked example: Solving a radical equation and checking for extraneous roots

Solve:

√(2x + 3) = x

The radical is already isolated, so square both sides:

(√(2x + 3))² = x²

2x + 3 = x²

Move every term to one side:

x² − 2x − 3 = 0

Factor the quadratic:

(x − 3)(x + 1) = 0

The resulting candidates are:

x = 3 or x = −1

Squaring both sides can create solutions that did not satisfy the original equation, so each candidate must be checked in that original equation.

For x = 3:

√(2(3) + 3) = √9 = 3

The left side equals the right side, so x = 3 is a solution.

For x = −1:

√(2(−1) + 3) = √1 = 1

But the right side is −1, and 1 ≠ −1. Therefore, x = −1 is an extraneous root. The only solution is x = 3.

Verification method: Compare the graphs of y = √(2x + 3) and y = x. The graphs intersect at (3, 3). They do not intersect at x = −1, confirming that 3 is the only solution.

Trap 1: Accepting every root after squaring. The quadratic equation has two roots, but the original radical equation has only one. Squaring removes sign information: both 1 and −1 become 1 when squared.

Trap 2: Checking the transformed equation instead of the original one. The value −1 satisfies x² − 2x − 3 = 0, but that equation was produced after squaring. A candidate must satisfy √(2x + 3) = x, not merely the quadratic derived from it.

Why students get Advanced Math questions wrong

Substituting into only part of a function during composition or transformation.

If f(x) = 2x² − 1 and g(x) = x + 3, then f(g(x)) = 2(x + 3)² − 1. It is not 2x² + 3 − 1. The entire expression x + 3 replaces every occurrence of x in f. Order also matters: f(g(x)) and g(f(x)) usually produce different results. For transformations, the direction inside the function is reversed: f(x − 4) shifts the graph 4 units right, while f(x) + 4 shifts it 4 units up.

Keeping roots introduced by a non-reversible operation.

Squaring both sides, multiplying by an expression containing a variable, or clearing a variable denominator can change the allowed solution set. For example, squaring x = −2 and x = 2 produces the same value even though a principal square root cannot equal −2. A value that satisfies the transformed equation is only a candidate until it is substituted into the original equation.

Treating exponential change as repeated addition.

A quantity that grows by 8% per period is multiplied by 1.08 each period. It does not increase by the same fixed amount each time. Starting from 500, two periods of 8% growth give 500(1.08)² = 583.2. Adding 40 twice gives 580, which incorrectly treats the change as linear. Equal differences indicate a linear pattern; equal ratios indicate an exponential pattern.

Mishandling negative signs in the discriminant.

For ax² + bx + c = 0, the discriminant is b² − 4ac. If the equation is x² + 6x + 10 = 0, then D = 6² − 4(1)(10) = 36 − 40 = −4. A negative discriminant means there are no real solutions. Writing 36 + 40, or treating −4ac as positive automatically, changes both the discriminant and the number of real solutions.

Final Advanced Math check

After solving, return to the original expression and ask: Did any step change the permitted inputs or create additional candidates? Test every result in the original equation, confirm that no denominator becomes zero, and check that every radical has a valid input. A value that works only after squaring, factoring, or clearing a denominator is not necessarily a solution.

Advanced Math.