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SAT Advanced Math
Study quadratics, nonlinear functions, equivalent expressions, and equations with free explanations and worked examples.
These are BuildSAT teaching examples. Use the study workspace to apply the concepts to official practice questions and review your progress.
Open interactive lessons →Equivalent expressions
Exponent rules and radicals
Simplify powers without confusing addition and multiplication.
Exponent laws
For the same nonzero base: aᵐaⁿ = aᵐ⁺ⁿ, aᵐ/aⁿ = aᵐ⁻ⁿ, and (aᵐ)ⁿ = aᵐⁿ. Also a⁰ = 1 and a⁻ⁿ = 1/aⁿ. These laws do not combine a sum such as a² + a³.
Roots and fractional exponents
For positive a, a^(1/2) = √a and a^(m/n) = ⁿ√(aᵐ). For real x, √(x²) = |x|, so the principal square root is never negative.
Exponents count repeated factors
For positive integer exponents, x³ means x·x·x. Multiplying x³ by x² produces five copies of x, explaining why exponents add. Raising x³ to the second power produces two groups of three copies, explaining why the exponents multiply in a power of a power.
Roots and negative powers extend the same consistent rules
A negative exponent represents a reciprocal: x⁻² = 1/x² for x ≠ 0. A fractional exponent represents a root, with real-domain restrictions. A square root is the nonnegative number whose square is the radicand; therefore √(x²) = |x|, even when x itself is negative.
Worked example
Simplify (x³)²/x⁴ for x ≠ 0.
- A power of a power multiplies exponents: x⁶.
- Division subtracts exponents: x⁶⁻⁴.
x²
Watch for this: A negative exponent changes reciprocal position, not the sign of the value.
Check your understanding
For positive x, x² · x⁵ equals
- x¹⁰
- 2x⁷
- x³
- x⁷
Show answer and explanation
Answer: D. Add exponents: 2 + 5 = 7, giving x⁷.
Equivalent expressions
Expand and factor expressions
Recognize equivalent forms and reveal useful structure.
Distribute and collect
Multiply every term in one factor by every term in the other. Combine terms only when they have the same variable powers.
Factor in reverse
Take out the greatest common factor first. For x² + bx + c, seek numbers that add to b and multiply to c. The difference of squares factors as a² − b² = (a − b)(a + b).
Factoring rewrites a sum as a product
Distribution turns a product such as 3(x + 2) into a sum, 3x + 6. Factoring reverses that process by finding shared factors. It can expose zeros, simplify rational expressions, or reveal the structure of an equation. An expression has no “solutions” until it is placed in an equation.
Check both the middle term and the constant
For (x + p)(x + q), distribution gives x² + (p + q)x + pq. That explains why p and q must add to the middle coefficient and multiply to the constant. When the leading coefficient is not 1, expand proposed factors to verify every coefficient.
Worked example
Factor x² + 7x + 12.
- Find two numbers with sum 7 and product 12.
- 3 and 4 meet both conditions.
(x + 3)(x + 4)
Watch for this: Checking one numerical input does not prove two expressions are identical.
Check your understanding
Which expression equals x² − 25?
- (x − 5)(x + 5)
- (x − 5)²
- (x + 5)²
- x(x − 25)
Show answer and explanation
Answer: A. This is a difference of squares: x² − 5² = (x − 5)(x + 5).
Nonlinear equations in one variable and systems of equations in two variables
Solve quadratic equations
Find all real solutions using the structure of a quadratic.
Zero-product rule
When a product equals zero, at least one factor must equal zero. Move every term to one side before using factoring.
Other methods
If factoring is difficult, complete the square or use x = (−b ± √(b² − 4ac))/(2a) for ax² + bx + c = 0. Taking a square root of x² = k requires both signs when k > 0.
Quadratics can have two real inputs for the same output
A squared input can produce the same result from a positive or negative value. A quadratic equation can therefore have two real solutions, one repeated solution, or no real solutions. Move all terms to one side so the other side is zero before applying the zero-product rule.
Choose a method that matches the equation’s form
Use square roots when a square is already isolated. Factor when small integer factors are visible. Use the quadratic formula when factoring is inconvenient. The ± in the formula or square-root step represents two candidates; keep both until the question’s conditions exclude one.
Worked example
Solve x² − 5x + 6 = 0.
- Factor: (x − 2)(x − 3) = 0.
- Set each factor equal to zero.
x = 2 or x = 3
Watch for this: √49 is 7, but solving x² = 49 gives two values of x.
Check your understanding
What are the real solutions to x² = 49?
- 7 only
- −7 only
- −7 and 7
- −49 and 49
Show answer and explanation
Answer: C. 7² = 49 and (−7)² = 49, so both are solutions.
Nonlinear functions
Quadratic graphs and useful forms
Read a vertex, zeros, and intercept from the right form.
Vertex form
In y = a(x − h)² + k, the vertex is (h, k). If a > 0 it is a minimum; if a < 0 it is a maximum. The axis of symmetry is x = h.
Factored and standard form
Factored form a(x − r)(x − s) reveals zeros r and s. In standard form ax² + bx + c, c is the y-intercept. Choose the form that reveals the requested feature.
The square explains the vertex
In a(x − h)² + k, the squared part is zero at x = h and nonnegative elsewhere. For a > 0, adding it to k can only increase the output, so (h, k) is the minimum. For a < 0, the contribution is nonpositive and the vertex is the maximum.
Different forms answer different questions
Factored form reveals roots because a zero factor makes the whole product zero. Standard form reveals the y-intercept because all variable terms vanish at x = 0. Vertex form reveals the extreme output. These are different descriptions of the same parabola, so choose the one that makes the requested feature visible.
Worked example
Find the minimum value of f(x) = 2(x − 4)² + 3.
- The squared quantity is at least zero.
- It equals zero at x = 4; the positive coefficient makes this a minimum.
Minimum output: 3, at x = 4.
Watch for this: The minimum value is the y-coordinate; the input where it occurs is the x-coordinate.
Check your understanding
What is the vertex of y = (x + 2)² − 5?
- (2, −5)
- (−2, −5)
- (−2, 5)
- (5, −2)
Show answer and explanation
Answer: B. Vertex form gives h = −2 and k = −5.
Nonlinear equations in one variable and systems of equations in two variables
Discriminant and number of solutions
Determine when a quadratic has two, one, or no real solutions.
Read the discriminant
For a real quadratic with a ≠ 0, D = b² − 4ac. Positive D gives two distinct real roots; zero D gives one repeated real root; negative D gives no real roots.
Parameter questions
When a coefficient is unknown, use the condition on D to form an equation or inequality. A graph can suggest the boundary, but algebra identifies it exactly.
The square root in the formula controls real solutions
The quadratic formula contains √(b² − 4ac). A positive number under the root gives two different real values from the ± signs. Zero makes both branches identical. A negative radicand has no real square root, so the equation has no real solution. This quantity is the discriminant.
Parameter conditions turn words into algebra
For “exactly one real solution,” set b² − 4ac = 0. For two distinct real solutions use > 0; for no real solutions use < 0. First rewrite the equation in standard form and confirm the quadratic coefficient is nonzero. If a parameter makes that coefficient zero, analyze the resulting linear equation separately.
Worked example
For what k does x² − 6x + k = 0 have exactly one real solution?
- Set D = 0: 36 − 4k = 0.
- Solve for k.
k = 9
Watch for this: “One real solution” can mean a tangent intersection. Do not estimate an exact parameter from a slider alone.
Check your understanding
How many distinct real solutions does x² + 4x + 4 = 0 have?
- Zero
- Two
- Four
- One
Show answer and explanation
Answer: D. D = 4² − 4(1)(4) = 0; the repeated root is −2.
Equivalent expressions
Rational expressions and restrictions
Simplify fractions while preserving excluded values.
Factor before canceling
Only common factors can cancel. In (x² − 9)/(x − 3), factor the numerator before simplifying. You cannot cancel separate terms across addition.
Keep the original restrictions
The original denominator cannot be zero, even after simplification. Multiplying by denominators can introduce candidates that must be checked.
A rational expression is a fraction of polynomials
The denominator must not equal zero. Those excluded inputs are part of the original expression’s domain and remain excluded after simplification. Two formulas can give the same values wherever the original is defined but differ in which inputs they permit. Always retain the original restrictions.
Cancel factors, not pieces of sums
In ab/ac, the factor a multiplies the entire numerator and denominator, so it can cancel when a ≠ 0. In (a + b)/(a + c), a is only a term inside a sum and cannot cancel that way. Factor first to expose genuine multiplicative factors.
Worked example
Simplify (x² − 9)/(x − 3).
- Factor the numerator into (x − 3)(x + 3).
- Cancel x − 3 for values where it is nonzero.
x + 3, with x ≠ 3.
Watch for this: A graph may hide a hole. Record denominator restrictions before using Desmos.
Check your understanding
For x ≠ 0, (x² + 2x)/x simplifies to
- x + 2
- x² + 2
- x + 2x
- 2
Show answer and explanation
Answer: A. Factor the numerator as x(x + 2), then cancel x to obtain x + 2.
Nonlinear equations in one variable and systems of equations in two variables
Radical and absolute-value equations
Find candidates, then check them in the original equation.
Absolute value is distance
|u| = a with a > 0 gives u = a or u = −a. If a < 0, there is no real solution.
Isolate a radical before squaring
Squaring both sides can introduce extraneous solutions because it loses sign information. Substitute every candidate back into the original equation and check the radicand domain.
Absolute value measures distance from zero
Distance is nonnegative. Thus |u| = 5 describes both u = 5 and u = −5, while |u| = −5 has no real solution. A shifted expression |x − h| measures how far x lies from h on the number line, so positive distances usually produce two possible positions.
Squaring preserves truth in one direction but can add answers
If two real quantities are equal, their squares are equal. The reverse can fail because opposite numbers share a square. When solving a radical equation, isolate the radical, note that its value is nonnegative, square as needed, and check every candidate in the original equation.
Worked example
Solve √(x + 5) = x − 1.
- The right side must be nonnegative, so x ≥ 1.
- Square: x + 5 = x² − 2x + 1; factor x² − 3x − 4 = 0.
- Candidates are 4 and −1; only 4 meets the original equation.
x = 4
Watch for this: A solution to a transformed equation is only a candidate until you check the original.
Check your understanding
What are the solutions to |x − 3| = 2?
- 5 only
- −1 and 1
- 1 and 5
- −2 and 2
Show answer and explanation
Answer: C. x − 3 = 2 or −2 gives x = 5 or 1.
Nonlinear functions
Exponential growth and decay
Interpret a starting value, multiplier, and time interval.
Multiplicative change
In y = a·bᵗ, a is the value at t = 0. A multiplier b > 1 gives growth; 0 < b < 1 gives decay. Growth at rate r uses b = 1 + r; decay uses b = 1 − r.
Respect the interval
If something doubles every 3 hours, use 2^(t/3) when t is in hours. Linear growth adds a constant amount; exponential growth multiplies by a constant factor.
Repeated percentage change is repeated multiplication
A 10% increase changes an amount A to A + 0.10A = 1.10A. Applying the increase again multiplies the new amount by 1.10, producing A(1.10)². This differs from repeatedly adding 10% of the original amount, which would be linear rather than exponential.
The exponent counts growth or decay periods
In A·b^(t/k), t/k is the number of periods of length k. If the amount halves every four days and t measures days, the multiplier 1/2 must be raised to t/4. Keep the time unit and period length aligned before interpreting the factor or solving for a value.
Worked example
A 200-gram sample loses 10% each hour. Model its mass after t hours.
- A 10% loss leaves 90%, or 0.9, each hour.
- Multiply the initial mass by that factor t times.
M(t) = 200(0.9)ᵗ
Watch for this: An exponent of t/5 means the stated multiplication happens every 5 time units.
Check your understanding
A population increases by 6% each year. What is its annual growth factor?
- 0.06
- 0.94
- 6
- 1.06
Show answer and explanation
Answer: D. The new amount is 100% + 6% = 106% = 1.06 of the old amount.
Nonlinear functions
Function notation and transformations
Evaluate a function and distinguish input shifts from output shifts.
Function notation
f(a) means substitute a for every input variable in the formula. It is a value, not multiplication by a letter f.
Inside versus outside
f(x) + k moves outputs up by k. f(x − h) shifts the graph right by h. −f(x) reflects across the x-axis; f(−x) reflects across the y-axis.
Function notation keeps input and output separate
In f(x), x is the input and f(x) is the output. To evaluate f(−2), replace every x with a parenthesized −2. For an expression such as f(x + 1), the entire x + 1 becomes the input; substitute it everywhere the original formula uses x.
Inside changes affect where an output occurs
For g(x) = f(x − 3), an old input of 2 occurs when x − 3 = 2, or x = 5. That is why subtracting 3 inside shifts the graph right by 3. An outside addition changes the output directly and moves the graph vertically.
Worked example
If f(x) = x² + 1, find f(3) and describe f(x) − 4.
- Substitute 3: 3² + 1 = 10.
- Subtracting 4 outside the function lowers every output by 4.
f(3) = 10; the new graph shifts down 4.
Watch for this: Parenthesize a negative input so exponent operations are applied correctly.
Check your understanding
If f(x) = 2x² − 1, what is f(−2)?
- −9
- 7
- −5
- 3
Show answer and explanation
Answer: B. 2(−2)² − 1 = 2(4) − 1 = 7.
Nonlinear equations in one variable and systems of equations in two variables
Nonlinear systems and intersections
Solve where a line and curve, or two curves, meet.
Set equal outputs equal
If y = f(x) and y = g(x), solve f(x) = g(x) for possible x-values. Substitute each into an original equation to get y.
Count all intersections
A line and a parabola can have zero, one, or two intersections. Desmos can locate them, but window settings can hide distant solutions.
Intersections represent shared input–output pairs
Two graphs meet where both equations are true for the same x and y. When both are written as y = an expression in x, set those expressions equal. That reduces the system to one equation whose solutions give possible x-coordinates; then compute each corresponding y-coordinate.
Finding an x-value is only part of the system
An x-coordinate alone is not an ordered-pair solution. Substitute it into an original equation, and check the pair in the other. A graph can locate candidates efficiently, but a limited window can hide intersections and a rounded coordinate may not be the exact value requested.
Worked example
Solve y = x² and y = x + 2.
- Set x² = x + 2, then factor (x − 2)(x + 1) = 0.
- The x-values are 2 and −1.
- Compute y for each.
Intersections: (2, 4) and (−1, 1).
Watch for this: Two equations must be satisfied simultaneously; a point on just one graph is insufficient.
Check your understanding
For y = x² and y = 4, what are the x-coordinates of the intersections?
- −2 and 2
- 4 only
- −4 and 4
- 2 only
Show answer and explanation
Answer: A. Set x² = 4; both x = −2 and x = 2 work.
Put the ideas into practice
Choose this domain in the question bank, review the explanation after each attempt, and use a mastery check to find what to revisit.
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