SAT · Advanced Math · TOPIC LESSON
Choose a method from the structure, keep domain restrictions, and verify every candidate.
BY THE END, YOU CAN
BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. This lesson supports SAT Math and Reading and Writing skill study; it does not reproduce official questions or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
The teaching passages, worked examples and exercises in this lesson are original. They are not official College Board questions or past-exam reproductions. Blueprint and observed-practice analytics describe separate factual metadata; those source questions are not displayed here.
01 / UNDERSTAND & APPLY
Nonlinear equations include squares, higher powers, absolute values, variable denominators and radicals. First record input restrictions, then look for a form that isolates a useful structure. A product equal to zero suggests factoring; an isolated square suggests square roots; a general quadratic may need completing the square or the quadratic formula.
A solution must satisfy the original equation. Squaring can introduce candidates, and multiplying by a denominator can conceal an excluded value. Verification is part of solving, not an optional final decoration.
02 / UNDERSTAND & APPLY
A product is zero exactly when at least one factor is zero. Use this property only after one side is zero; ab = 12 does not imply a = 12 or b = 12. A repeated factor gives a repeated root but only one distinct solution value.
ORIGINAL WORKED EXAMPLE
Solve (x − 4)²(x + 1)(2x − 3) = 0 over the reals.
Answer x = −1, 3/2 or 4
03 / UNDERSTAND & APPLY
For ax² + bx + c = 0 with a ≠ 0, factoring is fastest when factors are simple. Completing the square rewrites a quadratic as a shifted square. The quadratic formula x = (−b ± √(b² − 4ac))/(2a) works for every quadratic; the denominator divides the entire numerator.
When taking square roots of (x − h)² = k with k > 0, include both signs. If k = 0 there is one root, and if k < 0 there are no real roots.
ORIGINAL WORKED EXAMPLE
Solve x² − 6x + 1 = 0.
Answer x = 3 ± 2√2
ORIGINAL WORKED EXAMPLE
Solve 2x² + 3x − 1 = 0.
Answer x = (−3 ± √17)/4
04 / UNDERSTAND & APPLY
The discriminant D = b² − 4ac is positive for two distinct real roots, zero for one repeated real root, and negative for no real roots. This assumes a ≠ 0. If a parameter can make the quadratic coefficient zero, inspect that resulting linear equation separately.
ORIGINAL WORKED EXAMPLE
For which k does x² − 8x + k = 0 have exactly one real solution?
Answer k = 16
05 / UNDERSTAND & APPLY
The equation |u| = a has no solution if a < 0, one condition u = 0 if a = 0, and two cases u = a or u = −a if a > 0. Isolate the absolute value before splitting. Absolute value measures distance from zero; it does not mean “remove any minus sign anywhere.”
ORIGINAL WORKED EXAMPLE
Solve 3|2x − 5| + 1 = 22.
Answer x = −1 or 6
06 / UNDERSTAND & APPLY
Exclude every value making an original denominator zero before multiplying through by denominators. Then solve the resulting polynomial or linear equation and retain only candidates allowed in the original domain.
ORIGINAL WORKED EXAMPLE
Solve (x + 2)/(x − 1) = 3/(x − 1).
Answer No solution
07 / UNDERSTAND & APPLY
The principal square root √u requires u ≥ 0 and has a nonnegative value. Isolate a root before squaring. The other side must also be nonnegative. Squaring removes sign information, so check every candidate in the original equation.
ORIGINAL WORKED EXAMPLE
Solve √(x + 6) = x.
Answer x = 3
08 / UNDERSTAND & APPLY
Substitute the linear equation into the nonlinear equation, solve for one coordinate, and recover the other for each root. A line can miss a parabola, touch it once, or cross it twice. Distinct real roots after substitution correspond to intersections when all transformations are valid.
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ORIGINAL WORKED EXAMPLE
Solve y = x² and y = x + 2.
Answer (2,4) and (−1,1)
09 / UNDERSTAND & APPLY
In a multivariable formula, treat all symbols except the requested one as fixed parameters. Undo operations in reverse order, state nonzero denominators, and choose signs using the context. Solving for a length usually requires a nonnegative root.
ORIGINAL WORKED EXAMPLE
A cylinder has volume V = πr²h, with V > 0 and h > 0. Solve for radius r.
Answer r = √(V/(πh))
10 / UNDERSTAND & APPLY
Graphing both sides can locate candidates and show how many intersections are visible. It does not guarantee all roots are in the window, and it may hide a hole from a denominator restriction. Use exact factorization or root-count reasoning when possible; substitute approximate answers with suitable tolerance only when the task calls for a decimal.
Do not divide a zero-product equation by a factor containing x before checking its zero case. Doing so can erase precisely the solution being sought.
11 / UNDERSTAND & APPLY
Give all real solutions and enforce restrictions.
ORIGINAL PRACTICE · 1
x = 2 or −7
A zero product requires at least one zero factor.
ORIGINAL PRACTICE · 2
x = 1 or −7
x + 3 = ±4, giving two values.
ORIGINAL PRACTICE · 3
None
The discriminant is 4 − 60 = −56 < 0.
ORIGINAL PRACTICE · 4
x = 10 or −2
x − 4 equals 6 or −6.
ORIGINAL PRACTICE · 5
No solution
An absolute value cannot be negative.
ORIGINAL PRACTICE · 6
x = 1
Exclude −1; multiplying by x + 1 gives 2 = x + 1, hence x = 1, which checks.
ORIGINAL PRACTICE · 7
x = 3
Squaring gives 2x + 3 = 9; the candidate gives √9 = 3.
ORIGINAL PRACTICE · 8
(−3,9) and (3,9)
x² = 9 gives x = ±3, and both have y = 9.
ORIGINAL PRACTICE · 9
c > 9
Its discriminant 36 − 4c must be negative.
ORIGINAL PRACTICE · 10
h = 2A/b
Multiply by 2 and divide by b; the stated restriction allows the division.
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