Nonlinear equations and systems

This block builds from zero. If you can already look at 2x² − 5x − 12 = 0, factor it to (2x + 3)(x − 4) = 0, and say why that immediately gives you x = −3/2 and x = 4, skip to the next block — nothing here will be new.

An equation is linear when every variable appears to the first power and nowhere else: 3x + 7 = 19, y = 4x − 1. Its graph is a straight line. An equation is nonlinear when a variable appears any other way — squared or cubed (x² − 5x + 6 = 0), underneath a radical (√(x + 7) = x − 5), or in a denominator (6/(x − 3) = x). Those three forms are exactly the three families this skill covers, and each one has its own failure mode.

A solution is a number that makes the equation a true statement. That definition sounds too obvious to write down, and it is the most useful sentence in this lesson: because a solution is defined by substitution, you can always check one. Put the number back into the original equation, compute both sides, and see whether they match. On this skill, checking is not a nicety for the cautious — it is the operation that separates real roots from the ones the algebra manufactured.

Almost every nonlinear question routes through a quadratic: an equation you can write as ax² + bx + c = 0 with a ≠ 0. Standard form means exactly that — everything on one side, zero on the other, terms in descending order of power. Getting to standard form is step one of nearly every problem in this lesson, and the reason is not tidiness. It is the zero-product property.

The zero-product property: if two numbers multiply to zero, at least one of them is zero. If A × B = 0, then A = 0 or B = 0. That is a fact about the real numbers, and it is the only reason factoring solves anything. When (2x + 3)(x − 4) = 0, the product of two quantities is zero, so one of them must be zero — set each equal to zero and solve the two easy linear equations that result.

Nothing like this is true for any other number. If A × B = 12, you learn nothing about A or B individually; 12 can be 3 × 4 or 24 × 0.5 or −6 × −2. So (x − 5)(x + 2) = 8 cannot be attacked by setting x − 5 = 8. Multiply the left side out, subtract 8, and get a zero on the right before factoring is allowed to conclude anything.

Basic factoring, for the case a = 1: to factor x² + bx + c, find two numbers that multiply to c and add to b. For x² − 5x + 6, the pair is −2 and −3 (they multiply to 6, add to −5), so it factors as (x − 2)(x − 3). When a ≠ 1, find two numbers that multiply to a·c and add to b, split the middle term into those two pieces, and factor in pairs — the first worked example does this in full.

Two more pieces of notation. The symbol ± means "both the plus version and the minus version," and it always produces two candidate answers, not one. And √ means the principal (non-negative) square root: √9 is 3, not ±3. That is why the equation x² = 9 has two solutions, x = ±3, while the expression √9 has one value. Conflating those two facts is a top-five source of lost points on this skill.

The Digital SAT very rarely says "solve this equation" and stops there. It asks for the positive solution, the sum of the solutions, the value of a constant that makes something true, the number of real solutions, the y-coordinate of an intersection point, or the value of x + y. The algebra is a means; the last line of the question is the thing being scored.

So the skill has two halves, and they fail independently. The first is route selection — recognising in a few seconds which of four methods the numbers were built for, because three of them will be slow on any given problem and one will be fast. The second is target discipline — knowing, before you start, what number you are going to write down at the end. Strong students lose most of their points on the second half while executing the first half flawlessly.

Weak path: read the equation, start manipulating, produce the roots, look at the choices, pick the one that appears. Strong path: read the last clause of the question first and name the target ("they want the sum"), then look at the structure and name the route ("no linear term — square root method"), then execute, then check the answer against the target rather than against the choices. The choices are built to contain the roots you found, which is precisely why finding a root in the list is not evidence that you answered the question.

Route 1 — square root method. Use it when there is no plain x term, or when one side is already a squared binomial: x² = 49, (2x − 5)² = 49, 3(x + 1)² = 48. Take the square root of both sides and write ± on the numeric side, then solve the two resulting linear equations. This is the fastest route on the SAT and the one most often missed, because students expand (2x − 5)² into a mess they then have to factor back. If it is already squared, leave it squared.

Route 2 — factoring. Use it when the coefficients are small integers and the constant term factors easily. Set to zero first, always. For a = 1, find the pair that multiplies to c and adds to b. For a ≠ 1, use a·c grouping: find the pair that multiplies to a·c and adds to b, split the middle term, factor in pairs. Factoring is fast when it works and a trap when you force it — give it about fifteen seconds and then switch routes rather than hunting for a pair that does not exist.

Route 3 — the quadratic formula. x = (−b ± √(b² − 4ac)) / (2a). It always works, which is why it is the fallback for anything that will not factor, anything with ugly coefficients, and anything where the answer choices contain radicals. Two habits make it reliable: write a, b, and c down explicitly before substituting (especially their signs), and compute the discriminant b² − 4ac as its own separate number before assembling the fraction.

Route 4 — completing the square. Use it when the question wants the vertex, the minimum or maximum value, or the equation rewritten in the form y = a(x − h)² + k. To complete the square on x² + bx, add and subtract (b/2)²: x² + 6x + 13 becomes (x² + 6x + 9) + 4, which is (x + 3)² + 4, so the vertex is at (−3, 4). If the leading coefficient is not 1, factor it out of the first two terms before halving.

Route 0 — graph it. Bluebook includes a built-in Desmos graphing calculator on every Math question; that availability is College Board's own published statement, not a prep-industry claim. Type both sides as separate functions and read the intersection points, or type the equation as y = (left side) − (right side) and read the x-intercepts. This is genuinely strong for messy roots, for "how many solutions" questions, and for systems. It is weaker when the equation contains a letter parameter, because you then need a slider, and weakest when the answer is exact and irrational, because you will read a decimal and still have to match it to a radical. Whether graphing beats algebra on a given item is a judgment call; anyone quoting you a fixed number of seconds saved is estimating, not measuring.

Write it once and keep it separate from everything else: D = b² − 4ac, computed from a quadratic already in standard form. D > 0 means two distinct real solutions. D = 0 means exactly one real solution, a double root at x = −b/(2a). D < 0 means no real solutions.

The reason this is worth so many points is that the SAT almost never uses the word "discriminant." It says: the equation has exactly one real solution; the equation has no real solutions; the graph has two distinct x-intercepts; the graph does not intersect the x-axis; the line is tangent to the parabola; the system has exactly one solution; the system has no solution. Every one of those is a statement about the sign of b² − 4ac for the right quadratic, and translating the sentence is most of the work.

Worked in place: the graph of y = 3x² + 12x + c has no x-intercepts. x-intercepts are where y = 0, so the relevant quadratic is 3x² + 12x + c = 0, and "no x-intercepts" means D < 0: 144 − 12c < 0, so c > 12. Note what happens at the boundary — c = 12 gives D = 0, which is one intercept, not zero. "No solutions" and "exactly one solution" are adjacent conditions and the distractor built on confusing them appears constantly.

Two shortcuts worth having, because they let you answer without solving. For ax² + bx + c = 0, the sum of the solutions is −b/a and their product is c/a. When a question asks for the sum of the solutions of 2x² − 7x + 3 = 0, the answer is 7/2 and no factoring is required. Use these only on an equation already in standard form with the real a — reading b off an equation that still has terms on both sides is a reliable way to get a clean-looking wrong answer.

And the geometric reading, which makes the whole family one idea: b² − 4ac counts how many times the graph of y = ax² + bx + c meets the x-axis. After you substitute a line into a parabola, it counts how many times that line meets that parabola. Same number, same computation, different picture.

To solve a radical equation: isolate the radical on one side by itself, square both sides, then solve whatever is left — usually a quadratic. Then check every candidate in the original equation. The check is not optional and it is not a safety habit; it is a required step of the method, in the same way that the last step of a long division is a required step.

Mechanism: squaring is not reversible. If A = B then A² = B², but the converse fails — A² = B² only tells you A = ±B. So squaring both sides can turn a false statement into a true one (3 = −3 is false; 9 = 9 is true), which means the squared equation is a weaker statement than the one you started with and can carry solutions the original never had. Those are extraneous solutions: real roots of the equation you produced, not of the equation you were given.

There is a second, faster filter that catches most of them before any checking. √(anything) is the principal square root, so it is never negative. In √(x + 7) = x − 5, the left side is ≥ 0, so the right side must be too: x ≥ 5. That single inequality kills x = 2 without substituting anything. When a radical equals an expression, write down the sign constraint before you square — it is often the whole question.

Rational equations — a variable in a denominator — have the same structure with a different cause. Multiply both sides by the least common denominator to clear the fractions, solve the resulting polynomial, and then discard any root that makes an original denominator zero. Those values were never in play: the original equation is undefined there, so a number that the cleared equation accepts and the original cannot evaluate is not a solution.

The habit that makes this automatic: before doing any algebra on a rational equation, write down the excluded values. For 1/(x − 3) + 1/(x + 3) = 6/(x² − 9), note immediately that x ≠ 3 and x ≠ −3, because x² − 9 = (x − 3)(x + 3). Then, when the algebra hands you x = 3, you already know what to do with it, and "no solution" stops looking like a mistake.

One prohibition, absolute: never divide both sides by an expression containing a variable. From x² = 5x, dividing by x gives x = 5 and silently deletes x = 0, which is a perfectly good solution. Move everything to one side and factor instead: x² − 5x = 0 becomes x(x − 5) = 0, giving x = 0 and x = 5. Dividing by a variable assumes it is nonzero, and you were not told that.

Finally, the algebra that goes wrong most often inside these problems is squaring a two-term side. (A + B)² = A² + 2AB + B², and the middle term is where the points are. (x − 5)² is x² − 10x + 25, not x² + 25. (√x + 2)² is x + 4√x + 4, not x + 4 — which is why you isolate the radical first: so that the side you square has one term on it and no cross term exists.

A system of one linear and one nonlinear equation is solved by substitution, always. The linear equation gives you one variable in terms of the other; put that expression into the nonlinear equation and you are left with a single quadratic in a single variable. There is no case where elimination is easier and no case where guessing from the graph is more accurate.

The sequence, in order: solve the linear equation for y (or for x, whichever is already isolated), substitute into the nonlinear equation, collect every term onto one side so the equation is in standard form, then solve or take the discriminant. The collection step is where most errors enter, because the linear equation's x-term has to be combined with the parabola's x-term, and the resulting b is neither of the ones printed on the page.

Geometrically, the substituted quadratic is the equation for the x-coordinates of the intersection points, so its discriminant counts them: two intersections, one intersection (the line is tangent), or none. A question that says "the system has exactly one solution" and a question that says "the line is tangent to the graph" are the same question with the same computation.

Then finish the job. A system has solutions that are ordered pairs, so if the question asks for the solution, both coordinates matter; if it asks for y, substitute your x back into the easier of the two equations — the linear one; if it asks for x + y, you need both. Producing a correct x and writing it down when the question wanted y is not a comprehension failure, it is the single most common way this question type is missed.

Quadratic–quadratic systems appear less often and use the same move: set the two expressions for y equal, collect to one side, and the quadratic that results does the same job. If both leading terms are identical, they cancel and you are left with a linear equation — meaning the two parabolas meet exactly once, which is a fact worth recognising rather than deriving under time pressure.

If you already factor cleanly, know the discriminant, and check for extraneous roots, the points still leaving this domain are not comprehension. They are five hard variants and a short list of execution slips. Nothing below is a new concept — that is the point of the block.

Hard variant one — the parameter with two legal values. "The equation x² + (k − 2)x + 9 = 0 has exactly one real solution" does not have one answer for k; setting the discriminant to zero gives (k − 2)² = 36, so k = 8 or k = −4. The question will then either ask for the sum of all possible values of k, ask which of the following could be k, or supply a constraint (k > 0, b < 0) that eliminates one. Every one of those phrasings is a test of whether you produced both branches. When a squared quantity equals a number, stop and write both signs on the page before continuing.

Hard variant two — the equation that is only conditionally quadratic. In kx² + 12x + 9 = 0, if k is allowed to be zero the equation is 12x + 9 = 0, a linear equation with exactly one real solution — so "exactly one real solution" would have two answers, k = 4 and k = 0. This is why real items say "the quadratic equation" or "where k is a nonzero constant." That qualifier is not filler; when it is present it is closing this exact door, and when a question conspicuously omits it, check whether the degenerate case is one of the answers.

Hard variant three — extraneous by domain rather than by sign. The familiar extraneous root is the one that fails a sign check on a radical. The harder one is the rational equation whose only algebraic root is an excluded value: 1/(x − 3) + 1/(x + 3) = 6/(x² − 9) reduces honestly to 2x = 6, and x = 3 is exactly the value that makes two of the original denominators zero. The correct answer is that there is no solution, and "no solution" is a real, frequently correct answer rather than a sign that you made a mistake. Related and worth recognising: an expression like (x² − 16)/(x − 4) is not equal to x + 4 at x = 4 — it is undefined there, because 0/0 is undefined, not 1.

Hard variant four — questions that never require solving. "What is the sum of the solutions of 3x² − 12x + 5 = 0?" is −b/a = 4, full stop. "What is the product?" is c/a = 5/3. For a double root, the single solution is x = −b/(2a). Recognising these saves ninety seconds on items designed to eat two minutes, and the trap is grabbing the wrong one of the pair — the product when the question said sum — or reading b off an equation that has not been collected into standard form yet.

Hard variant five — the quadratic in disguise. x⁴ − 5x² + 4 = 0 is quadratic in x²: substitute u = x², solve u² − 5u + 4 = 0 to get u = 1 and u = 4, then convert back — x² = 1 gives x = ±1 and x² = 4 gives x = ±2, four real solutions. The same move handles 2x + 5√x − 3 = 0 with u = √x: 2u² + 5u − 3 = 0 factors to (2u − 1)(u + 3) = 0, so u = 1/2 or u = −3, and u = −3 is impossible because √x ≥ 0 — leaving √x = 1/2 and x = 1/4. Two things go wrong here: forgetting to convert back from u to x, and forgetting that the substitution carries its own domain constraint.

Execution slips, in the order they cost points. First, the sign of c inside the discriminant: when c is negative, b² − 4ac becomes b² PLUS a positive number, and a slip here flips a two-solution answer into a no-solution answer. Second, the collected b in a system — after substitution the linear coefficient is neither the one on the parabola nor the one on the line, and reading either one off the page produces a clean, plausible, wrong number. Third, answering about the wrong quantity: x when they asked for y, both roots when they asked for the positive one, the roots when they asked for their sum. Fourth, the second branch of a ±. Fifth, arithmetic on a negative substituted into a square: (−3)² is 9, and the number of otherwise perfect solutions lost to a squared negative is not small.

One format detail that only matters on grid-in items: when a student-produced response has more than one correct answer, you enter only one of them — that is College Board's own instruction in Bluebook, not a strategy. So an item whose answer is "m = 0 or m = 12" is entered as either number. On multiple choice, the same item would ask for the sum, or say "which of the following could be," and the difference in phrasing is the whole difficulty.

Time. Each Math module is 22 questions in 35 minutes — about 95 seconds per question, which is arithmetic on College Board's published module structure rather than advice from anyone. The recommendation is Meridian's: nonlinear items deserve more than the average because the checking step is where the points are, and the time comes from the linear and arithmetic items that should be finished in forty seconds. The overspend that wastes it is hunting for a factorisation that does not exist; give factoring fifteen seconds and switch to the formula or to a graph. The underspend that costs points is skipping the substitution check on a radical or rational equation because the algebra felt clean.