Solving Nonlinear Equations and Systems Questions on the SAT

What this skill covers on the SAT

Nonlinear equations in one variable and systems of equations in two variables is one of the three skills College Board groups under Advanced Math, next to equivalent expressions and nonlinear functions. Those two ask you to rewrite expressions or describe how a function behaves. This one asks you to solve: find every x that makes an equation true, or every point where two graphs meet.

Advanced Math is about 35% of SAT Math, or 13 to 15 questions, shared among those three skills. The Math section is two modules of 22 questions, 35 minutes each, and it adapts: your module 1 performance decides which module 2 you get. About 75% of Math questions are four-option multiple choice and about 25% are student-produced response (SPR), where you type an answer with no choices to check against. The built-in Desmos graphing calculator is available for the entire section, and you can bring an approved calculator of your own.

College Board’s Skills Insight ties statements about this skill to Math section score bands. They’re quoted below, minus notes that limit some skills to the SAT, PSAT/NMSQT and PSAT 10. Read them as a ladder: each step up adds a harder kind of question.

The 550–600 statement also covers exponential functions and key features of graphs, which belong to our nonlinear functions guide. This page handles its linear-plus-quadratic system. And look at the top step. At 680–800 the solving itself is routine. What’s hard is a constant like k that you must choose so the equation or system has a stated solution or a stated number of solutions. Section 06 covers it.

Quadratics: four ways to solve

The quadratic formula solves every quadratic equation, but it’s rarely the fastest route. Start by moving every term to one side so the equation reads ax² + bx + c = 0. Then pick the tool that fits.

Factor when the numbers are friendly. x² + 2x = 15 becomes x² + 2x − 15 = 0, then (x + 5)(x − 3) = 0. The zero-product property says a product is zero only when at least one of its factors is zero, so x = −5 or x = 3. It only works with zero on the other side: x(x + 2) = 15 does not mean x = 15 or x + 2 = 15.

Take square roots when there’s no separate x-term, or all the x’s sit inside one squared group. (x − 4)² = 9 gives x − 4 = ±3, so x = 7 or x = 1. The ± is where students lose points: every positive number has two square roots.

Complete the square only when it helps: a = 1 and b is even, especially if the choices look like −3 ± 2√5. For x² + 6x = 11, add (6 ÷ 2)² = 9 to both sides: (x + 3)² = 20, so x = −3 ± √20 = −3 ± 2√5. In other cases, go straight to the formula.

Use the quadratic formula when nothing factors: x = (−b ± √(b² − 4ac)) / (2a). Work out the part under the root first. It tells you how many solutions to expect and whether they’ll be integers, fractions or radicals. Step through one below.

A good habit: try to factor for about fifteen seconds, then switch to the formula. Whichever method you use, the check is quick. Substitute back, or confirm that the two solutions add to −b/a and multiply to c/a.

How many real solutions? Use b² − 4ac

College Board’s framework labels one of its sample questions for this skill “Determine the conditions under which a quadratic equation has zero, one, two, or infinitely many real solutions.” You don’t have to solve anything. Write the equation as ax² + bx + c = 0 and compute the discriminant, b² − 4ac.

• Positive: two real solutions. The parabola y = ax² + bx + c crosses the x-axis twice.
• Zero: exactly one real solution, x = −b/(2a). The vertex sits on the x-axis.
• Negative: no real solutions. The graph never reaches the x-axis.

And infinitely many? That can only happen when the x² terms cancel once you simplify. If both sides turn out identical (x(3x + 1) − 2 = 3x² + x − 2, for example), every term cancels and every real x works. If only the x² terms cancel, you’re left with a linear equation, which our linear equations guide covers.

The usual mistake is a sign, so work out 4ac with its signs before you subtract it. For −x² + 4x − 7 = 0, 4ac = 4(−1)(−7) = +28, so the discriminant is 16 − 28 = −12: no real solutions. Only when a and c have opposite signs does −4ac turn positive. Classify the equations below without solving them.

Two shortcuts come from the same formula. If ax² + bx + c = 0 has solutions r and s, then r + s = −b/a and rs = c/a. They save time when a question asks for the sum or product instead of the solutions. Use them carefully. The equation must equal zero before you read off a, b and c. The shortcuts say nothing useful about real solutions when the discriminant is negative. And when the discriminant is zero, the one solution is −b/(2a), not −b/a.

Absolute value, radical and rational equations

Absolute value: isolate it, then split into two cases. 3|x + 1| − 4 = 8 becomes |x + 1| = 4, so x + 1 = 4 or x + 1 = −4. That gives x = 3 or x = −5. If the isolated absolute value equals a negative number, as in |x − 2| + 5 = 1, stop there: an absolute value is never negative, so the equation has no solution.

Radical: isolate the radical, square both sides, solve, then check every candidate in the original equation. Squaring can create solutions the original equation doesn’t have, called extraneous solutions, because it erases the difference between a number and its negative.

Rational: before you start, write down the excluded values, the x’s that make a denominator zero. Multiply every term by the common denominator, solve, and throw out anything excluded. Take x/(x − 3) = 3/(x − 3) + 2. The excluded value is 3. Multiplying by (x − 3) gives x = 3 + 2(x − 3), so x = 2x − 3 and x = 3. That’s the excluded value, so this equation has no solution.

Polynomials in factored form: a product equal to zero hands you the solutions. (x − 4)(x + 1)(2x − 3) = 0 gives x = 4, x = −1 and x = 3/2. Don’t expand it. If the polynomial isn’t factored yet, pull out a common factor first: 2x³ − 8x² = 0 becomes 2x²(x − 4) = 0, so x = 0 or x = 4.

Systems: where a line meets a curve

A system of one linear and one nonlinear equation asks for every (x, y) that satisfies both: the points where a line meets a parabola or another curve. Substitution is almost always the right move. Set the two expressions for y equal, or replace y in the nonlinear equation with the line’s expression. You end up with one quadratic in x, and everything from sections 02 and 03 applies. If you need a refresher on substitution itself, see systems of linear equations.

The discriminant of that quadratic counts the intersection points. Positive means the line crosses the curve twice. Zero means it touches at exactly one point, which makes it tangent. Negative means they never meet. In College Board’s framework, the sample question labeled “Solve systems of linear and nonlinear equations in two variables” uses a horizontal line that meets a parabola at exactly one point. A horizontal line y = k can only do that at the vertex, so k equals the vertex’s y-value.

Desmos is fast here. Type both equations and click where the graphs cross to read the coordinates. If the question asks how many solutions there are, count the crossings, then zoom out once to make sure none is off-screen.

Find the unknown constant

This is the 680–800 skill: solve problems “in order to find the value of an unknown constant.” The equation contains a letter such as k, c or m that is fixed but unknown, and the question gives you a condition. Here are three common conditions and how to use each.

A given solution. Substitute it. If x = 3 is a solution of x² + kx − 12 = 0, then 9 + 3k − 12 = 0, so k = 1. The product shortcut gives the other solution: 3 · s = −12, so s = −4. Check: x² + x − 12 = (x + 4)(x − 3).

A number of solutions. Write the discriminant in terms of the constant and apply the condition: = 0 for one, > 0 for two, < 0 for none. For x² − 6x + c = 0 to have exactly one real solution, 36 − 4c = 0, so c = 9. For a system, substitute first and then use the discriminant of the quadratic you get. y = x² + 4 and y = kx meet exactly once when x² − kx + 4 = 0 has k² − 16 = 0, so k = 4 or k = −4. Both lines are tangent, so check whether the question restricts the sign.

Infinitely many solutions. Expand both sides and match coefficients term by term. ax² + 6x − 5 = 4x² + bx − 5 is true for every x only when a = 4 and b = 6.

Desmos helps with these too. Type the equations with the constant as a letter, such as y = kx, accept the slider Desmos offers for k, and drag until the graphs show the condition you need, such as touching at a single point. The slider gets you close. Finish with algebra, because a slider rarely lands exactly on a value like 2√5.

Desmos or algebra?

Desmos is available on every Math question, so the real question is when it’s faster. Here’s the rule of thumb we teach.

Graph it when the coefficients are decimals or ugly, when the question asks how many solutions there are, when you need the intersection points of a system, or when you want to rule out extraneous roots. For a one-variable equation, graph each side as its own function, such as y = √(x + 7) and y = x + 1, and read the x-coordinate where they cross. The extraneous x = −3 from section 04 never shows up on the graph.

Use algebra when the answer has to be exact (choices like 2 ± √5), when the answer is an expression in a constant, or when the condition is about a constant and a slider would only get you close. Do both on hard questions: solve with algebra, then spend ten seconds confirming on a graph.

One more habit: when you check in Desmos, type the original equation, not your rearranged version. If you made a mistake while rearranging, the graph of your version will simply agree with the mistake.

Spot the traps in the choices

Most wrong answers on this skill come from a short list of predictable mistakes. Learn the list and you can often eliminate a choice on sight.

• Keeping an extraneous solution. Squaring both sides, or multiplying by an expression that contains x, can add solutions. On radical and rational equations, check every candidate in the original equation. A choice that lists both candidates is often the trap.
• Dividing by x. Dividing both sides of x² = 5x by x leaves x = 5 and loses x = 0. Move everything to one side and factor instead: x(x − 5) = 0.
• A sign slip in b² − 4ac. b² is never negative, even when b is. When a and c have opposite signs, −4ac is positive, so the discriminant can’t be negative.
• Dropping the ±. x² = 16 has two solutions. A choice with only the positive one is waiting for you.
• Answering the wrong quantity. Underline what the question asks for: x², 2x + 1, the sum or product of the solutions, the positive solution, or x + y in a system. A choice equal to x when the question wants x² is a common trap.
• Trusting an SPR answer without a check. On student-produced response questions there are no choices to catch an error, so substitute your answer back before you type it. If the equation has two solutions, re-read the question for the condition that picks one (the positive solution, the greater one, x > 0) and enter that one. Keep fractions exact until the end, because rounding 7/3 to 2.3 halfway through changes the answer.

Practice until it’s automatic

College Board’s description of Advanced Math says students should “fluently solve” these equations, so speed matters as well as accuracy. Work each drill on paper before you reveal it. The ones marked SPR are written like student-produced response questions: one number to type, with nothing to check it against.

When these feel easy, move to mixed practice, where nonlinear equations sit among other Advanced Math and Algebra questions and you have to recognize the type yourself. For the toughest versions of the find-the-constant questions, see our guide to the hardest SAT Math questions.

Practice questions

Here are three Nonlinear Equations questions from our problem sets, from easier to harder, in the same format as the Digital SAT. Answer each one to see the full explanation.

More practice

Six original questions, from easy to hard. Solve each one on paper and check it by substitution before you pick. Practice 5 and 6 are find-the-constant problems, the kind the 680–800 Skills Insight statement describes.

Frequently asked questions