Digital SAT Math · Advanced Math

Nonlinear equations

Digital SAT Math · Advanced Math

Once the variable is squared, rooted or sitting in a denominator, “do the same thing to both sides” stops being a plan. There are four routes out — factor, take a square root, run the formula, or refuse to solve and just count the solutions — and the expensive mistake on this skill is not bad algebra, it is spending ninety seconds on the wrong route. Two of the routes can also change the answer set behind your back: one quietly deletes a solution, the other quietly invents one. This page is about picking the route in five seconds and knowing which of the two thefts to watch for.

On the test


Domain	Advanced Math (score report)
What it looks like	A quadratic, a radical, a rational equation, or a polynomial already in factored form — sometimes with a letter standing in for a coefficient
Often asked	“What is the solution to…?”, “How many distinct real solutions…?”, “For what value of $k$ …?”, “What is the positive solution…?”
Format	Multiple choice and student-produced response; the live digital sets on this subskill run one grid-in in five
Calculator	Desmos solves almost anything here, and is slower than factoring on clean numbers — the box below draws the line

Recognition cues: an $x^2$ with an $x$ still around, a square root sign wrapped around the variable, a variable in a denominator, an equation printed as a product equal to zero, the phrases distinct real solutions, exactly one solution, no real solutions, and any sentence that ends in “where $k$ is a constant”.

What this page owns

The College Board files “nonlinear equations in one variable” and “systems of equations in two variables” under one heading, and most of the web copies that heading onto one page. We split them. Here there is one variable and one equation, and the job is a value of $x$ or a count of them. Two neighbours: when the rewrite is the answer and nothing is ever solved, that is equivalent expressions; when a parabola has to meet a line and you need a point $(x, y)$ , that is nonlinear systems. The factoring is identical in all three places — what changes is what you do once the expression is factored.

Pattern recognition

Five shapes account for nearly every item:

Quadratic set to zero — $ax^2 + bx + c = 0$ , or something you can push into that form in one line.
Already factored — a product of factors equal to zero, sometimes with one factor squared and one carrying a coefficient.
Radical — the variable under a square root, with the whole equation begging to be squared.
Rational — the variable in a denominator, with values of $x$ that were never allowed in the first place.
Count or parameter — nothing is solved: you are asked how many solutions exist, or what constant makes the count come out to one, two or none.

Route before you solve — five seconds, four questions. Is it a product equal to zero? → set each factor to zero, and remember a coefficient inside a factor still has to be divided out. Is the variable in exactly one squared expression, with no loose $x$ ? → take the square root of both sides and write both signs. Does the constant term factor obviously? → factor. If two integers multiplying to $ac$ and adding to $b$ do not appear within about ten seconds, stop hunting. Does the question ask “how many” or “for what value of $k$ ”? → do not solve at all. Compute $b^2 - 4ac$ . Everything left over goes to the quadratic formula, or to Desmos. The four questions are in that order on purpose: the routes above the line finish in one line, and the formula is the route you take when the cheap ones fail, not the route you start with.

Method

Get one side to zero. Every route except the pure square-root one needs $= 0$ first, because the whole point of a factored form is that a product is zero only when a factor is zero. $x(x+2) = 8$ does not mean $x = 8$ or $x + 2 = 8$ ; expand, move the 8, then factor.
Route it using the four questions in the box above.
Never divide both sides by the variable. $3x^2 = 12x$ divided by $3x$ gives $x = 4$ and silently throws away $x = 0$ , because dividing by $x$ assumes $x \neq 0$ . Move everything to one side and factor out the $x$ instead: $3x(x - 4) = 0$ keeps both.
Never lose the $\pm$ . $(x - 3)^2 = 49$ has two solutions, because both $7$ and $-7$ square to 49. Write $x - 3 = \pm 7$ on its own line before you finish the arithmetic.
Check whichever roots the method could have invented. Squaring both sides and multiplying out a denominator can both produce numbers that solve the new equation and not the original one. Substitute every root into the equation as printed — not into your rearranged version — and into the situation, if there is one.
Answer the question that was asked. Positive solution, greatest solution, sum of the solutions, the number of solutions, the value of $k$ — the test almost never wants the plain list of roots, and the list is always one of the wrong choices.

What you are looking at	Route	Finishes in
product of factors $= 0$	set each factor to zero	one line
one squared expression, no loose $x$	square root of both sides, keep $\pm$	one line
clean integer coefficients, small $ac$	factor, then zero product	two lines
ugly coefficients, or nothing factors	quadratic formula (or Desmos)	three lines
“how many solutions”, “for what $k$ ”	discriminant $b^2 - 4ac$ — do not solve	one line
radical, or a variable in a denominator	isolate, clear, solve, then check	plus a check

The discriminant is a counting tool, not a solving tool. For $ax^2 + bx + c = 0$ with the equation already set to zero,

b^2 - 4ac > 0 \Rightarrow \text{two distinct real solutions} \qquad = 0 \Rightarrow \text{exactly one} \qquad < 0 \Rightarrow \text{none}

Three things go wrong with it, in this order of frequency. The equation was not set to zero first, so $a$ , $b$ and $c$ were read off the wrong arrangement — $x(x-6) = -9$ has $c = 9$ , not $c = 0$ and not $c = -9$ . The 4 or the $a$ went missing: it is $4ac$ , so a leading coefficient of 3 matters as much as the constant does. The sign was misread — a negative discriminant means the solutions are not real, which on this test means there is no answer to grid in, not that the answer is zero. And the reverse direction is the archetype worth practising: when the question fixes the count and hides a letter in $a$ , $b$ or $c$ , set $b^2 - 4ac$ equal to zero (for one solution) or write it as an inequality (for two, or for none) and solve for the letter. You still never solve for $x$ .

One route loses solutions, one route gains them, and they are different routes. Dividing by a variable loses: the lost root is always $x = 0$ , and the tell is an item where 0 is one of the choices and your answer set has only one member. Squaring both sides, and multiplying through by a denominator, gain: the invented root satisfies the equation you built but not the equation you were given. The reason is worth carrying — squaring destroys sign information, so $\sqrt{\phantom{x}}\,$ equations can end up “solved” by a value that makes the right-hand side negative, and a square root is never negative. A denominator does the same thing for a different reason: $x = 5$ can look like a perfect root of the cleared equation while making the original fraction $\tfrac{0}{0}$ . Losing is prevented by technique (factor, never divide). Gaining is prevented only by checking.

Desmos on this skill — what it is for and when it costs you time. Type the equation with its equals sign exactly as printed and Desmos marks every real solution as a grey dot on the $x$ -axis; click a dot to read it. That makes it the fastest route for anything with ugly coefficients, for radical and rational equations (the graph simply has no point where a solution is extraneous, so the checking is done for you), and for confirming a count — no crossings means no real solutions. Two places it is the wrong tool. Clean factoring is faster than typing: $x^2 - 8x + 15 = 0$ is done in four seconds in your head and takes longer than that to enter. Parameters do not graph: an equation containing $k$ has nothing to plot until you decide what $k$ is, so “for what value of $k$ does this have exactly one solution” is a discriminant question no matter how good your calculator is — a slider will get you close and will not tell you the exact value the grid-in wants. One more habit: when Desmos shows a solution like $x = 2.5$ , the grid-in accepts $5/2$ too, and a fraction is safer than a rounded decimal.

Worked example 1 — the two thefts, side by side

Stem. Find all solutions of $4x^2 = 20x$ , and all solutions of $(x + 6)^2 = 81$ .

Step 1 — the first equation has a loose $x$ , so it is a factoring problem. Move everything to one side rather than dividing:

4x^2 - 20x = 0 \qquad \Rightarrow \qquad 4x(x - 5) = 0

A product is zero exactly when one of its factors is zero, so $4x = 0$ or $x - 5 = 0$ , giving $x = 0$ and $x = 5$ . Two solutions.

Step 2 — the second equation has no loose $x$ , so it is a square-root problem. The variable appears only inside the squared expression, so undo the square directly, and carry both signs:

x + 6 = \pm 9 \qquad \Rightarrow \qquad x = 3 \quad \text{or} \quad x = -15

Check. First equation, $x = 0$ : $4(0) = 0$ and $20(0) = 0$ ✓. $x = 5$ : $4(25) = 100$ and $20(5) = 100$ ✓. Second equation, $x = 3$ : $(9)^2 = 81$ ✓. $x = -15$ : $(-9)^2 = 81$ ✓. Answers: $x = 0$ or $5$ ; and $x = 3$ or $-15$ .

Trap watch. Dividing the first equation by $4x$ leaves the tidy-looking $x = 5$ and destroys $x = 0$ — and $x = 0$ is a perfectly good solution of the original, so nothing about the answer looks wrong afterwards. In the second equation, taking only the positive square root leaves $x = 3$ and loses $x = -15$ ; the negative root is the one the test puts in the choices. Note also that expanding the second equation into $x^2 + 12x - 45 = 0$ and factoring is not wrong, just slower — the route matters, the answer does not change.

Worked example 2 — squaring invents a root, so squaring obliges you to check

Stem. What is the solution to $\sqrt{3x + 4} = x - 2$ ?

Step 1 — the radical is already alone, so square both sides. Square the whole right side as a product, not term by term:

3x + 4 = (x-2)^2 = x^2 - 4x + 4

Step 2 — set to zero and factor.

0 = x^2 - 7x = x(x - 7)

so $x = 0$ or $x = 7$ . Note that even here the $x$ is factored out rather than divided away — the candidate $x = 0$ has to be produced before it can be judged.

Step 3 — check both candidates in the equation as printed. At $x = 0$ : the left side is $\sqrt{4} = 2$ and the right side is $0 - 2 = -2$ . Since $2 \neq -2$ , $x = 0$ is extraneous — a square root symbol means the non-negative root, so the left side can never equal a negative number. At $x = 7$ : the left side is $\sqrt{25} = 5$ and the right side is $7 - 2 = 5$ ✓.

Check. Only $x = 7$ survives, and it survives in the original equation, not in the squared one. Answer: $x = 7$ .

Trap watch. Three ways this goes wrong. Squaring $x - 2$ as $x^2 + 4$ instead of $x^2 - 4x + 4$ produces $x^2 - 3x = 0$ and the wrong pair $\{0, 3\}$ . Skipping the check reports both roots, and “0 and 7” is always one of the choices. And checking in the squared equation proves nothing at all — both candidates satisfy that one by construction, which is exactly why it cannot be used as the check.

Worked example 3 — a constant chosen to control the count

Stem. In the equation $x^2 + 6x + c = 0$ , $c$ is a constant. (a) For what value of $c$ does the equation have exactly one real solution? (b) For what values of $c$ does it have no real solutions?

Step 1 — recognise that nothing is being solved for $x$ . The question is about how many solutions exist, so the tool is the discriminant and the unknown is $c$ . Read the coefficients off the equation as printed, which is already set to zero: $a = 1$ , $b = 6$ , $c = c$ .

b^2 - 4ac = 36 - 4c

Step 2 — part (a): exactly one solution means the discriminant is zero.

36 - 4c = 0 \qquad \Rightarrow \qquad c = 9

Step 3 — part (b): no real solutions means the discriminant is negative. Same expression, an inequality instead of an equation:

36 - 4c < 0 \qquad \Rightarrow \qquad 36 < 4c \qquad \Rightarrow \qquad c > 9

Check. With $c = 9$ the equation is $x^2 + 6x + 9 = (x+3)^2 = 0$ , whose only solution is $x = -3$ — exactly one ✓. With $c = 10$ , $36 - 40 = -4 < 0$ , no real solutions ✓. With $c = 8$ , $36 - 32 = 4 > 0$ , and indeed $x^2 + 6x + 8 = (x+2)(x+4)$ gives the two solutions $-2$ and $-4$ ✓. Answers: (a) $c = 9$ ; (b) $c > 9$ .

Trap watch. Dropping the 4 turns $36 - 4c$ into $36 - c$ and answers 36. Reading “no real solutions” as “the discriminant is zero” answers 9 to part (b), which is the one value of $c$ for which the equation does have a solution. And the largest time sink is not an error at all — running the quadratic formula on a letter, getting $x = \frac{-6 \pm \sqrt{36-4c}}{2}$ , and then having to notice the discriminant anyway.

Practice

Answer before opening the explanation. Two items are student-produced response, matching the grid-in share of the live digital sets on this subskill; one item hands you a parabola and no equation, because the number of solutions of $f(x) = c$ is a fact about a picture as much as about a formula. Every wrong choice below is one named misconception — a solution divided away, a $\pm$ collapsed to a $+$ , an extraneous root kept, a discriminant read without its 4. When you miss one, log the misconception, not the item number.

12 questions — 10 multiple choice, 2 student-produced response. Every wrong choice has its own explanation.

Question 1 Warm-up

What are all values of x that satisfy the equation x² - 8x + 15 = 0?

Show the answer Choice C

Why it is right

The equation is already set to zero, so factor the left side. Two integers must multiply to the constant 15 and add to the coefficient -8; those integers are -3 and -5, giving x² - 8x + 15 = (x - 3)(x - 5). A product is zero exactly when one of its factors is zero, so x - 3 = 0 or x - 5 = 0, which means x = 3 or x = 5. Substituting back confirms both: 9 - 24 + 15 = 0 and 25 - 40 + 15 = 0.

Why each other choice fails

Choice A: Reads the constants inside the factors (x - 3)(x - 5) as the answers without flipping their signs. Setting x - 3 = 0 gives x = +3, not -3; at x = -3 the left side is 9 + 24 + 15 = 48.
Choice B: Chooses two numbers that add to 8 but ignores the requirement that they multiply to 15, since 2 · 6 = 12. At x = 2 the left side is 4 - 16 + 15 = 3, not 0.
Choice D: Chooses a factor pair of 15 without checking the sum: 1 + 15 = 16, not 8. At x = 1 the left side is 1 - 8 + 15 = 8, so this value does not satisfy the equation.

Question 2 Standard

What are all values of x that satisfy the equation 3x² = 12x?

Show the answer Choice D

Why it is right

Move every term to one side rather than dividing by the variable: 3x² - 12x = 0. Factoring out the common factor 3x gives 3x(x - 4) = 0, and a product is zero exactly when a factor is zero, so 3x = 0 or x - 4 = 0. That gives the two solutions x = 0 and x = 4. Both check in the original equation: at x = 0 both sides are 0, and at x = 4 the left side is 3(16) = 48 and the right side is 12(4) = 48.

Why each other choice fails

Choice A: Treats the equation as 3x² = 12, dropping the variable from the right side, and then takes x² = 4. At x = 2 the left side is 12 but the right side is 24.
Choice B: Divides both sides by 3x. That step assumes x ≠ 0 and therefore deletes the solution x = 0, which does satisfy the original equation since both sides are then 0.
Choice C: Flips a sign while collecting terms, factoring 3x(x + 4) = 0, which comes from 3x² + 12x = 0. At x = -4 the left side is 48 and the right side is -48.

Question 3 Standard

What are the solutions to the equation x² - 6x + 4 = 0?

Show the answer Choice B

Why it is right

No integer pair multiplies to 4 and adds to -6, so use the quadratic formula with a = 1, b = -6 and c = 4. The discriminant is b² - 4ac = 36 - 16 = 20, so x = (6 ± √20) / 2. Since √20 = 2√5, this is (6 ± 2√5) / 2, and the 2 in the denominator divides both terms of the numerator, giving x = 3 ± √5. Checking the sum of the two roots gives 6, which matches -b/a as expected.

Why each other choice fails

Choice A: Uses b rather than -b in the numerator, producing (-6 ± √20) / 2. The formula begins with the opposite of the coefficient of x, so a coefficient of -6 contributes +6.
Choice C: Divides only the rational part of the numerator by 2, turning (6 ± 2√5) / 2 into 3 ± 2√5. The denominator divides every term of the numerator, including the radical part.
Choice D: Adds 4ac instead of subtracting it, computing a discriminant of 36 + 16 = 52 and then (6 ± √52) / 2 = 3 ± √13. The discriminant is b² - 4ac, and here 4ac is positive.

Question 4 Standard Student-produced response

What is the greatest value of x that satisfies the equation (x - 3)² = 49?

Your answer

Show the answer 10

Why it is right

The variable appears only inside the squared expression, so undo the square directly instead of expanding. Both 7 and -7 square to 49, so x - 3 = 7 or x - 3 = -7, which gives x = 10 or x = -4. The question asks for the greatest of these, so the answer is 10. Checking in the original equation, (10 - 3)² = 7² = 49, and the other root also checks since (-4 - 3)² = (-7)² = 49.

Answers students type instead

4: Subtracts 3 from 7 instead of adding it. The equation reads x - 3 = 7, so x is 3 more than 7, not 3 less.
7: Stops at the intermediate step x - 3 = 7 and reports 7 instead of solving for x. The 3 still has to be added to both sides.
-4: The other solution. Both roots satisfy the equation, but the question asks for the greatest value of x, and -4 is the lesser of the two.

Question 5 Standard

What are all values of x that satisfy the equation (2x - 5)(x + 3)² = 0?

Show the answer Choice A

Why it is right

The left side is already a product equal to zero, so set each distinct factor to zero. From 2x - 5 = 0 comes 2x = 5 and therefore x = 5/2. From (x + 3)² = 0 comes x + 3 = 0 and therefore x = -3; the exponent 2 repeats that factor but does not create a second, different value of x. So the solutions are x = -3 and x = 5/2, and substituting either one makes one factor zero and the whole product zero.

Why each other choice fails

Choice B: Reads the constant inside (x + 3) as the solution without changing its sign. Setting x + 3 = 0 gives x = -3; at x = 3 that factor equals 6, so the product is not zero.
Choice C: Flips the sign when solving 2x - 5 = 0. Adding 5 to both sides gives 2x = 5, so x is positive; at x = -5/2 the first factor equals -10.
Choice D: Forgets to divide by the coefficient 2, reading 2x - 5 = 0 as x = 5. At x = 5 the first factor is 10 - 5 = 5 and the second is 64, so the product is 320.

Question 6 Harder

What is the solution set of the equation √(2x + 11) = x + 4?

Show the answer Choice C

Why it is right

Squaring both sides gives 2x + 11 = (x + 4)² = x² + 8x + 16. Setting that to zero gives 0 = x² + 6x + 5 = (x + 1)(x + 5), so the candidates are x = -1 and x = -5. Squaring can invent roots, so test both in the original equation. At x = -1 the left side is √9 = 3 and the right side is 3, so it works. At x = -5 the left side is √1 = 1 and the right side is -1; a square root symbol denotes the non-negative root, so 1 ≠ -1 and x = -5 is extraneous. Only x = -1 survives.

Why each other choice fails

Choice A: Reports both candidates without checking them in the original equation. Squaring both sides is a one-way step: everything that solved the original still appears, but new roots can be created.
Choice B: Keeps the extraneous candidate and discards the valid one. At x = -5 the equation reads 1 = -1, which is false because the square root symbol never denotes a negative value.
Choice D: Comes from squaring the right side term by term as x² + 16 rather than x² + 8x + 16. That produces x² - 2x + 5 = 0, whose discriminant 4 - 20 is negative, so it wrongly suggests no solution exists.

Question 7 Harder

How many distinct real solutions does the equation 3x² + 12x + 12 = 0 have?

Show the answer Choice B

Why it is right

The equation is already set to zero with a = 3, b = 12 and c = 12, so the discriminant is b² - 4ac = 144 - 4(3)(12) = 144 - 144 = 0. A discriminant of zero means the quadratic has exactly one distinct real solution. Solving confirms it without changing the answer: 3x² + 12x + 12 = 3(x² + 4x + 4) = 3(x + 2)², which is zero only at x = -2. The repeated factor is why the single solution is sometimes described as a double root.

Why each other choice fails

Choice A: Comes from forgetting to square b, computing 12 - 4(3)(12) = -132 and reading the negative value as no real solutions. The discriminant squares the coefficient of x first.
Choice C: Drops the leading coefficient from the product 4ac, computing 144 - 4(12) = 96 and concluding the discriminant is positive. The 3 in front of x² belongs in that product.
Choice D: Treats counting as impossible without solving, which is exactly what the discriminant avoids. One subtraction settles the count, and no value of x ever has to be found.

Question 8 Harder

The graph of y = f(x) is shown in the xy-plane, where f is a quadratic function. The equation f(x) = c, where c is a constant, has exactly one real solution. What is the value of c?

Figure 1. The graph of y = f(x), a quadratic function with minimum value -4.

Show the answer Choice A

Why it is right

The solutions of f(x) = c are the x-coordinates where the horizontal line y = c meets the curve. A line drawn above the lowest point of this upward-opening parabola meets it twice, a line below it misses entirely, and exactly one meeting point occurs only at the height of the vertex itself. The marked vertex is (1, -4), so the line y = -4 touches the curve at exactly that one point and c = -4. This is the graphical form of a zero discriminant: one solution means the curve is tangent to the horizontal line.

Why each other choice fails

Choice B: Reads the y-intercept, where the curve crosses the y-axis at (0, -3), instead of the minimum. The horizontal line y = -3 sits above the vertex and meets the curve at two points.
Choice C: Assumes one solution means the graph meets the x-axis. Here y = 0 crosses the curve twice, at the marked points (-1, 0) and (3, 0), so c = 0 gives two solutions.
Choice D: Reports the x-coordinate of the vertex rather than its y-coordinate. The constant c is a value of the output f(x), so it is read on the vertical axis.

Question 9 Harder

A rectangular patio has an area of 96 square feet. Its length, in feet, is 4 more than twice its width. What is the width of the patio, in feet?

Show the answer Choice C

Why it is right

Let w be the width in feet. The length is 2w + 4, so the area gives w(2w + 4) = 96, or 2w² + 4w - 96 = 0. Dividing every term by 2 gives w² + 2w - 48 = 0, which factors as (w + 8)(w - 6) = 0, so w = -8 or w = 6. A width cannot be negative, so the only value that fits the situation is w = 6. Checking: the length is 2(6) + 4 = 16 feet, and 6 · 16 = 96 square feet.

Why each other choice fails

Choice A: This root solves the equation but not the problem: substituting w = -8 gives (-8)(-12) = 96, yet a patio cannot have a width of -8 feet. A root that contradicts the situation must be rejected.
Choice B: Takes the size of the rejected root and drops its minus sign. The factor (w + 8) gives w = -8, not +8; a width of 8 feet would make the length 20 feet and the area 160 square feet.
Choice D: Reports the length instead of the width. The length is 2(6) + 4 = 16 feet, which is the other dimension of the same correct rectangle.

Question 10 Hardest

The equation x⁴ - 13x² + 36 = 0 has four distinct real solutions. What is the sum of the two positive solutions?

Show the answer Choice B

Why it is right

Only even powers of x appear, so the equation is a quadratic in x². Writing u = x² turns it into u² - 13u + 36 = 0, which factors as (u - 4)(u - 9) = 0, so u = 4 or u = 9. Each value of u then gives two values of x: x² = 4 gives x = 2 or x = -2, and x² = 9 gives x = 3 or x = -3. The four solutions are -3, -2, 2 and 3, so the two positive ones are 2 and 3 and their sum is 5.

Why each other choice fails

Choice A: Adds all four solutions, -3 + (-2) + 2 + 3, which is zero by symmetry. The question asks only for the two positive ones.
Choice C: Multiplies the two positive solutions, 2 · 3 = 6, instead of adding them. The stem asks for a sum.
Choice D: Stops at the substitution and adds the two values of u, 4 + 9 = 13. Those are values of x², not of x, so each still has to be square-rooted.

Question 11 Hardest Student-produced response

In the equation x² + kx + 36 = 0, k is a positive constant. The equation has exactly one distinct real solution. What is the value of k?

Your answer

Show the answer 12

Why it is right

Exactly one distinct real solution means the discriminant is zero, so no value of x ever has to be found. With a = 1, b = k and c = 36, the discriminant is k² - 4(1)(36) = k² - 144. Setting k² - 144 = 0 gives k² = 144, so k = 12 or k = -12, and the stem says k is positive, which leaves k = 12. Checking: x² + 12x + 36 = (x + 6)², whose only solution is x = -6.

Answers students type instead

6: Drops the 4 from the product 4ac, solving k² = 36 instead of k² = 144. With k = 6 the discriminant is 36 - 144 = -108, so the equation would have no real solutions at all.
144: Reports k² rather than k. The discriminant equation gives k² = 144, and the square root still has to be taken.
-12: The other root of k² = 144. It does make the discriminant zero, but the stem states that k is positive, so this value is ruled out by the question.

Question 12 Hardest

What is the solution set of the equation (x² - 2x)/(x - 5) = 15/(x - 5)?

Show the answer Choice D

Why it is right

Both sides share the denominator x - 5, which is undefined at x = 5, so x = 5 is excluded before any algebra happens. Multiplying both sides by x - 5 gives x² - 2x = 15, or x² - 2x - 15 = 0, which factors as (x - 5)(x + 3) = 0 with candidates x = 5 and x = -3. The candidate x = 5 is exactly the excluded value, so it is extraneous. Only x = -3 remains, and it checks: the left side is (9 + 6)/(-8) = -15/8 and the right side is 15/(-8) = -15/8.

Why each other choice fails

Choice A: Assumes that once the extraneous root is thrown out nothing is left. The quadratic has two roots, and only one of them is excluded; x = -3 makes the denominator -8 and satisfies the equation.
Choice B: Picks the candidate that the domain forbids. At x = 5 both denominators are zero, so neither side of the original equation has a value and x = 5 cannot be a solution.
Choice C: Solves the cleared equation correctly and never returns to the original. Multiplying by x - 5 is legal only when x ≠ 5, so that candidate has to be discarded afterwards.

Common mistakes

Dividing both sides by the variable — $3x^2 = 12x$ becomes $x = 4$ , and $x = 0$ is gone. Move everything to one side and factor the $x$ out; a factor can be set to zero, a divisor cannot.
Losing the negative square root — $(x-3)^2 = 49$ gives $x - 3 = \pm 7$ , so $x = 10$ and $x = -4$ . Write the $\pm$ on its own line before doing the arithmetic.
Splitting a product that is not equal to zero — from $x(x+2) = 8$ , concluding $x = 8$ or $x + 2 = 8$ . The zero product property needs a zero; expand first, then move the 8.
Keeping an extraneous root — after squaring or after clearing a denominator, every candidate has to be tested in the original equation. Candidates that fail are not “second answers”.
Checking in the wrong equation — substituting into the squared or cleared version, which every candidate satisfies by construction. The check only counts against the equation as printed.
Squaring a binomial term by term — $(x-2)^2$ written as $x^2 + 4$ . It is a product: $(x-2)(x-2) = x^2 - 4x + 4$ .
Reading the discriminant off an unarranged equation — $x(x-6) = -9$ is $x^2 - 6x + 9 = 0$ , so $c = 9$ . Set the equation to zero before naming $a$ , $b$ and $c$ .
Losing the 4 or the $a$ in $4ac$ — for $3x^2 + 12x + 12 = 0$ the discriminant is $144 - 4(3)(12) = 0$ , not $144 - 4(12) = 96$ . One dropped factor changes the count.
Forgetting the coefficient inside a factor — $2x - 5 = 0$ gives $x = \tfrac{5}{2}$ , not $x = 5$ . A repeated factor, $(x+3)^2$ , contributes the same single value of $x$ , not two different ones.
Keeping a root the situation forbids — a width, a count of people or an elapsed time cannot be negative. The negative root solves the algebra and still is not the answer.
Solving when only a count was asked — “how many distinct real solutions” is one subtraction, and grinding through the quadratic formula to answer it costs a minute you will want later.
Answering the wrong part of the question — reporting both roots when the item asked for the positive one, the sum, the greatest, or the value of a constant. Reread the last six words before entering anything.

FAQ

Do I have to memorize the quadratic formula? Yes. The Digital SAT provides a reference sheet of geometry formulas and the quadratic formula is not on it. It is $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ , and the two things students actually get wrong are the sign of $-b$ and forgetting that the $2a$ divides the whole numerator, radical included. That said, if you route well you will use it on maybe one item per module.

When is Desmos faster than factoring? When the coefficients are not small integers, when the equation is radical or rational, and when you want a count. It is slower for anything that factors on sight, and it cannot help at all with a letter in place of a coefficient. Typing an equation with an equals sign gives you the solutions as grey dots on the $x$ -axis — you do not need to rearrange it into $y =$ form first.

How do I know whether to factor or use the formula? Give factoring about ten seconds: look for two integers whose product is $ac$ and whose sum is $b$ . If they do not surface in that time, they probably are not integers, and the formula is the faster route. Deciding quickly matters more than deciding correctly — both routes give the right answer, only one wastes a minute.

An equation has an $x^2$ but no constant term. Is it special? It is the single most valuable pattern on this page. $ax^2 + bx = 0$ always factors as $x(ax + b) = 0$ , so one solution is always $x = 0$ and the other is $-b/a$ , and it takes about three seconds. It is also the exact situation where dividing by $x$ throws a solution away.

What is an extraneous solution, really? A number that solves an equation you created, not the equation you were given. Squaring both sides and multiplying by a denominator are both one-way operations: everything that solved the original still solves the new equation, but the new equation can have extra solutions that the original does not. That is why the check runs against the original and why extraneous roots appear on radical and rational items and essentially never on plain quadratics.

“No real solutions” — do I grid in 0? No. Zero is a solution, “no real solutions” is the absence of one, and they are different answers. On multiple choice this appears as a choice worded that way; a grid-in will not ask a question whose answer does not exist.

How many of these show up on a test? Nonlinear equations and systems together are among the most heavily weighted parts of Advanced Math, and free guides put the pair at roughly five to eight questions across the two modules. Systems live on their own page; the one-variable half of that band is what this page drills.

Grid-in format questions? Enter the exact value. If a question asks for “the positive solution” or “the greatest value of $x$ ”, there is exactly one number to enter — entering both roots, or the wrong one of the two, is the most common grid-in loss on this skill. Fractions may be entered as fractions ( $5/2$ rather than $2.5$ ), and negatives are allowed.

Coming next in this domain: nonlinear systems, rational equations and expressions, radicals and rational exponents.

Editorial team · last reviewed 2026-08-06

Nonlinear equations

On the test

What this page owns

Pattern recognition

Method

Worked example 1 — the two thefts, side by side

Worked example 2 — squaring invents a root, so squaring obliges you to check

Worked example 3 — a constant chosen to control the count

Practice

Question 1 Warm-up

Why it is right

Why each other choice fails

Question 2 Standard

Why it is right

Why each other choice fails

Question 3 Standard

Why it is right

Why each other choice fails

Question 4 Standard Student-produced response

Why it is right

Answers students type instead

Question 5 Standard

Why it is right

Why each other choice fails

Question 6 Harder

Why it is right

Why each other choice fails

Question 7 Harder

Why it is right

Why each other choice fails

Question 8 Harder

Why it is right

Why each other choice fails

Question 9 Harder

Why it is right

Why each other choice fails

Question 10 Hardest

Why it is right

Why each other choice fails

Question 11 Hardest Student-produced response

Why it is right

Answers students type instead

Question 12 Hardest

Why it is right

Why each other choice fails

Common mistakes

FAQ

Related skills