The third most common SAT Math question type (8.7% of the bank): solving equations that involve a square, a root or a denominator, and systems where one equation is curved. Covers moving everything to one side, factoring versus the quadratic formula, the discriminant as a solution counter, extraneous roots from squaring, excluded values from denominators, and substituting a line into a curve — with six worked examples, four traps and three checks.
Subject: SAT Prep · 61 slides · applied lesson
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Title
SAT Math · Type 3 of 19
8.7% of the question bank — 145 of 1675 questions
Objectives
This type asks you to solve rather than to read. The equation contains a square, a root, or a variable in a denominator, or you are handed a system in which one equation is curved. The mathematics is standard; what makes this type dangerous is that two of its standard moves can invent solutions that do not exist.
The one habit that separates a reliable score on this type from an erratic one: the answer is not final until it has been substituted back into the equation you were given.
Naruhodo Tutoring — SAT Math question bank export (sat-question-index.json) — 145 tagged questions of this type in the site's bank
Section
Section 1
Concept
This type asks you to solve rather than to read. The equation contains a square, a root, or a variable in a denominator, or you are handed a system in which one equation is curved. The mathematics is standard; what makes this type dangerous is that two of its standard moves can invent solutions that do not exist.
You will see it phrased in these ways:
That third phrasing is the one worth preparing for. How many questions never require you to find the solutions at all, and answering them with the discriminant turns a long question into a ten-second one.
Picture it
The same three-panel card as the survey deck, so the shorthand carries over: what identifies it, what you write first, and what is built to catch you.
Figure (svg): Nonlinear equations and nonlinear systems: the tell, the move, and the trap
The red panel is the reason this type is harder than it looks. Every other question type punishes you for doing the algebra wrong; this one can punish you for doing it perfectly.
Prediction
Two of the standard moves on this type are not reversible. Spot the one that is safe.
Predict first
Solving which of these carries no risk of an extraneous root?
Correct: x squared minus 5x plus 6 equals 0
Why: Factoring a polynomial is a reversible operation: every root you find genuinely satisfies the original. The other three each require a move that is not reversible — squaring both sides in two cases, and multiplying by a denominator in the third. Squaring can create a solution because it destroys sign information, and clearing a denominator can create one that makes the denominator zero. Those two operations are exactly when a check-back is mandatory.
Concept
Four shapes, one routine, but different hazards attached to each.
| shape | the move | the hazard |
|---|---|---|
| Quadratic equation | set to zero, factor or use the formula | losing a root by dividing through by x |
| Radical equation | isolate the root, then square both sides | squaring invents solutions — must check back |
| Rational equation | multiply through by the denominator | a root that makes a denominator zero is excluded |
| Line and curve system | substitute the line into the curve | forgetting to find the second coordinate |
Notice that three of the four hazards are about what happens after you have the answer. That is unusual, and it is why the verify step on this type is not optional.
Definition probe
No solving. Just name the opening move.
Sort into buckets
What is the first thing you do to each?
Discrimination
Six equations. Which ones can produce a solution that fails in the original?
Sort into buckets
Does solving this risk an extraneous root?
Warm-up
Try it before the rules section.
Discussion prompt
How many real solutions does x squared plus 4x plus 7 equals 0 have? Answer without solving it.
Hint: There is a formula that counts solutions rather than finding them.
Answer:
None. The discriminant is b squared minus 4ac, which here is 16 minus 28, or negative 12.
A negative discriminant means the parabola never reaches the x-axis, so there are no real solutions.
You did not need the quadratic formula, only the part under its radical — and that part is the whole answer to any how-many question.
This matters because the SAT asks how many far more often than students expect, and the discriminant answers it in one subtraction.
Pattern
Three steps. The third is the one nobody does, and it is where the marks leak.
Get everything onto one side so the equation reads something equals zero.
Why: Factoring depends on the zero-product rule, which only works when the product actually equals zero. For a system, substitute first so you have a single variable.
Solve: factor if the numbers are friendly, otherwise use the quadratic formula.
Why: Try factoring for about fifteen seconds. If no pair of integers works, stop and use the formula rather than persevering.
Substitute every candidate back into the ORIGINAL equation and discard any that fail.
Why: Squaring and clearing denominators can invent solutions. The original equation is the only authority on whether a candidate is real.
Step 3 takes about ten seconds and it is the difference between this type being reliable and being a coin flip.
Section
Section 2
Concept
A product equals zero exactly when one of its factors equals zero — and this reasoning fails against any other number.
This is why step 1 exists. Setting to zero is not tidiness — it is the precondition that makes factoring valid.
Concept
Try to factor for about fifteen seconds; if no integer pair works, switch to the quadratic formula.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The plus-or-minus in the formula is not decoration. Dropping it discards a solution, and the discarded one is always among the answer choices.
Prediction
The discriminant, used as a counter.
Predict first
How many real solutions does 2x squared minus 4x plus 5 equals 0 have?
Correct: none
Why: The discriminant is b squared minus 4ac, which is 16 minus 4 times 2 times 5, or 16 minus 40, giving negative 24. A negative discriminant means the parabola never reaches the x-axis, so there are no real solutions. A quadratic can never have infinitely many solutions unless every coefficient is zero, which would not be a quadratic at all.
Concept
The quantity under the radical, b squared minus 4ac, decides how many real solutions exist.
| discriminant | real solutions | what the graph does |
|---|---|---|
| positive | two | crosses the x-axis twice |
| zero | exactly one | touches the x-axis at its vertex |
| negative | none | never reaches the x-axis |
This rule converts one of the longest-looking questions on the test into a single subtraction.
Prediction
Both candidates came from correct algebra. Only one is real.
Predict first
Solving the square root of (x plus 6) equals x gives candidates x equals 3 and x equals negative 2. Which are actual solutions?
Correct: only x equals 3
Why: Check both in the original. For x equals 3: the square root of 9 is 3, and the right side is 3, so it holds. For x equals negative 2: the square root of 4 is 2, but the right side is negative 2, and 2 does not equal negative 2. The square root symbol always returns the non-negative root, so it can never equal a negative number — which is exactly why squaring manufactured this second candidate.
Concept
Squaring destroys sign information, so a candidate that satisfies the squared equation may fail the original.
The practical rule: after squaring, substitute back and keep only what survives. Never skip this because the algebra looked clean.
Concept
Any value that makes a denominator zero is excluded, even if it emerges from correct algebra.
This is the rational-equation twin of the extraneous root. Different cause, same fix: check the candidates against the original.
Prediction
A one-line temptation that costs a solution.
Predict first
How many solutions does x squared equals 7x have?
Correct: two: 0 and 7
Why: Moving everything to one side gives x squared minus 7x equals 0, which factors as x(x minus 7) equals 0. The zero-product rule then gives x equals 0 and x equals 7. Dividing both sides by x at the start would produce only x equals 7, silently discarding the zero solution — and one is always offered as a choice for exactly that reason.
Concept
Replace y in the curve with the line's expression, solve the resulting single-variable equation, then find the partner coordinates.
Half-finished is the standard error here. A solution to a two-variable system is a pair, and a question asking for the sum of the y-values cannot be answered from the x-values alone.
Concept
Dividing by x destroys the solution x equals zero.
This is the mirror image of the extraneous-root problem: squaring adds false solutions, while dividing by a variable removes true ones.
Prediction
The discriminant run backwards.
Predict first
For what value of k does x squared plus kx plus 9 equals 0 have exactly one real solution, with k positive?
Correct: 6
Why: Exactly one solution means the discriminant is zero: k squared minus 4 times 1 times 9 equals 0, so k squared equals 36 and k is 6 or negative 6. The question asks for the positive value, so k equals 6. Checking: x squared plus 6x plus 9 is (x plus 3) squared, which touches the axis once at x equals negative 3.
Two truths and a lie
Three statements about solving nonlinear equations are correct. The one left standing is false.
Eliminate the wrong options
Which statement is FALSE?
Survives elimination: c
Why: This is the central idea of the whole type. Correct algebra does not guarantee valid solutions, because some legal operations are not reversible. Squaring both sides and multiplying by a denominator both produce equations that have MORE solutions than the one you started with. That is why the routine ends by substituting back into the original — not to catch arithmetic mistakes, but to catch solutions that the method itself created.
Check
The right side is not zero. Handle that first.
Check your understanding
What are the solutions of (x minus 2)(x plus 6) equals 20?
Answer: A
Why: Expand the left side to x squared plus 4x minus 12, then subtract 20 to get x squared plus 4x minus 32 equals 0. That factors as (x minus 4)(x plus 8), giving x equals 4 and x equals negative 8. Checking: (4 minus 2)(4 plus 6) is 2 times 10, which is 20.
Choice B is the trap and it is the one that requires no work, which is exactly why it tempts under time pressure.
Section
Section 3
Worked example
Solve x squared minus 7x plus 16 equals 6.
Figure (svg): A parabola crossing the x-axis at 2 and 5
Subtract 6 from both sides: x squared minus 7x plus 10 equals 0.
Why: The zero-product rule requires a zero on the right; factoring against 6 would prove nothing.
Find two numbers multiplying to 10 and adding to negative 7: negative 2 and negative 5.
Why: For a leading coefficient of 1, factoring is exactly this search.
So (x minus 2)(x minus 5) equals 0, giving x equals 2 or x equals 5.
Why: A product is zero when either factor is zero.
The whole difficulty was step 1. A student who factors the left side while the right side still reads 6 will produce nonsense from otherwise correct work.
Verify: check x equals 2 in the original: 4 minus 14 plus 16 is 6.
Why: The original equation is satisfied, and x equals 5 gives 25 minus 35 plus 16, also 6.
Answer: x = 2 and x = 5
Worked example
Solve 2x squared plus 3x minus 4 equals 0.
Figure (svg): Three parabolas showing two, one and no real solutions according to the discriminant
Try factoring: no integer pair multiplies to negative 8 and adds to 3 in the required way. Abandon it.
Why: Fifteen seconds is enough; persevering past that costs more than the formula does.
Identify a equals 2, b equals 3, c equals negative 4, and compute the discriminant: 9 plus 32, which is 41.
Why: b squared minus 4ac, being careful that minus 4 times 2 times negative 4 is plus 32.
Apply the formula: x equals negative 3 plus or minus the square root of 41, all over 4.
Why: The plus-or-minus produces both solutions; dropping it loses one.
An irrational answer is a signal you were meant to use the formula. If the SAT wanted a tidy answer it would have supplied factorable numbers.
Verify: check the discriminant is positive, so two real solutions are expected.
Why: 41 is positive, and the formula duly produced two distinct values — consistent.
Answer: x = (negative 3 plus or minus the square root of 41) over 4
Worked example
Solve the square root of (x plus 6) equals x.
Figure (svg): Two candidates checked against the original equation, one valid and one rejected
Square both sides: x plus 6 equals x squared.
Why: The radical is already isolated, so squaring is legal immediately.
Move to zero: x squared minus x minus 6 equals 0, which factors as (x minus 3)(x plus 2).
Why: Two numbers multiplying to negative 6 and adding to negative 1 are negative 3 and 2.
Candidates are x equals 3 and x equals negative 2. Substitute both into the ORIGINAL.
Why: Squaring is not reversible, so candidates are not yet solutions.
Notice that negative 2 will be among the answer choices, and so will both 3 and negative 2 as a pair. The check is the only thing separating them.
Verify: check x equals negative 2: the square root of 4 is 2, not negative 2, so it fails.
Why: A square root is never negative, so this candidate is extraneous and must be discarded.
Answer: x = 3 only
Step zero
Only the opening line.
Discussion prompt
A question reads: solve (x minus 3)(x plus 4) equals 8. A student immediately writes x equals 3 and x equals negative 4. What went wrong, and what should the first line have been?
Hint: What does the zero-product rule actually require?
Answer:
The zero-product rule was applied against 8. It only works against zero — there are many pairs of numbers whose product is 8, so nothing follows from the factored form here.
The correct first line is to expand and move everything across: x squared plus x minus 12 equals 8, so x squared plus x minus 20 equals 0.
That factors as (x plus 5)(x minus 4) equals 0, giving x equals negative 5 and x equals 4.
Note that the true answers share no values with the wrong ones. Applying the rule against a non-zero number does not give an approximate answer; it gives an unrelated one.
Worked example
Solve 12 over x equals x minus 1.
Figure (svg): A number line marking the two solutions and the excluded value at zero
Note the exclusion first: x cannot be 0, because it sits in a denominator.
Why: Recording the exclusion before solving means you know what to watch for afterwards.
Multiply both sides by x: 12 equals x squared minus x.
Why: Clearing the denominator turns this into a polynomial equation.
Move to zero: x squared minus x minus 12 equals 0, factoring as (x minus 4)(x plus 3).
Why: Two numbers multiplying to negative 12 and adding to negative 1 are negative 4 and 3.
Here the exclusion did not bite, but noting it cost nothing. When a question is built so that one root IS the excluded value, this habit is the only thing that catches it.
Verify: check neither 4 nor negative 3 is the excluded value 0, and test x equals 4: 12 over 4 is 3, and 4 minus 1 is 3.
Why: Both candidates are permitted and the first one satisfies the original equation.
Answer: x = 4 and x = negative 3
Worked example
Solve the system: y equals x squared minus 3 and y equals 2x.
Figure (svg): A parabola and a line crossing at two points
Both are solved for y, so set them equal: x squared minus 3 equals 2x.
Why: Substitution is immediate when both equations give y explicitly.
Move to zero: x squared minus 2x minus 3 equals 0, factoring as (x minus 3)(x plus 1).
Why: Two numbers multiplying to negative 3 and adding to negative 2 are negative 3 and 1.
x equals 3 and x equals negative 1. Now find each y using y equals 2x: y equals 6 and y equals negative 2.
Why: A solution to a two-variable system is an ordered pair, so each x needs its partner.
Stopping at the x-values is the standard incompleteness here, and questions frequently ask for the sum of the y-coordinates precisely to catch it.
Verify: check (3, 6) in the curve: 3 squared minus 3 is 6.
Why: The point satisfies both equations, and (negative 1, negative 2) does too since 1 minus 3 is negative 2.
Answer: (3, 6) and (negative 1, negative 2)
Faded example
From memory. These four lines are the spine of the type.
Fill in the blanks
Before factoring, move everything to one side so the equation equals zero. To count solutions without finding them, compute the discriminant. After squaring both sides you must check back in the original. And dividing both sides by x will lose the solution x = 0.
Why: The last two lines are a matched pair worth noticing: squaring ADDS solutions that are not real, while dividing by a variable REMOVES solutions that are. Both errors are invisible in the algebra and only appear when you compare candidates against the original equation.
Worked example
For what positive value of k does x squared minus kx plus 25 equals 0 have exactly one real solution?
Figure (svg): Three parabolas showing two, one and no real solutions
Exactly one real solution means the discriminant is zero.
Why: A repeated root is where the parabola touches the axis without crossing.
Compute it: b is negative k, so b squared is k squared, and 4ac is 100. Set k squared minus 100 equal to 0.
Why: Squaring negative k gives positive k squared, so the sign of k does not survive.
So k squared equals 100 and k is 10 or negative 10; the positive value is 10.
Why: The question restricted k to positive values.
Recognising the result as a perfect square is the fastest sanity check. Exactly-one-solution questions always produce one.
Verify: check k equals 10: x squared minus 10x plus 25 is (x minus 5) squared, with the single root 5.
Why: A perfect square trinomial has exactly one repeated root, confirming the answer.
Answer: k = 10
Fill the middle
The counter, in all three cases.
Fill in the blanks
If b squared minus 4ac is positive there are two real solutions. If it equals zero there is one. If it is negative there are none. To make an equation have exactly one solution, set the discriminant equal to zero.
Why: The fourth blank is the one the SAT actually tests. Questions asking for the value of a constant that gives exactly one solution are common, and every one of them is solved by setting the discriminant to zero and solving for that constant.
Estimation
A rough value catches a sign error under the radical.
Predict first
The solutions of x squared minus 6x plus 4 equals 0 are 3 plus or minus the square root of 5. Roughly where do they sit?
Correct: about 0.8 and about 5.2
Why: The square root of 5 is about 2.24, so the solutions are about 3 minus 2.24 and 3 plus 2.24, giving 0.76 and 5.24. Knowing that the root of 4 is 2 and the root of 9 is 3 places the root of 5 just above 2, which is enough. This estimate also confirms the structure: the two solutions must sit symmetrically either side of 3, which is the vertex x-coordinate.
Check
Do the algebra, then do the substitution.
Check your understanding
What is the solution set of the square root of (3x plus 4) equals x?
Answer: A
Why: Squaring gives 3x plus 4 equals x squared, so x squared minus 3x minus 4 equals 0, factoring as (x minus 4)(x plus 1). The candidates are 4 and negative 1. Checking x equals 4: the square root of 16 is 4, which matches. Checking x equals negative 1: the square root of 1 is 1, but the right side is negative 1, so it fails. Only 4 survives.
Choice B is not a careless answer — it is the answer a careful student gets by skipping one ten-second step. That is what makes this trap expensive.
Section
Section 4
Trap
The trap. You solve the square root of (2x plus 8) equals x, square both sides, factor cleanly, and get x equals 4 and x equals negative 2.
Both came from flawless algebra. You select the choice offering both.
But the square root symbol returns the non-negative root, so it can never equal negative 2. Substituting gives the square root of 4, which is 2, against a right-hand side of negative 2. It fails.
The fix. Treat squaring as a step that produces candidates, not solutions.
The original equation is the only authority. Substitute every candidate into it before choosing.
Trap
The trap. The equation reads (x minus 3)(x plus 4) equals 8, and you write x minus 3 equals 8 and x plus 4 equals 8.
Or, more commonly, you read the brackets and answer 3 and negative 4 as though the right side were zero.
Neither is valid. Two numbers can multiply to 8 in unlimited ways, so knowing the product tells you nothing about either factor.
The fix. The rule is about zero specifically, because zero is the only number with the property that a product can only reach it by one of its factors being it.
Expand, collect, and move everything to one side first — then factor the new expression.
Error analysis
A student's work. The algebra is legal at every line and the answer is still incomplete.
Annotate
On: \( x^2 = 5x \;\Rightarrow\; \frac{x^2}{x} = \frac{5x}{x} \;\Rightarrow\; x = 5 \)
The general principle: any operation that is not reversible changes the solution set. Factoring and moving terms are always safe; squaring and dividing by a variable are not.
Trap
The trap. From x squared equals 49 you write x equals 7 and move on.
But negative 7 squared is also 49, so there are two solutions. The same slip happens in the quadratic formula when the plus-or-minus is written once and then only the plus branch is computed.
This is a loss of a correct answer rather than the gain of a wrong one, and it is invisible unless you look for it.
The fix. Taking a square root of both sides of an equation introduces two branches, always.
Write both immediately, before continuing, so the second cannot be forgotten while you work on the first.
Elimination
The equation is the square root of (x plus 12) equals x.
Eliminate the wrong options
Which choice gives the complete solution set? Three can be ruled out by reasoning rather than algebra.
Survives elimination: b
Why: Squaring gives x plus 12 equals x squared, so x squared minus x minus 12 equals 0, factoring as (x minus 4)(x plus 3). The candidates are 4 and negative 3. Because a square root is never negative, any candidate that makes the right-hand side negative is automatically extraneous — so negative 3 goes without any substitution. That single observation eliminates three of the four choices immediately.
Trap
The trap. A system gives x equals 3 and x equals negative 1, and you select the choice containing 3.
But a solution to a system in two variables is an ordered pair. The question may want the point, the y-value, the sum of the y-values, or the product.
The x-values are an intermediate result that feels like a destination because it is where the algebra ends.
The fix. After finding each x, immediately substitute it into the linear equation — it is the easier of the two — to get its partner y.
Then re-read the stem to see which piece of the pair it wanted.
Counterexample
A claim that holds most of the time.
Discussion prompt
A student says: if I do the algebra correctly, every answer I get is a solution. Give a counterexample and explain the mechanism.
Hint: Think of an operation that is legal but not reversible.
Answer:
Counterexample: the square root of x equals negative 3. Squaring both sides gives x equals 9, which is correct algebra.
But substituting back, the square root of 9 is 3, not negative 3. So the original equation has no solutions at all, while the algebra produced one.
The mechanism: squaring is a many-to-one operation. It maps both 3 and negative 3 to 9, so it erases the distinction the original equation depended on.
The same happens with multiplying by a denominator, which can introduce a root that makes that denominator zero.
The general lesson: correct algebra guarantees you have not LOST any solutions, but it does not guarantee you have not GAINED any.
Edge cases
The discriminant counts real solutions. Where does that description need care?
Discussion prompt
A question asks how many solutions x squared plus 1 equals 0 has. The discriminant is negative 4. Is the answer none — and is that always the right word?
Hint: The SAT says real solutions for a reason.
Answer:
On the SAT, the answer is none, and the test is careful to ask about real solutions.
A negative discriminant means the quadratic formula would require the square root of a negative number, which has no real value.
Graphically, the parabola sits entirely above or entirely below the x-axis and never touches it.
The edge case worth knowing: a discriminant of exactly zero gives one solution, but it is sometimes described as a repeated or double root, because the two branches of the formula collapse onto the same value. If a question offers both one and two, the answer is one.
A second edge: the discriminant only applies to quadratics. For a radical or rational equation, counting solutions means solving and then checking, with no shortcut available.
Check
No solving required at all.
Check your understanding
For what value of c does 3x squared minus 12x plus c equals 0 have exactly one real solution?
Answer: A
Why: Exactly one real solution means the discriminant is zero: b squared minus 4ac equals 144 minus 12c, and setting that to zero gives 12c equals 144, so c equals 12. Checking: 3x squared minus 12x plus 12 is 3(x squared minus 4x plus 4), which is 3(x minus 2) squared, touching the axis once at x equals 2.
Every exactly-one-solution question is this question. Set the discriminant to zero and solve for whichever letter is unknown.
Section
Section 5
Matching
Six equations, six things that can go wrong. No solving.
Match the pairs
Why: Six hazards, and only two of them are about the algebra being hard. The other four are about the algebra being easy and the answer still being wrong — which is the character of this whole question type.
Sorting
Each pairing is an equation and a candidate produced by correct algebra.
Sort into buckets
Does the candidate survive the original equation?
Item (f) is worth dwelling on: the algebra produced a candidate for an equation that has no solutions whatsoever. That is the clearest possible demonstration that candidates are not solutions.
Comparison
Fill the blanks from memory. Both reach the same answer; they cost different amounts.
Comparison matrix
| method | when to use it | what goes wrong |
|---|---|---|
| Factoring | when integers multiply to c and add to b | wasting a minute when no integer pair exists |
| Quadratic formula | when factoring fails, or the answer looks irrational | sign errors under the radical, or dropping the plus-or-minus |
| Discriminant only | when the question asks how many, not which | using it when the equation is not quadratic |
| Graphing in Desmos | when the numbers are ugly and you want the count | misreading a near-tangent as a crossing |
The third row is the one students never use and should. If the stem says how many, you are not being asked to solve anything.
Trade off
Fill in the price of each convenience.
Comparison matrix
| shortcut | what it saves | what it can cost |
|---|---|---|
| Dividing both sides by x | one line of factoring | the entire solution x equals 0 |
| Squaring both sides | removes the radical immediately | invents roots that fail the original |
| Skipping the check-back | about ten seconds | the whole question, on radical and rational equations |
| Reading zeros off a factored form | all the algebra | nothing — but only when the right side is zero |
Only the last row is a free shortcut, and even it has a precondition. The other three all trade correctness for a few seconds, which on this type is never a good trade.
Real world
One minute on why extraneous roots exist at all.
Discussion prompt
Why would anyone care about a solution that satisfies a transformed equation but not the original? Where does this matter outside a maths test?
Answer:
Because the transformation is often how the problem gets solved at all. Squaring, clearing denominators and taking logarithms are the standard ways to make an equation tractable.
In physics and engineering the same thing appears as a non-physical solution: the algebra of a projectile's flight gives two times, and one of them is negative — mathematically valid, physically meaningless.
Distance and speed problems produce them constantly, because distance is defined with a square root and squaring loses the sign.
The general discipline is the same everywhere: the model you transformed is not the model you were asked about, so answers must be tested against the original conditions.
That is exactly what the SAT is checking when it offers you both roots as a choice.
Ranking
Order from fewest real solutions to most.
Put in order
Why: The first has discriminant negative 16, so no real solutions. The second is (x minus 3) squared, a perfect square with the single repeated root 3. The third factors as (x minus 2)(x plus 2), giving two. The fourth factors as x(x minus 2)(x plus 2), giving three: 0, 2 and negative 2. Note that the fourth is the one where dividing by x would cost you a root.
Concept
This type is 8.7 per cent of the section, and its errors are procedural rather than conceptual — which makes it very responsive to drilling one habit.
| session | what you do | why |
|---|---|---|
| 1 | Twenty quadratics, always moving to zero first. Factor where possible, formula where not. | Builds the set-to-zero reflex and calibrates when to abandon factoring. |
| 2 | Fifteen radical and rational equations, checking every candidate back in the original. | This is the session that fixes extraneous roots. Do not skip the check even when it is obviously fine. |
| 3 | Ten how-many questions using only the discriminant, solving none of them. | Separates counting from solving, which is the shortcut most students never adopt. |
| 4 | Ten line-and-curve systems, writing every answer as an ordered pair. | Prevents the stopping-at-x error and connects this type to type 5. |
Naruhodo Tutoring — SAT Math question bank export (sat-question-index.json) — filter the bank to this skill tag — 145 questions, roughly a third of each difficulty
Explain it to yourself
Close the deck.
Discussion prompt
Without looking, name the two operations that change the solution set, say which direction each changes it, and give the fix for both.
Hint: One adds, one removes.
Answer:
Squaring both sides ADDS solutions that do not satisfy the original, because squaring erases sign information.
Dividing by a variable REMOVES the solution where that variable is zero, because the division silently assumed it was not zero.
Multiplying by a denominator behaves like squaring: it can introduce a root that makes the denominator zero.
The fix for the first and third is the same: substitute every candidate into the original equation and discard failures.
The fix for the second is to never do it: move everything to one side and factor, which keeps the zero root visible.
Explain it
Two minutes, out loud.
Discussion prompt
A friend says checking answers is a waste of time because their algebra is good. How do you convince them otherwise for this specific question type?
Answer:
Agree with the premise and reject the conclusion: their algebra probably is fine, and that is not the issue.
Show them the square root of x equals negative 3. Ask them to solve it — they will square and get x equals 9, correctly.
Then ask them to check it. The square root of 9 is 3, not negative 3. The equation has no solutions, yet their perfect algebra produced one.
Make the point explicit: the check is not looking for their mistakes, it is looking for the method's mistakes. Squaring adds roots as a matter of course.
Then give the trigger, so it is not a general resolution to be careful: check whenever you square or clear a denominator, and only then.
Commit first
Commit before you check.
Predict first
How many real solutions does the system y equals x squared plus 2 and y equals x have?
Correct: none
Why: Setting them equal gives x squared plus 2 equals x, so x squared minus x plus 2 equals 0. The discriminant is 1 minus 8, which is negative 7 — negative, so there are no real solutions. Geometrically the parabola sits entirely above the line and they never meet. The discriminant answers a system question just as it answers a single-equation one, once you have substituted.
Connect it up
Blank paper.
Draw it
Draw a decision tree for solving nonlinear equations. Start with the equation shape at the top: quadratic, radical, rational, or system. For each branch, write the opening move — set to zero, isolate and square, note exclusions and clear, substitute. Then draw all four branches into one shared box labelled CHECK BACK IN THE ORIGINAL, and beside that box write which branches genuinely need it and why. In a separate corner, draw the discriminant with its three outcomes and a small parabola for each. Finally, list the two operations that change the solution set, with an arrow up for squaring and an arrow down for dividing by a variable.
Exit ticket
One question before you close the deck.
Predict first
You have squared both sides of a radical equation and factored cleanly to get two candidates. What do you do next?
Correct: Substitute both into the original equation
Why: Squaring is not reversible, so the candidates satisfy the squared equation by construction — testing them there proves nothing. Only the original equation can distinguish a real solution from one the squaring manufactured. Selecting both is the designed trap. And taking the positive one is a rule of thumb that happens to work often and fails whenever the right-hand side is an expression rather than a bare variable.
Recap
One type, one discipline: candidates are not solutions until the original equation says so.
| never do this | do this instead |
|---|---|
| Factor while the right side is still 6 | Move it across so the right side is zero |
| Trust candidates from a squared equation | Substitute each one into the original |
| Divide both sides by x | Factor out the x and keep the zero root |
| Write x equals 7 from x squared equals 49 | Write plus or minus 7 |
| Solve a how-many question | Compute the discriminant and stop |
| Stop at the x-values of a system | Find each partner y and re-read the stem |
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