The content half of SAT Math prep for a student already scoring high: the three quadratic forms and the discriminant, counting solutions of linear and nonlinear systems, percent change and exponential growth, function notation and transformations, two-way tables and statistical claims, circles and the complementary-angle identity — each presented as the one trap that still costs points, with the tell that gives it away.
Subject: SAT Prep · 67 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
SAT Prep · content session
One trap per topic — the places a 700-plus score still leaks
Objectives
At this level the remaining misses are almost never a topic you have not met. They are a small set of recognisable traps, one or two per content area, that are engineered to reward the fast wrong answer.
College Board — Digital SAT Suite Assessment Specifications, Math section content domains — the four content domains this deck covers
Section
Section 1
Warm-up
Two minutes with your last practice test, if you have it to hand.
Discussion prompt
Take the last five math questions you got wrong. For each one, was it a topic you did not know, or something else? Name the something else.
Hint: Be specific. 'Careless' is not a category you can do anything with.
Answer:
Almost every answer at this level is: I solved for the wrong quantity, I picked the answer that was one step short, or I made an arithmetic slip while hurrying.
Those three have completely different fixes, and none of the fixes is 'learn more maths'.
Knowing which of the three is yours is worth more tomorrow morning than any new topic.
Khan Academy — Official Digital SAT Prep, Math — the official practice platform reports misses by skill for exactly this reason
Picture it
Rough, but the ordering is what matters and it holds up across students.
Figure (svg): Five bars ranking the causes of remaining misses, with misreading the question and falling for a designed trap far ahead of not knowing the topic
Which is why this deck is organised as one trap per topic rather than as a content review.
Concept
The wrong options are not random. On a well-made item, one of them is exactly what you get if you stop one step early, and another is what you get if you solve for the other variable.
So an answer appearing in the list is not evidence you are right. It is the default, and it is engineered.
College Board — Digital SAT Suite Assessment Specifications, Math section content domains — distractor design in the assessment specifications
Sorting
Same four categories from the chart.
Sort into buckets
Sort each description.
The first two are the same size as each other and together they are most of it.
Pattern
Four questions, in this order, before every answer you commit to. It costs about four seconds and it is where the last thirty points live.
On a section where you have finished with minutes to spare, this check is the highest-value thing you can do with those minutes.
College Board — Digital SAT Suite — and see the answer-review deck for the full three-pass system
Check
Read it twice before answering.
Check your understanding
If 3x minus 7 equals 14, what is the value of 6x minus 14?
Answer: A
Why: The target is exactly twice the given expression, so double both sides: 6x minus 14 is 2 times 14, which is 28. Solving for x first also works and gives x equal to 7, then 42 minus 14 equals 28.
Section
Section 2
Concept
A quadratic can be written three ways, and the test chooses the form that hides what it is asking about.
\[ y = x^2 - 6x + 5 \qquad \text{standard} \]
\[ y = (x - 1)(x - 5) \qquad \text{factored} \]
\[ y = (x - 3)^2 - 4 \qquad \text{vertex} \]
Standard gives the y-intercept. Factored gives the roots. Vertex gives the minimum and the axis of symmetry. Converting is the whole skill.
OpenStax, Algebra and Trigonometry 2e Ch. 5 — the three forms and the conversions between them
Matching
If you can do this instantly, half of the Advanced Math domain becomes fast.
Match the pairs
Why: The last one has two routes, and the second is faster when the quadratic is already factored: the axis sits exactly halfway between the roots. For the example in this deck the roots are 1 and 5, and the axis is at 3.
Worked example
Convert the standard form to vertex form and read off the minimum.
Take half the coefficient of the linear term, and square it
Why: Half of negative six is negative three, and its square is nine. That nine is what turns the first two terms into a perfect square.
\[ x^2 - 6x + 9 = (x-3)^2 \]
Add and subtract that nine so nothing changes
Why: Adding nine and subtracting nine leaves the expression equal to what it was.
\[ y = (x^2 - 6x + 9) - 9 + 5 \]
\[ y = (x-3)^2 - 4 \]
Read the vertex straight off
Why: The vertex is at three, negative four, and since the leading coefficient is positive that is a minimum.
Verify: by substituting three into the original
Why: Nine minus eighteen plus five is negative four. The two forms agree at the vertex, which is the fastest possible check.
Picture it
Roots from the factored form, vertex from the vertex form, and the axis halfway between the roots.
Figure (svg): A parabola crossing the horizontal axis at one and five with its vertex marked at three comma negative four and a dashed axis of symmetry
Notice the axis sits at 3, the average of 1 and 5. That shortcut is worth knowing on test day.
Trap
Completing the square, in a hurry.
\[ y = x^2 - 6x + 5 = (x-3)^2 + 5 \]
Add the nine and carry on
Why: The nine was added but never taken back, so the whole expression is nine too large.
Report a minimum of 5
Why: The check catches it instantly: substituting three into the original gives negative four, not five.
Add it and subtract it in the same line.
\[ y = (x^2 - 6x + 9) - 9 + 5 = (x-3)^2 - 4 \]
Write the minus nine before simplifying anything
Why: Writing it immediately is what stops it being forgotten. Do it as one motion.
Always substitute the vertex input back
Why: Ten seconds, and it catches every version of this error.
Concept
The part under the square root in the quadratic formula decides the number of real solutions on its own.
\[ b^2 - 4ac \]
Positive gives two, zero gives exactly one, and negative gives none. Questions asking for the value of a constant that makes a quadratic have exactly one solution are asking you to set it to zero.
Picture it
The sign of the discriminant is the same information as how the curve meets the axis.
Figure (svg): Three parabolas showing a positive discriminant crossing the axis twice, a zero discriminant touching once, and a negative discriminant never crossing
Worked example
For what value of k does the equation with leading coefficient k, linear coefficient 8 and constant 4 have exactly one real solution?
Recognise the phrase
Why: Exactly one real solution means the discriminant is zero. No solving is needed.
\[ b^2 - 4ac = 0 \]
Substitute the three coefficients
Why: Here a is k, b is eight and c is four.
\[ 64 - 16k = 0 \]
\[ k = 4 \]
Verify: by substituting k back
Why: The equation becomes four x squared plus eight x plus four, which is four times the square of x plus one. Its only root is negative one, so there is exactly one solution.
Picture it
The verified quadratic touches the axis at exactly one point.
Figure (svg): A parabola touching the horizontal axis at negative one and rising on both sides, illustrating a single repeated root
Prediction
Do not solve it.
Predict first
How many real solutions does the equation with a equal to 2, b equal to 3 and c equal to 5 have?
Correct: None.
Why: The discriminant is nine minus forty, which is negative thirty-one. A negative discriminant means the parabola never reaches the horizontal axis, so there are no real solutions. This takes about four seconds and questions of this type are designed to eat a minute if you solve.
Worked example
Find every point where the line and the parabola intersect.
\[ y = x^2 - 4 \qquad \text{and} \qquad y = 3x \]
Set the two expressions equal
Why: At an intersection both give the same output for the same input, which is what lets you eliminate y.
\[ x^2 - 4 = 3x \]
Move everything to one side and factor
Why: Never divide by x here — doing so loses a solution.
\[ x^2 - 3x - 4 = 0 \Rightarrow (x-4)(x+1) = 0 \]
Find the matching outputs
Why: Substitute each input into the easier of the two equations, which is the line.
\[ (4, 12) \quad \text{and} \quad (-1, -3) \]
Verify: in the harder equation too
Why: Four squared minus four is twelve, and one minus four is negative three. Both points satisfy the parabola as well, so both are genuine.
Picture it
A line and a parabola meet at zero, one or two points, and the algebra says which.
Figure (svg): A parabola and a straight line crossing at the points four comma twelve and negative one comma negative three
If the combined quadratic has a negative discriminant, the line misses the parabola entirely — which is another way the same question gets asked.
Fill the middle
Setting a line equal to a parabola. The setup and the answer are given.
Fill in the blanks
x^2 - 4 = 3x \;\Rightarrow\; x^2 - 3x - 4 = 0 \;\Rightarrow\; (x-4)(x+1) = 0
Why: Subtracting 3x from both sides gives the standard form, and the factors must multiply to negative four and add to negative three, which gives negative four and positive one. The roots are the inputs of the intersection points.
Section
Section 3
Concept
Two linear equations either cross once, never cross, or are the same line. Slope and intercept decide which, and you can tell without solving anything.
| slopes | intercepts | solutions |
|---|---|---|
| different | anything | exactly one |
| same | different | none |
| same | same | infinitely many |
OpenStax, Algebra and Trigonometry 2e Ch. 9 — systems of linear equations
Picture it
Parallel and identical look similar in the algebra and are opposite in the answer.
Figure (svg): Two coordinate planes, the first with two parallel lines that never meet and the second with two coincident lines drawn on top of each other
Worked example
For what value of k do these two equations describe the same line?
\[ 3x + ky = 8 \qquad \text{and} \qquad 6x + 10y = 16 \]
Compare the coefficients of the first variable
Why: Six is twice three, so the second equation must be exactly twice the first, term for term.
Apply the same factor to the middle term
Why: Twice k must be ten.
\[ 2k = 10 \Rightarrow k = 5 \]
Check the constant too
Why: Twice eight is sixteen, which matches. If it had not matched, no value of k would give infinitely many solutions and the answer would be none instead.
Verify: by doubling the first equation
Why: Doubling gives six x plus ten y equals sixteen, which is the second equation exactly. Same line, so infinitely many solutions.
Picture it
Getting the variable coefficients to line up is only two thirds of the job.
Figure (svg): Three bars comparing the first equation doubled against the second equation term by term, with the constant term highlighted as the deciding check
If the first two match and the constant does not, the answer flips from infinitely many to none. The test asks both versions.
Trap
Solving a system by elimination and reaching a statement with no variables.
\[ 0 = 0 \]
Conclude there is no solution
Why: This statement is always true, so every point on the line works.
Answer zero solutions
Why: Exactly backwards, and it is a coin flip students lose about half the time.
Read what the leftover statement actually says.
\[ 0 = 0 \;\Rightarrow\; \text{infinitely many} \]
\[ 0 = 7 \;\Rightarrow\; \text{no solution} \]
Ask whether the statement is true
Why: A true statement means the equations agreed all along, so every solution of one solves the other. A false one means they contradict.
Say it out loud
Why: Zero equals zero is true, so the system is fine. Zero equals seven is false, so nothing works.
Check
Solve it on paper before you click.
Check your understanding
The system is 2x plus 3y equals 6 and 4x plus 6y equals 15. How many solutions does it have?
Answer: A
Why: Doubling the first equation gives 4x plus 6y equals 12, which contradicts 15. Same slopes, different intercepts, so the lines are parallel and never meet.
Section
Section 4
Concept
Increasing by twenty percent means multiplying by 1.2. Decreasing by twenty percent means multiplying by 0.8. Working with multipliers rather than with added amounts removes most of the errors in this domain.
\[ A = P(1 + r)^t \]
Khan Academy — Official Digital SAT Prep, Math — problem solving and data analysis, percentages
Prediction
A price rises by twenty percent, then falls by twenty percent.
Predict first
Where does it end up relative to where it started?
Correct: Four percent lower.
Why: The two multipliers are 1.2 and 0.8, and their product is 0.96. The percentages are taken from different bases: the rise is twenty percent of the original, and the fall is twenty percent of the larger new amount, so the fall is bigger in absolute terms.
Picture it
The second twenty percent is taken from a bigger number than the first.
Figure (svg): Three bars showing a starting value of one hundred, rising to one hundred and twenty, then falling to ninety-six rather than back to one hundred
The same trap appears as a discount followed by a tax, and as a population that grows then shrinks by the same percentage.
Worked example
A population of 2400 grows by three percent each year. What is it after ten years?
Turn the percentage into a multiplier
Why: Three percent growth means multiplying by 1.03 once per year.
\[ A = 2400(1.03)^{10} \]
Evaluate the power
Why: The tenth power of 1.03 is about 1.3439.
\[ A \approx 2400 \times 1.3439 = 3225 \]
Notice it is not thirty percent
Why: Ten years of three percent is not thirty percent, because each year's growth is computed on the previous total.
Verify: against the simple-interest estimate
Why: Thirty percent of 2400 would be 720, giving 3120. The compound answer is larger, at 3225, which is the right direction. If your answer had come out below 3120, something is wrong.
Picture it
Each bar is three percent taller than the one before it.
Figure (svg): Eleven bars rising from two thousand four hundred to three thousand two hundred and twenty-five over ten years of three percent growth
That widening is the whole difference between exponential and linear, and it is what the test is checking.
Discrimination
The wording tells you which model, and picking wrong makes every later step wrong.
Sort into buckets
Which model does each describe?
Section
Section 5
Concept
Whatever sits inside the brackets replaces every appearance of the variable in the definition. Nothing more mysterious than that.
\[ f(x) = 2x + 1, \qquad g(x) = x^2 \]
\[ f(g(3)) = f(9) = 19 \]
Work from the inside out, always. Reversing the order gives a different answer and the reversed answer will be among the choices.
OpenStax, Algebra and Trigonometry 2e Ch. 3 — function composition
Prediction
Using the same two functions.
Predict first
What is g(f(3))?
Correct: 49.
Why: Inside first: f(3) is 7, and then g(7) is 49. Composing the other way gives 19, and that value will be sitting in the answer choices waiting for anyone who worked outside-in.
Worked example
Given a function f, describe the graph of the function that subtracts three inside the brackets and adds two outside.
\[ g(x) = f(x - 3) + 2 \]
Handle the outside change first, because it behaves as expected
Why: Adding two outside moves every output up by two.
Now the inside change, which behaves backwards
Why: Subtracting three inside moves the graph three to the right, not to the left.
Say why the inside is backwards
Why: To get the output that f used to give at an input of zero, the new function needs an input of three. So every feature arrives three units later.
\[ (0, 0) \;\longrightarrow\; (3, 2) \]
Verify: with a single point
Why: If f is the squaring function then f of zero is zero, and g of three is f of zero plus two, which is two. The vertex moved from the origin to three, two, exactly as predicted.
Picture it
The dashed curve is the original; the solid one is the transformation.
Figure (svg): Two parabolas, the original with its vertex at the origin and the transformed one with its vertex at three comma two, joined by a dashed arrow
Trap
Describing the graph of the function with three subtracted inside the brackets.
Read the minus sign as 'move left'
Why: It looks like a subtraction, so leftwards feels right.
Shift left three
Why: Every point is now six units from where it should be, and both the correct and the reflected answers are among the choices.
Ask which input now produces the old output.
\[ g(3) = f(0) \]
Notice that the new function needs a larger input to do what f did
Why: Needing a larger input means the feature appears further to the right.
Check with one point rather than memorising
Why: The memorised rule fails under pressure; a single substituted point never does.
Matching
Four transformations of the same function.
Match the pairs
Why: Outside the brackets acts on the output and behaves as written: plus is up, minus is a vertical flip. Inside the brackets acts on the input and behaves backwards: plus is left, minus is a horizontal flip.
Section
Section 6
Concept
The whole difficulty is which total goes on the bottom of the fraction. The words given that, of those who, and among tell you.
| Passed | Failed | Total | |
|---|---|---|---|
| Group A | 18 | 6 | 24 |
| Group B | 21 | 15 | 36 |
| Total | 39 | 21 | 60 |
Khan Academy — Official Digital SAT Prep, Math — two-variable data and probability
Worked example
Two questions that look almost identical and have different answers.
Given that a student is in Group B, what is the probability they passed?
Why: Given that fixes the group, so the denominator is the Group B total.
\[ \frac{21}{36} = \frac{7}{12} \approx 0.583 \]
Given that a student passed, what is the probability they are in Group B?
Why: Now the condition is passing, so the denominator is the passing total.
\[ \frac{21}{39} = \frac{7}{13} \approx 0.538 \]
Notice the numerator never changed
Why: Both questions concern the same twenty-one people. Only the population being conditioned on moved.
Verify: that both are below one and near a half
Why: Twenty-one out of thirty-six and twenty-one out of thirty-nine both sit just above and just below sixty percent, which is the right size. A denominator of sixty would have been the unconditional version, and that is the third distractor.
Picture it
The twenty-one people who are in Group B and passed, measured against three different populations.
Figure (svg): Three bars showing twenty-one out of thirty-six, twenty-one out of thirty-nine and twenty-one out of sixty as three different probabilities from the same cell
Concept
Standard deviation measures how far the values sit from their own mean. Two data sets with the same mean can have very different spreads, and the test asks you to compare them by eye rather than compute.
A margin of error narrows when the sample gets larger. It says nothing about whether the sample was collected fairly, which is what the evaluating-claims questions are really about.
College Board — Digital SAT Suite Assessment Specifications, Math section content domains — problem solving and data analysis
Two truths and a lie
A survey of 200 randomly selected students at one school finds a mean study time of 9.4 hours per week, with a margin of error of 0.6 hours.
Eliminate the wrong options
Which conclusion is defensible?
Survives elimination: A
Why: Two rules cover almost every question in this domain: the interval is about the mean of the population the sample was drawn from, and random selection is what licenses generalising, but only to that population.
Check
Solve it on paper before you click.
Check your understanding
Using the table, what is the probability that a randomly chosen student who failed is in Group A?
Answer: A
Why: The condition is that the student failed, so the population is the 21 students who failed. Six of them are in Group A.
Section
Section 7
Worked example
Find the centre and radius of the circle given in expanded form.
\[ x^2 + y^2 - 6x + 8y = 0 \]
Group the terms by variable
Why: Keep the x terms together and the y terms together, with the constant on the right.
Complete the square for each variable separately
Why: Half of negative six squared is nine; half of eight squared is sixteen. Add both to the right as well.
\[ (x-3)^2 + (y+4)^2 = 0 + 9 + 16 \]
\[ (x-3)^2 + (y+4)^2 = 25 \]
Read the centre and radius
Why: The centre is three, negative four, and the radius is the square root of twenty-five, which is five.
Verify: by testing a point on the circle
Why: The origin satisfies the original equation, and its distance from three, negative four is the square root of nine plus sixteen, which is five. The origin is on the circle, as it must be.
Picture it
Centre and radius, both hidden in the expanded form.
Figure (svg): A circle of radius five centred at three comma negative four with the radius drawn as a segment
That sign flip is the trap here, and it is the same reasoning as the horizontal shift in the transformation section.
Trap
Reading the completed square form.
\[ (x-3)^2 + (y+4)^2 = 25 \]
Report the centre as negative three, positive four
Why: The signs were copied from the page rather than read from the form.
Also report the radius as twenty-five
Why: The right-hand side is the square of the radius, not the radius.
Both values are the opposite of what they look like.
\[ \text{centre } (3, -4), \quad r = 5 \]
The centre is whatever makes each bracket zero
Why: x minus three is zero at three; y plus four is zero at negative four.
The radius is the square root of the right-hand side
Why: Twenty-five is r squared, so r is five.
Concept
In a right triangle the two non-right angles add to ninety degrees, and each one's sine is the other one's cosine.
\[ \sin(a) = \cos(90^{\circ} - a) \]
Questions that hand you a sine and ask for a cosine of a different angle are almost always testing exactly this, and they can be answered without finding either angle.
OpenStax, Algebra and Trigonometry 2e Ch. 7 — right triangle trigonometry
Picture it
A three-four-five triangle, with the two acute angles labelled.
Figure (svg): A right triangle with legs three and four and hypotenuse five, with the two acute angles marked as complementary
So if the sine of one acute angle is three fifths, the cosine of the other is three fifths too, with no angle ever computed.
Check
Solve it on paper before you click.
Check your understanding
In a right triangle, the sine of angle a is 0.6. What is the cosine of the other acute angle?
Answer: A
Why: The two acute angles are complementary, and the sine of one equals the cosine of the other. So the cosine of the other angle is also 0.6, with no computation at all.
Section
Section 8
Concept
The built-in graphing calculator is fastest when you already know the answer and want it confirmed, and slowest when you are using it to avoid thinking.
The one genuinely time-saving move: type each side of an equation as its own function and read the intersection. That solves anything, including the equations that resist algebra.
Desmos graphing calculator, the version built into Bluebook — the same calculator as the one in Bluebook
Picture it
The quadratic from section two, solved by intersection instead of by factoring.
Figure (svg): The Bluebook Desmos panel with two equations entered and their intersection points listed alongside the graph
Worth thirty seconds when the algebra is ugly. Worth nothing when it is not — typing takes longer than factoring a friendly quadratic.
Trade off
Fill in the missing judgements.
Comparison matrix
| situation | by hand | on the calculator |
|---|---|---|
| factorable quadratic | about 15 seconds | about 30 seconds of typing |
| ugly decimals in a system | slow and error-prone | fast and exact |
| checking an answer you already have | re-doing the work | fast, and independent of your method |
| a question about the form of an expression | the only route | no help at all |
The last row matters: questions asking which expression is equivalent cannot be graphed your way out of, and they are common at the hard end.
Pattern
This is the card to reread in the morning. One line per trap.
| topic | the trap | the tell |
|---|---|---|
| any question | solving for the wrong quantity | the last line asks for 2x, not x |
| completing the square | adding without subtracting | the vertex does not check out |
| quadratics | solving when the discriminant would do | the phrase exactly one solution |
| systems | reading zero equals zero as no solution | no variables left in the statement |
| percent change | applying both percentages to the original | up then down by the same percent |
| transformations | shifting the wrong way | the change is inside the brackets |
| two-way tables | the wrong denominator | the words given that or among |
| circles | copying the centre's signs | the bracket reads x minus h |
College Board — Digital SAT Suite — and the pacing deck for how to fit all of this into the clock
Elimination
You solved for x correctly and selected an answer that turned out to be wrong.
Eliminate the wrong options
Which single habit would have caught this?
Survives elimination: A
Why: This whole category of miss is invisible to every other fix, because the work was correct. The only thing that catches it is comparing the answer against what was actually asked.
Exit ticket
Answer honestly — this decides the last thirty minutes of revision.
Predict first
Of the eight traps on the card, which one have you personally fallen for most recently?
Correct: Whichever you named is the one to reread tonight, and the only one.
Why: The night before a test is not the time for breadth. One trap, reread until you could describe it to someone else, is worth more than a skim of all eight — and it is far better for sleep, which matters more tomorrow than any of this.
Connect it up
Fifteen minutes, tonight, then stop.
Draw it
Write the eight traps out from memory, one line each, with the tell that gives each one away. Circle the two you had to look up.
Read the two circled ones once more in the morning. Nothing else. Then go in and use the four-second check on every question.
Recap
Content is not the constraint at your level. Recognition is, and recognition is what this deck drilled.
| question type | the fast route | the answer here |
|---|---|---|
| vertex of x squared minus 6x plus 5 | complete the square | (3, -4) |
| one solution, k x squared plus 8x plus 4 | discriminant equals zero | k = 4 |
| line meets parabola | set equal, factor | (4, 12) and (-1, -3) |
| infinitely many solutions | scale term by term | k = 5 |
| 2400 at 3 percent for 10 years | multiplier to the power | about 3225 |
| circle from expanded form | complete both squares | centre (3, -4), r = 5 |
College Board — Digital SAT Suite — official practice, if you want one more set tonight — but one, not three
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