Strong test-takers know what kind of question they’re holding before they start. Learn the 16 types, then name the type of fresh practice questions and solve them with the right first move.
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16
Types
Equivalent expressions Rewrite it, don’t solve it
Combine & distribute
Bands 1 to 2
(3x2+2x−5)−(x2−4x+1)
Distribute the minus sign: 3x2+2x−5−x2+4x−1
Group like terms: (3x2−x2)+(2x+4x)+(−5−1)
Answer2x2+6x−6
Equivalent expressions trace the same curve. Slide x and watch the values agree.
At x=1: original =2, simplified =2 (equal), the mistake gives −6.
Spot it“equivalent to”parentheses everywhereadd, subtract or multiply expressions
First moveDistribute first, then group like terms.
Watch outA minus sign before parentheses flips every term inside.
Factor & rewrite
Band 2
Factor x2−x−12
Find two numbers that multiply to −12 and add to −1: −4 and 3
Write the factors: (x−4)(x+3)
Answer(x−4)(x+3)
Move the two roots. The factored form and the x-intercepts always agree.
x2−x−12=(x−4)(x+3). The graph crosses the x-axis at 4 and −3.
Spot it“a factor of”difference of squaresvertex form
First movePull out the common factor, or find two numbers that multiply to c and add to b.
Watch outExpand your answer to check it.
Match the constants
Band 7
(2x+3)(x+4)=2x2+kx+12 for all x. Find k.
Expand: 2x2+8x+3x+12=2x2+11x+12
Match the x terms: kx=11x
Answerk=11
Slide k until the red curve lands exactly on the blue one. That is the value that works for all x.
(2x + 3)(x + 4)2x² + 6x + 12
k=6 gives a different curve, so the equation is not true for all x.
Spot it“true for all x”“where a, b, c are constants”asks for the value of a, b, k or ab
First moveExpand one side, then match like powers term by term.
Watch outx² matches x², x matches x, numbers match numbers.
Exponents, fractions & radicals
Bands 5 to 7SAT only
For x>0, write 3x5 with an exponent.
A cube root is the exponent 31
(x5)1/3=x5/3
Answerx5/3
Change the root and the power. The curve is x raised to (power / root).
x^(5/3)y = x
3x5=x5/3. At x=4 both give 10.1. The power goes on top, the root on the bottom.
Spot it√ or a fractional exponentx in a denominator“for positive x”
First moveTurn roots into fractional exponents. Use a common denominator for fractions.
Watch outIn x^(a/b), the top is the power and the bottom is the root.
Equations & systems Find the x (or the how-many)
Solve a quadratic
Band 4
Solve x2−2x−8=0
Factor: (x−4)(x+2)=0
Set each factor to 0: x=4 or x=−2
Answerx=4 or x=−2
Solutions are where the graph meets the x-axis. Move the roots and compare their sum.
x2−2x−8=0 has solutions x=4 and x=−2. Their sum is 2, which is −ab.
Spot it“positive solution”“sum / product of the solutions”x² in the equation
First moveSet it equal to 0, then factor or use the quadratic formula.
Watch outSum of solutions = −b/a and product = c/a, so you may not need to solve.
How many solutions?
Band 7
x2−6x+c=0 has exactly one real solution. Find c.
One solution means the discriminant is 0: b2−4ac=0
(−6)2−4(1)c=0⇒36−4c=0
Answerc=9
Slide c. The discriminant tells you how many times the parabola meets the x-axis.
x2−6x+12=0: b2−4ac=36−4(12)=−12. Negative: no real solutions (the graph never reaches the axis).
Spot it“no real solutions”“exactly one solution”line touches a parabola at one point
First moveb² − 4ac: positive is two, zero is one, negative is none.
Watch outLine meets a parabola once? Set the equations equal, then discriminant = 0.
Radical, rational & absolute value
Bands 2 to 4Not on PSAT 8/9
Solve x+4=x−2
Square both sides: x+4=x2−4x+4, so x2−5x=0 and x=0 or x=5
Check x=0: 4=2, but x−2=−2. Extraneous.
Answerx=5
Squaring creates a second, fake solution. The hollow point is the extraneous one.
√(x + 4)−√(x + k) (appears when you square)y = x − 2
Squaring x+4=x−2 gives roots 0,5. Real solution: x=5. The hollow point sits on the lower branch, where x−2 is negative, so a square root can never equal it.
Spot it√ around the xx in a denominator| |
First moveIsolate it, then square both sides, multiply by the denominator, or split into ±.
Watch outPlug answers back in. Squaring can create fake solutions.
Systems & intersections
Band 4
y=x2−2 and y=x+4. Find the solutions.
Set them equal: x2−2=x+4, so x2−x−6=0
(x−3)(x+2)=0, so x=3 or x=−2
Put each into the line: y=7 and y=2
Answer(3,7) and (−2,2)
A solution of a system is a point on both graphs. Move the line and count.
y = x² − 2y = x + 4
Set equal: x2−2=x+4, so x2−x−6=0 (discriminant 25). Two intersection points, so two solutions.
Spot it“system of equations”“ordered pair”“intersect”
First moveSubstitute one equation into the other.
Watch outRe-read what it asks for last: x, y, or xy?
System with unknown constants
Bands 5 to 7
2ax+by=22 and ax+2by=17. What is the value of ax+by?
Add the equations: 3ax+3by=39
Divide by 3: ax+by=13
Answer13
Spot it“a and b are constants”matching coefficients like 2ax + by and ax + 2byasks for ax + by, not x and y
First moveAdd or subtract the equations to build the combination the question asks for.
Watch outYou often never need a, b, x, or y by themselves.
Rearrange a formula
Bands 3 to 7
Solve V=31πr2h for h.
Multiply both sides by 3: 3V=πr2h
Divide both sides by πr2
Answerh=πr23V
Spot it“in terms of”“correctly expresses”a formula with 3+ letters
First moveUndo operations in reverse order to isolate the target.
Watch outDo the same thing to every term on both sides.
Functions Graphs, tables, models, f(x)
Quadratic: vertex & max/min
Bands 4 to 6Not on PSAT 8/9
h(t)=−4t2+16t+3 models a ball's height. Find the maximum height.
The parabola opens down, so the vertex is the maximum
t=−2(−4)16=2
h(2)=−16+32+3=19
AnswerMaximum height 19, at t=2
The vertex is the maximum. Change a and the peak time and see where it lands.
h(t)=−4t2+16t+3. The vertex is at t=−2(−4)16=2, and the maximum height is 19.
Spot it“maximum” or “minimum”“vertex”height after t seconds
First moveVertex form shows the vertex. Otherwise x = −b/2a.
Watch outThe max VALUE is the y of the vertex. The input is the x.
Exponential growth & decay
Bands 4 to 6Not on PSAT 8/9
A population of 500 grows 20% each year. Write the function.
Growth multiplies by 1+0.20=1.2 each year
Start value 500, base 1.2, exponent t years
AnswerP(t)=500(1.2)t
Change the percent. Above 0% it grows, below it decays. The base is 1 plus the rate.
P(t)=500(1.2)t. Growing 20% a year: base 1+0.2.
Spot it“doubles every”“% per year”an initial value
First movey = a(1 ± r)^t. The start is a, the base is 1 ± the rate.
Watch outBase is 1.03, not 0.03. The time unit must match the exponent.
Read a graph or table
Bands 1 to 6
A table shows p(2)=0. What must be a factor of p(x)?
A value of x where p(x)=0 is a zero (an x-intercept)
A zero at x=2 means (x−2) is a factor
Answer(x−2)
Move the zero. Wherever the table shows p(x) = 0, that x gives a factor.
xp(x)−3−55−2−24−1−90−41−320311p(2)=0, so the graph crosses the x-axis at x=2 and (x−2) is a factor.
Spot it“graph shown”a table of values“x-intercept”, “zero”, “factor”
First movey-intercept: x = 0. x-intercept: y = 0. A zero at a means (x − a) is a factor.
Watch outAn x-intercept has y = 0, not x = 0.
Evaluate, compose & shift
Bands 3 to 6Not on PSAT 8/9
How is the graph of g(x)=(x−3)2+2 related to y=x2?
(x−3) moves the graph 3 units right
+2 moves the graph 2 units up
Answer3 right, 2 up
Move h and k. The dashed curve is y = x². The vertex tells you the shift.
y = x²y = (x − h)² + k
y=(x−3)2+2: 3 right, 2 up. Note that (x−h) moves the graph right when h is positive.
Spot it“what is f(3)?”f(x) + 2 or f(x + 2)“translated up / down”
First moveSubstitute exactly what is inside the parentheses.
Watch outf(x) + 2 moves up. f(x + 2) moves left.
Interpret in context
Band 4
C(w)=25w+60 is a gym's total cost after w weeks. What do 25 and 60 mean?
25 multiplies w: dollars added each week
60=C(0): the cost before any weeks (the sign-up fee)
Answer25 dollars per week; 60 dollar sign-up fee
Change the numbers. The multiplier is the rate (the slope). The constant is where the graph starts.
C(w)=25w+60. The 25 is how much the cost rises each week. The 60 is the cost at w=0.
Spot it“best interpretation”“which statement”“what does the number mean”
First moveName each variable with its units, then say what the number multiplies.
Watch outAnswer in the problem’s units, not in “math words.”
Build it from the situation
Bands 4 to 6Not on PSAT 8/9
A rectangle's length is 3 more than its width. Its area is 40. Write an equation.
Let the width be w. The length is w+3
Area = length × width
Answerw(w+3)=40
The solution of the equation is where the area curve reaches the target area.
area = w(w + 3)target area 40
Width w, length w+3. Area =w(w+3)=40. Solving gives w≈5.
Spot ita rectangle, area or sequence story“which equation represents”no equation given
First moveDefine a variable, translate each phrase, then test one number.
Watch outTry a simple number in your equation before you choose.