Digital SAT Prep · Digital SAT Reading, Writing & Math
Advanced Math
Nonlinear equations, quadratics, polynomials, exponentials and functions.
Seven concepts on the nonlinear toolkit the Digital SAT Advanced Math domain expects — quadratics, vertex form, the quadratic formula, Vieta's shortcuts, exponentials and function notation. Each card states the idea, works a real SAT-style problem, and flags the trap that costs marks.
- Digital SAT Prep
- Hard level
- 7 concepts
- 5 practice questions
1The discriminant counts real roots
The discriminant D = b^2 - 4ac tells you how many real roots a quadratic ax^2+bx+c=0 has before you solve it. When D>0 the parabola crosses the x-axis twice, when D=0 it just touches once, and when D<0 it never reaches the axis. On the SAT, exactly one solution is the signal to set D=0 and solve for the unknown parameter — you never need to find the root itself.
Figure. When c = 9 the parabola y = x^2 - 6x + 9 touches the x-axis at x = 3 and nowhere else — the graph of D = 0. Lift c above 9 and the whole curve sits above the axis; drop c below 9 and it crosses twice.
How it works
- Read the coefficientsWrite the equation as ax^2+bx+c=0 and identify a, b, and c, including signs.
- Compute DEvaluate D = b^2 - 4ac. A negative middle term still squares to a positive contribution.
- Match sign to rootsTwo roots if D>0, one repeated root if D=0, none if D<0. For a parameter, set D equal to the count you need.
| Sign of D | Real roots | Graph |
|---|---|---|
| D > 0 | two distinct | crosses the axis twice |
| D = 0 | one repeated | touches the axis once |
| D < 0 | none | stays above or below the axis |
Find c for exactly one root
For what value of c does x^2 - 6x + c = 0 have exactly one real solution?
- One root means D = 0; here a=1, b=-6D = 36 - 4c
- Set 36 - 4c = 04c = 36
- Solve for cc = 9
Pro tip. Exactly one solution always means D=0, not that you should factor or use the quadratic formula. The repeated root is x = -b/(2a) = 3 here, but the SAT question only wanted c.
If x^2 - 6x + c = 0 has no real solutions, which must be true of c?
- c < 9
- c = 9
- c > 9
- c = 0
No real roots means D < 0, so 36 - 4c < 0, hence c > 9. At c = 9 the discriminant is zero and there is one repeated root.
2Factoring before the formula
When ax^2+bx+c has small integer coefficients, factoring is faster than the quadratic formula. Find two numbers that multiply to ac and add to b, split the middle term, then pull out a common factor from each pair. Each factor set to zero gives a root. If no clean pair exists after a quick scan, switch to the formula — do not grind through a messy factorisation under time pressure.
Figure. The graph crosses the axis at x=2 and x=3 — exactly the roots from (x-2)(x-3)=0. A parabola that factors over the integers always has x-intercepts at those integer roots.
How it works
- Target the productFor x^2+bx+c, hunt two integers whose product is c and sum is b. For ax^2+bx+c with a\neq 1, the product target is ac.
- Split and groupRewrite the middle term using your pair, group into two binomials, and factor each group.
- Zero-product ruleIf (x-r_1)(x-r_2)=0, then x=r_1 or x=r_2. Each bracket gives one root.
Solve by factoring
What are the solutions to x^2 - 5x + 6 = 0?
- Need two numbers: product 6, sum -5-2 and -3
- Factor as (x-2)(x-3)=0roots x=2 or x=3
- Check: (2)^2-5(2)+6 and (3)^2-5(3)+60 in both cases
Pro tip. On the SAT, try factoring first whenever the constant term has only small factors. The question bank repeats the same product-sum pairs — 6 with sum 5, 6 with sum -5, 15 with sum -8 — so a ten-second scan often finishes the problem.
Which factorisation is equivalent to x^2 - 9 = 0?
- (x-3)(x-3)=0
- (x-3)(x+3)=0
- (x+9)(x-1)=0
- (x-9)(x+1)=0
x^2-9 is a difference of squares: (x-3)(x+3)=0, giving x=\pm 3. The repeated factor (x-3)^2 would yield only x=3, and the other pairs do not expand to x^2-9.
3Vertex form and the extreme value
Completing the square rewrites y=ax^2+bx+c as y=a(x-h)^2+k, exposing the vertex (h,k) directly. The axis of symmetry is the vertical line x=h, and a controls direction: a>0 gives a minimum at (h,k), a<0 a maximum. On the SAT, minimum or maximum value questions almost always want the output at the vertex, not the input that produces it — find h=-b/(2a) only when you need to evaluate f(h).
Figure. The lowest point sits at (4,-1) — below the axis because -1<0. Every minimum-value question on a upward-opening parabola asks for that y-coordinate.
How it works
- Locate the axisFor ax^2+bx+c, the axis of symmetry is x = -b/(2a). That x-value is h in vertex form.
- Evaluate at the vertexSubstitute x=h into the original expression. The output is k, the minimum or maximum.
- Read the sign of aa>0 means the parabola opens up and k is a minimum; a<0 means k is a maximum.
Minimum value of a quadratic
The function f is defined by f(x)=x^2-8x+15. What is the minimum value of f(x)?
- Axis: x = -(-8)/(2\cdot 1)x = 4
- Evaluate: f(4) = 4^2 - 8(4) + 1516 - 32 + 15 = -1
- Completing the square check: (x-4)^2 - 1minimum -1 at x=4
Pro tip. The answer to a minimum-value question is the y-coordinate at the vertex, here -1, not the x-coordinate 4. If the options list only x-values, you still need to evaluate f at that x.
For g(x)=-2(x+1)^2+5, which statement is true?
- The vertex is (1,5) and 5 is a minimum
- The vertex is (-1,5) and 5 is a maximum
- The vertex is (-1,5) and 5 is a minimum
- The axis of symmetry is y=5
Vertex form a(x-h)^2+k gives vertex (h,k)=(-1,5). Here a=-2<0, so the parabola opens down and 5 is a maximum, not a minimum.
4The quadratic formula
When factoring fails, x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} always works for ax^2+bx+c=0 with a\neq 0. The \pm produces two roots when D>0, one when D=0, and complex roots when D<0 — though the Digital SAT sticks to real roots. Simplify the radical before dividing: factor out perfect squares so the final answer stays in lowest radical form.
Figure. When integer factoring fails, the quadratic formula is the reliable next move — same roots, different route.
How it works
- Standard formMove everything to one side so the equation reads ax^2+bx+c=0. Identify a, b, c with signs.
- Compute D and \sqrt{D}Evaluate D=b^2-4ac. Simplify the square root — e.g. \sqrt{12}=2\sqrt{3}.
- Substitute and reducePlug into x=\dfrac{-b\pm\sqrt{D}}{2a} and cancel any common factor in the fraction.
Irrational roots
Solve x^2 + 4x + 1 = 0.
- a=1, b=4, c=1; D = 16 - 4D = 12
- \sqrt{12} = 2\sqrt{3}; x = \dfrac{-4 \pm 2\sqrt{3}}{2}x = -2 \pm \sqrt{3}
- Two distinct roots because D=12>0x = -2+\sqrt{3} or x = -2-\sqrt{3}
Pro tip. Always divide -b and \sqrt{D} by 2a together — here dividing top and bottom by 2 turns -4\pm 2\sqrt{3} into -2\pm\sqrt{3}. Skipping that cancellation is the most common algebra slip on radical answers.
How many real solutions does x^2 + x + 5 = 0 have?
- Two
- One
- None
- Infinitely many
D = 1 - 20 = -19 < 0, so there are no real solutions. The quadratic formula would involve \sqrt{-19}, which is not a real number.
5Sum and product without solving
For ax^2+bx+c=0 with roots r_1 and r_2, Vieta's relations give r_1+r_2=-b/a and r_1 r_2=c/a — no factoring or formula required. The SAT often asks for a sum or product directly, or for a new quadratic whose roots are a translate of the original. Expand (x-r_1)(x-r_2) and match coefficients when you need the equation back from its roots.
Figure. Vieta maps coefficients straight to sum and product. Reach for these when the stem asks for r1+r2 or r1 r2 without the roots themselves.
How it works
- Read a, b, cFrom ax^2+bx+c=0, identify the three coefficients including signs.
- Apply VietaSum =-b/a. Product =c/a. Divide by a even when a\neq 1.
- Build a new equation if neededGiven roots p and q, the monic quadratic is (x-p)(x-q)=0, i.e. x^2-(p+q)x+pq=0.
| Quantity | Formula | From x^2-5x+6=0 |
|---|---|---|
| Sum of roots | -b/a | 5 |
| Product of roots | c/a | 6 |
| Monic equation from roots p,q | x^2-(p+q)x+pq=0 | (x-2)(x-3)=0 |
Sum and product
For 2x^2 - 10x + 12 = 0, find the sum and product of the roots without solving.
- a=2, b=-10, c=12coefficients identified
- Sum = -(-10)/25
- Product = 12/26
Pro tip. Divide by a first mentally: dividing the whole equation by 2 gives x^2-5x+6=0, where the sum 5 and product 6 are visible from the coefficients. The full Vieta formulas are the general version of that shortcut.
If the roots of x^2 + kx + 12 = 0 have product 12 and sum -7, what is k?
- k = 7
- k = -7
- k = 12
- k = -12
Vieta gives sum =-b/a=-k. Setting -k=-7 yields k=7. The product c/a=12 is already consistent. Answering k=-7 treats the sum as equal to k instead of -k.
6Exponential growth and decay
A quantity modeled by y=a\,b^x starts at y=a when x=0 and changes by the fixed factor b each time x increases by 1. Growth happens when b>1; decay when 0<b<1. In percent form, b=1+r for growth rate r and b=1-r for decay — subtract 1 from the base to read the percent. Doubling time and half-life questions reduce to finding when b^x equals 2 or \tfrac{1}{2}.
Figure. Each unit step to the right multiplies the height by 1.08 — that fixed ratio is what makes the curve exponential rather than linear. The starting height 500 is the coefficient a.
How it works
- Identify a and bMatch the equation to y=a\,b^x. The coefficient a is the initial value; the base b is the per-step multiplier.
- Classify growth or decayb>1 means growth; 0<b<1 means decay. Compare b to 1, not to 0.
- Convert to a percent rateWrite b=1+r. Then r as a percent is the per-period change — e.g. b=1.08 means 8% growth per period.
Reading the growth factor
A population is modeled by P = 500(1.08)^t, with t in years. What does 1.08 represent?
- Form P = a\,b^t with a=500, b=1.08initial population 500
- b = 1 + r gives r = 1.08 - 1r = 0.08
- Interpret as a percent per year8% annual growth
Pro tip. In a\,b^t, subtract 1 from the base to read the percent rate: 1.08\Rightarrow 8\% growth. A base of 0.85 would mean 15% decay, because 0.85=1-0.15.
If 2^{x+3}=64, what is 3^x?
- 27
- 81
- 9
- 243
Since 64=2^6, match exponents to get x+3=6, so x=3. Then 3^x=3^3=27. The trap is stopping at x=3 without evaluating the expression the question actually asked for.
7Function notation and zeros
f(x) is the output when x is substituted into the rule — so f(3) means replace every x with 3 and simplify. The zeros of f are the x-values where f(x)=0, the same as the x-intercepts of its graph. Factored form f(x)=(x-r_1)(x-r_2) exposes zeros directly: set each factor to zero. For nested expressions like f(g(x)), evaluate the inner function first, then feed that output into f.
Figure. Function notation is substitution: feed the input into the rule and simplify. Zeros are the inputs that make the output zero — the x-intercepts of the graph.
How it works
- Evaluate f(a)Substitute x=a everywhere in the rule. Follow order of operations — exponents before multiplication.
- Find zerosSet f(x)=0 and solve. In factored form, each factor (x-r)=0 gives a zero r.
- Compose inside outFor f(g(x)), compute g(x) first, then apply f to that result.
Evaluate and locate zeros
If f(x)=2x^2-3, what is f(4), and what are the zeros of f?
- f(4) = 2(4^2) - 32(16)-3 = 29
- Set 2x^2-3=0x^2 = \tfrac{3}{2}
- Take square rootsx = \pm\sqrt{\tfrac{3}{2}}
Pro tip. Squaring comes before multiplying by 2 in f(4): 2(4^2)-3, not (2\cdot 4)^2-3. For zeros of 2x^2-3, the graph crosses the axis symmetrically at \pm\sqrt{3/2}.
If g(x)=x^2+2x, what is g(-2)?
- 0
- 4
- -4
- 8
g(-2)=(-2)^2+2(-2)=4-4=0. The trap is forgetting that (-2)^2=4, not -4.
Notes
- Quadratic Solutions: A quadratic ax^2+bx+c=0 is solved by factoring, completing the square, or the quadratic formula; the discriminant b^2-4ac tells how many real roots exist.
- Vertex Form: y=a(x-h)^2+k has vertex (h,k); the axis of symmetry is x=h and a controls direction and width.
- Exponential Growth/Decay: y = a\cdot b^x grows when b>1 and decays when 0<b<1; a is the initial value.
- Function Notation and Roots: The zeros of f(x) are the x-intercepts; f(x)=(x-r_1)(x-r_2) shows roots directly as r_1 and r_2.
Formulas
- Quadratic formula: x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}
- Discriminant: D = b^2 - 4ac (two real roots if D>0, one if D=0, none if D<0)
- Sum and product of roots: r_1+r_2 = -\dfrac{b}{a}, r_1 r_2 = \dfrac{c}{a}
- Vertex of a parabola: x = -\dfrac{b}{2a}
- Exponential model: y = a\,b^x with growth factor b = 1 + r
Exam traps & shortcuts
- Use the discriminant to answer 'how many solutions' questions without solving the whole quadratic.
- When a quadratic must have exactly one solution, set b^2 - 4ac = 0 and solve for the unknown parameter.
- For root questions, recall r_1+r_2=-b/a and r_1 r_2=c/a to find sums or products without computing each root.
Reference tables
Every line here should be recoverable from the concepts above, not merely recalled.
| Tool | Formula | Watch for |
|---|---|---|
| Quadratic formula | x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} | Simplify \sqrt{D} before dividing by 2a |
| Discriminant | D = b^2 - 4ac | Two / one / none real roots when D>0 / =0 / <0 |
| Vertex x | x = -\dfrac{b}{2a} | Minimum-value questions want f(x) there, not x alone |
| Vertex form | y = a(x-h)^2 + k | Vertex (h,k); axis x=h |
| Sum of roots | r_1+r_2 = -\dfrac{b}{a} | Works without finding either root |
| Product of roots | r_1 r_2 = \dfrac{c}{a} | Divide by a, not just c |
| Exponential model | y = a\,b^x | Percent rate: b=1+r; subtract 1 from b |
| Zeros from factors | f(x)=(x-r_1)(x-r_2) | Zeros are r_1 and r_2 directly |
Recap
Read only this the night before.
- Discriminant
- D=b^2-4ac counts roots before you solve. Exactly one solution means D=0 — solve for the parameter, not the root.
- Factoring
- Product-sum pair, then zero-product rule. No clean pair in ten seconds? Reach for the formula.
- Vertex
- x=-b/(2a) locates the extreme; the minimum or maximum value is f at that x, usually written as k in vertex form.
- Formula
- x=\dfrac{-b\pm\sqrt{D}}{2a} always works. Simplify the radical, then cancel with 2a.
- Vieta
- Sum =-b/a, product =c/a. The SAT asks for these without making you find the roots first.
- Exponentials
- y=a\,b^x: a is the start, b the multiplier. b=1.08 means 8% growth per step.
- Functions
- f(a) means substitute. Zeros are x-intercepts. Compose inside out: g first, then f.
Practise Advanced Math
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