In this guide
The anatomy of a quadratic
A quadratic equation is any equation where the highest power of the variable is 2. The standard form is:
ax² + bx + c = 0
Its graph is a parabola — a symmetric U-shaped curve. Everything the SAT asks about quadratics is really a question about this curve: where it crosses the x-axis (the roots or solutions), where its highest or lowest point sits (the vertex), and which direction it opens. When a is positive the parabola opens up and has a minimum; when a is negative it opens down and has a maximum.
The single most useful habit is matching the form of the quadratic to the question. Factored form hands you the roots. Vertex form hands you the vertex. Standard form hands you the y-intercept. Rather than grinding through algebra, ask: which form already contains the answer?
Method 1: Factoring
Factoring is the fastest route when the numbers are friendly. You rewrite the quadratic as a product of two binomials, then use the fact that if a product equals zero, at least one factor must be zero.
Solve x² + 5x + 6 = 0.
Find two numbers that multiply to +6 (the constant) and add to +5 (the middle coefficient). Those numbers are 2 and 3.
Write the factors: (x + 2)(x + 3) = 0
Set each factor to zero: x + 2 = 0 gives x = −2, and x + 3 = 0 gives x = −3.
The solutions are x = −2 and x = −3. These are exactly where the parabola crosses the x-axis.
The sum-and-product trick — find two numbers that multiply to c and add to b — is worth drilling until it's instant. It also runs in reverse: if a question tells you the roots, you can rebuild the equation.
Method 2: The quadratic formula
When a quadratic won't factor cleanly, the quadratic formula always works. Memorize it — it is not on the SAT reference sheet:
x = ( −b ± √(b² − 4ac) ) / 2a
Solve 2x² − 4x − 3 = 0.
Identify a = 2, b = −4, c = −3.
Substitute carefully (watch the signs): x = ( 4 ± √(16 + 24) ) / 4 = ( 4 ± √40 ) / 4
Simplify √40 = 2√10, so x = ( 4 ± 2√10 ) / 4 = 1 ± (√10)/2
Two irrational solutions — the kind factoring could never have found.
Time-saver
Before reaching for the formula, glance at the numbers. If a and c are small and the middle term looks tidy, try factoring first — it's usually faster. Save the formula for when factoring stalls.
Vertex form and the parabola
Vertex form is the SAT's favorite because it reveals the turning point of the parabola at a glance:
y = a(x − h)² + k
Here (h, k) is the vertex — the minimum point if the parabola opens up, the maximum if it opens down. Notice the sign trap: the vertex form y = (x − 3)² + 4 has its vertex at (3, 4), not (−3, 4). The x-coordinate flips sign coming out of the parentheses.
You can also find the vertex straight from standard form. The axis of symmetry — and therefore the x-coordinate of the vertex — sits at:
x = −b / 2a
Plug that x back into the equation to get the y-coordinate. This shortcut answers "what is the maximum height" and "how many seconds until the ball peaks" word problems without any graphing.
The discriminant: counting solutions
Sometimes the SAT doesn't want the solutions — it wants to know how many real solutions exist. The piece under the square root in the quadratic formula, b² − 4ac, is called the discriminant, and it tells you everything:
- b² − 4ac > 0: two distinct real solutions. The parabola crosses the x-axis twice.
- b² − 4ac = 0: exactly one real solution. The parabola just touches the x-axis at its vertex.
- b² − 4ac < 0: no real solutions. The parabola never touches the x-axis.
When a question gives a quadratic with an unknown coefficient and asks for the value that produces exactly one solution, set the discriminant equal to zero and solve. That's the entire problem.
Practice this topic
Quadratics reward pattern recognition, and pattern recognition comes from volume. Work enough of them that you can name the right method within a few seconds of reading the question.
- Drill the Advanced Math question bank with worked explanations.
- Review the full Advanced Math topic guide.
- Keep the formula sheet open while you practice.