SAT Math · Advanced Math

The quadratic formula

Q: What does the discriminant tell you?

The discriminant b² − 4ac tells you how many real solutions a quadratic has. Positive means two, zero means one, and negative means none, because you cannot take the real square root of a negative number.

Q: Is the quadratic formula on the SAT reference sheet?

No. The SAT reference sheet covers geometry formulas like areas, volumes, and special right triangles. The quadratic formula is not on it, so you need to memorize it before test day.

Q: When should you use the quadratic formula instead of factoring?

Factor when the numbers work out quickly, like x² − 5x + 6 = (x − 2)(x − 3). Use the formula when the quadratic does not factor neatly, when the answers involve square roots, or when you need the discriminant.

Q: How do you find the vertex of a parabola?

The vertex has x-coordinate −b ÷ 2a. Substitute that value back into the equation to get the y-coordinate. For y = x² − 2x − 3, x = 1 and y = −4, so the vertex is (1, −4).

Q: How can Aristotle help with SAT advanced math?

Aristotle is a voice AI SAT tutor. Your child talks through a quadratic out loud, and Aristotle asks guiding questions instead of giving the answer, starting from the step where their reasoning broke. It tracks 113 SAT Math skills, including nonlinear equations.

Solve any ax² + bx + c = 0, and use the discriminant to count solutions without solving.

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Updated October 30, 2026

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The solutions of $x^2 - 2x - 3 = 0$ are where the graph crosses the $x$ -axis: $-1$ and 3.

What is the quadratic formula?

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

It solves any $ax^2 + bx + c = 0$ straight from the three coefficients, including the equations that refuse to factor. On a graph, the two answers are the $x$ -values where the parabola crosses the $x$ -axis.

y = x² − 4x + 3 crosses at x = 1 and x = 3.

It is a center, plus or minus a step

x = \underbrace{-\dfrac{b}{2a}}_{\text{center}} \pm \underbrace{\dfrac{\sqrt{b^2 - 4ac}}{2a}}_{\text{step}}

A parabola is symmetric, so its two crossings sit the same distance either side of its axis of symmetry. The first term finds that axis. The root term is the step: go left for one answer, right for the other.

Axis x = 1, step 2: answers 1 − 2 = −1 and 1 + 2 = 3.

The discriminant decides how many answers

b^2 - 4ac \;\begin{cases} > 0 & \text{2 solutions} \\ = 0 & \text{1 solution} \\ < 0 & \text{0 real solutions} \end{cases}

The step is a square root, and the root is the only part that can fail. Positive inside: two crossings. Zero: the step vanishes and the vertex just touches the axis. Negative: no real step, so the parabola never reaches the axis.

The same parabola slid up: two crossings, then one, then none.

Where the formula comes from

Answer each question in your head, then scroll for the answer.

Think it through

Question 1
In $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , why are there usually two answers?
Answer
The $\pm$ : one answer adds the square root, the other subtracts it.
Question 2
What if $b^2 - 4ac$ is 0?
Answer
Adding or subtracting 0 gives the same number, so there is one solution: the vertex touches the $x$ -axis.
Question 3
And if it is negative?
Answer
No real square root, so no real solutions. The graph never crosses the $x$ -axis.

Try it yourself

Work it out first, then check your answer. SAT-style questions, not official College Board questions.

Question 1

Solve $2x^2 + 3x - 2 = 0$ .

Question 2

How many real solutions does $x^2 - 2x + 5 = 0$ have?

Question 3

For what value of $k$ does $x^2 + 6x + k = 0$ have exactly one real solution?

Question 4

What is the sum of the solutions of $x^2 - 6x - 7 = 0$ ?

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The formulas

Quadratic formula

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Solves any quadratic. Set the equation equal to 0 first.

Discriminant

$D = b^2 - 4ac$

Positive: two solutions. Zero: one. Negative: none.

Vertex

$x = -\dfrac{b}{2a}$

The vertex's $x$ -coordinate, and the axis of symmetry. Plug it in to get $y$ .

Sum and product of roots

$r_1 + r_2 = -\dfrac{b}{a} \qquad r_1 r_2 = \dfrac{c}{a}$

Use these when a question asks for the sum or product, not the solutions.

What the discriminant tells you
$b^2 - 4ac$	Real solutions	The graph
positive	two	crosses the $x$ -axis twice
zero	one	touches the $x$ -axis at its vertex
negative	none	never reaches the $x$ -axis

Try it

Shape the parabola

Change a, b, and c and watch the solutions appear and disappear.

b−2

c−3

Equationy = x² − 2x − 3

Discriminantb² − 4ac = 16, so two real solutions

Solutionsx = −1, x = 3

Vertex(1, −4)

Common mistakes

Not setting the equation to zero

$x^2 + 4x = 5$ becomes $x^2 + 4x - 5 = 0$ first, so $c = -5$ .

Dropping a negative sign

When $b$ is negative, $-b$ is positive, and $b^2$ is still positive. Write the signs in parentheses: $(-6)^2 = 36$ .

Dividing only part of the top by 2a

The whole numerator, $-b \pm \sqrt{b^2 - 4ac}$ , is divided by $2a$ , not just the square root.

Using a = 0 or the wrong a

In $3 - x^2 + 2x = 0$ , $a$ is $-1$ , not 3. Reorder before reading $a$ , $b$ , $c$ .

How it shows up on the SAT

Advanced Math is about 35% of SAT Math. Expect a quadratic that will not factor, a discriminant question, or the sum of the solutions.

What Aristotle tracks in nonlinear equations in one variable and systems of equations in two variables

8 SAT Math skills, including:

Solving quadratic equations in different forms
Solving absolute value, rational, and radical equations
Solving factored polynomial equations in one variable
Solving systems with linear and nonlinear equations

What is the quadratic formula?

For any equation ax² + bx + c = 0, the solutions are x = (−b ± √(b² − 4ac)) ÷ 2a. The ± gives the two solutions, one using plus and one using minus.

What does the discriminant tell you?

Is the quadratic formula on the SAT reference sheet?

When should you use the quadratic formula instead of factoring?

How do you find the vertex of a parabola?

How can Aristotle help with SAT advanced math?

More SAT concepts

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Read a line's slope and y-intercept straight from y = mx + b, and write one from a graph or two points.

Equation of a circle

Read the center and radius from (x − h)² + (y − k)² = r², and complete the square when the equation is expanded.

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