SAT Math Quadratic Equations with Methods and SAT Strategy

Factoring Method for Solving Quadratic Equations (Fastest SAT Strategy)

When the SAT gives you a quadratic with clean, small integers, factoring is almost always the quickest path to the answer. The goal is to rewrite ax² + bx + c as a product of two binomials, then set each factor equal to zero.

Standard approach for x² + bx + c = 0

Find two numbers that multiply to c and add to b. Write the equation as (x + p)(x + q) = 0, then solve x = −p and x = −q.

Solve x² − 5x + 6 = 0

• Find two numbers that multiply to 6 and add to −5, those are −2 and −3.
• So (x − 2)(x − 3) = 0, which gives x = 2  or  x = 3

Key tip

On the SAT, if the quadratic has a leading coefficient other than 1, look for factoring by grouping or check whether the answer choices are simple fractions. If the roots are messy, switch to the quadratic formula immediately.

When does factoring NOT work? When the discriminant is not a perfect square, the numbers will not factor neatly. That is your signal to move on.

Using the Quadratic Formula to Solve Any Quadratic Equation

The quadratic formula solves any quadratic equation, no matter how messy the numbers are.

x = (−b ± √(b² − 4ac)) / 2a

This applies whenever the equation is in the form ax² + bx + c = 0.

When to use it on the SAT

• Factoring does not produce clean integers within about 20 seconds.
• The problem gives decimal or fractional coefficients.
• You need the exact value of the roots to match an answer choice.

How to avoid arithmetic errors under time pressure

Arithmetic mistakes inside the square root are the most common way students lose points here. Use this three-step process every time.

Step 1 – Write out a, b, and c explicitly before substituting anything.

Step 2 – Calculate b² − 4ac as a separate operation. Write the number down.

Step 3 – Substitute the result into the full formula.

Solve 2x² − 3x − 2 = 0

• a = 2, b = −3, c = −2
• b² − 4ac = (−3)² − 4(2)(−2) = 9 + 16 = 25
• x = (3 ± √25) / 4 = (3 ± 5) / 4  →  x = 2  or  x = −1/2

Key tip

Always write the sign of b carefully. A negative b means −b becomes positive in the formula. This single step causes more errors than any other part of the calculation.

Completing the Square to Solve Quadratic Equations

Completing the square to solve quadratics appears less often than the other two methods on the SAT. When it does appear, the question is almost always asking about vertex form or the minimum or maximum value of a quadratic.

Convert ax² + bx + c into a(x − h)² + k

Solve x² + 6x + 5 = 0

Step 1 – Move constant
x² + 6x = −5

Step 2 – Add square term
x² + 6x + 9 = 4

Step 3 – Factor
(x + 3)² = 4

Step 4 – Square root
x + 3 = ±2

Step 5 – Solve
x = −1 or x = −5

Key tip

On the SAT, completing the square is most useful when a question asks for the vertex of a parabola or asks “what is the minimum value of f(x)?” Convert to vertex form a(x − h)² + k, and the vertex is (h, k). The minimum value is k when a > 0.

f(x) = a(x − h)² + k  →  Vertex at (h, k)

Reading Parabola Features from a Graph

Several SAT questions give you a graph and ask about key features. Here is what to look for.

Vertex

The vertex is the highest or lowest point on the parabola. If the parabola opens upward (a > 0), the vertex is the minimum. If it opens downward (a < 0), the vertex is the maximum. In vertex form, h is the x-coordinate and k is the y-coordinate of the vertex.

Axis of Symmetry

x = h   or equivalently   x = −b / 2a

Key tip

On the SAT, if you know the two x-intercepts (roots) of a parabola, the axis of symmetry is exactly halfway between them – x = (x₁ + x₂) / 2

Roots (x-intercepts)

The roots are the points where the parabola crosses the x-axis. A parabola can have two roots, one root (touching the x-axis), or no real roots. Reading roots from a graph means identifying the x-values where y = 0.

y-intercept

Plug x = 0 into the equation. In standard form ax² + bx + c, the y-intercept is always (0, c).

The Discriminant Formula, Answer Questions Without Solving

The discriminant is the expression inside the square root of the quadratic formula. You do not need to finish the quadratic formula to use it.

Discriminant = b² − 4ac

b² − 4ac > 0  →  Two distinct real roots (parabola crosses x-axis at two points)

b² − 4ac = 0  →  Exactly one real root (vertex touches the x-axis)

b² − 4ac < 0  →  No real roots (parabola does not cross the x-axis)

Key tip

SAT questions often ask “how many times does the graph intersect the x-axis?” or “for what value of k does the equation have exactly one solution?” Both are answered using the discriminant, no full solving required.

Find k for exactly one real solution

For x² − 6x + k = 0 to have exactly one real solution

Set b² − 4ac = 0 

(−6)² − 4(1)(k) = 0  →  36 − 4k = 0  →  k = 9

Advanced SAT Quadratic Equation Problems (Solved Step-by-Step)

Question 1

The function f is defined by f(x) = ax² + bx + c, where the vertex is (4, −2) and it passes through (2, 10). A second function g is defined by g(x) = f(x − 3) + 5. What is the y-coordinate of the y-intercept of the graph of g?

A. 145  B. 150  C. 153  D. 160

Solution

Step 1 – Build f(x) using vertex form.

Since the vertex is (4, −2)

f(x) = a(x − 4)² − 2

Step 2 – Find a using the point (2, 10).

10 = a(2 − 4)² − 2

12 = a(−2)² = 4a   →   a = 3

f(x) = 3(x − 4)² − 2

Step 3 – Find g(0) to get the y-intercept of g.

g(0) = f(0 − 3) + 5 = f(−3) + 5

Step 4 – Evaluate f(−3).

f(−3) = 3(−3 − 4)² − 2 = 3(−7)² − 2 = 3(49) − 2 = 147 − 2 = 145

Step 5 – Final answer.

g(0) = 145 + 5 = 150

Answer – B. 150

Key rule

For complex transformations, first find the specific equation of the base function using the vertex and a given point. Then, to find the y-intercept of any transformed function, always substitute x = 0 into the new expression.

Why the other choices are wrong

A (145) is the value of f(−3) but forgets to add the vertical shift of +5. C (153) results from a sign error when squaring the negative number inside the bracket. D (160) results from incorrectly applying the horizontal shift inside the square.

Think10x.ai video explaining vertex form and function transformations (Finding y-intercept)