Overview
Quadratic equations have the form ax² + bx + c = 0 and are among the most tested equation types in SAT Advanced Math. The SAT presents quadratic equations in word problems, algebraic manipulation tasks, and system contexts. Mastery requires fluency with at least three solution methods, plus the ability to choose the fastest approach under time pressure.
Key Points
1. Standard Form and Four Solution Methods
A quadratic equation in standard form is:
ax² + bx + c = 0 (a ≠ 0)
| Method | When to Use | Speed |
|---|---|---|
| Factoring | Integer coefficients, factors obvious | Fastest |
| Quadratic Formula | Always works, no factoring needed | Reliable |
| Completing the Square | Converts to vertex form; needed for derivations | Slower |
| Desmos (Digital SAT) | Complex coefficients, visual check | Instant |
2. Factoring Methods
GCF (Greatest Common Factor) first — always check:
2x² + 6x = 0 → 2x(x + 3) = 0 → x = 0 or x = -3
Trinomial factoring (a=1):
x² + 5x + 6 = 0 → (x + 2)(x + 3) = 0 → x = -2 or x = -3
(find two numbers that multiply to c and add to b)
Difference of squares:
x² - 9 = 0 → (x + 3)(x - 3) = 0 → x = ±3
3. Quadratic Formula
x = (-b ± √(b² - 4ac)) / 2a
Always identify a, b, c from standard form before substituting. The ± gives both solutions.
4. Discriminant (b² - 4ac)
| Discriminant | Number of Real Solutions | Graph |
|---|---|---|
| b² - 4ac > 0 | Two distinct real solutions | Crosses x-axis at 2 points |
| b² - 4ac = 0 | One repeated real solution | Tangent to x-axis |
| b² - 4ac < 0 | No real solutions | Does not cross x-axis |
The SAT frequently asks how many solutions a quadratic has — compute the discriminant without fully solving.
5. Vieta’s Formulas (Sum and Product of Roots)
For ax² + bx + c = 0 with roots r₁ and r₂:
Sum of roots: r₁ + r₂ = -b/a
Product of roots: r₁ · r₂ = c/a
These allow quick answers when the SAT asks “what is the product of the solutions” without solving the equation.
6. U-Substitution for Higher-Degree Equations
When an equation has the structure of a quadratic but with a different variable expression:
x⁴ - 5x² + 6 = 0
Let u = x², then: u² - 5u + 6 = 0 → (u-2)(u-3) = 0
u = 2 or u = 3 → x² = 2 or x² = 3 → x = ±√2 or ±√3
Pitfalls and Common Mistakes
Pitfall 1: Sign error in the quadratic formula Students write +b instead of –b, or forget to negate the whole expression. Fix: Write the formula out completely before substituting. Double-check the sign of b.
Pitfall 2: Dividing only part of the numerator by 2a
WRONG: x = (-6 ± √8) / 2 · 3 computed as x = -6 ± (√8 / 2) · 3
RIGHT: the entire numerator (-6 ± √8) is divided by 2a = 6
Fix: Use parentheses: x = (-6 ± √8) / (2·3).
Pitfall 3: Misreading the discriminant Students compute b² – 4ac correctly but misinterpret: “discriminant = 0” means ONE solution, not “no solution.” Fix: Memorize the three cases. Zero discriminant → one repeated root (the parabola is tangent to the x-axis).
Pitfall 4: Stopping at x-values when the problem asks for something else The SAT frequently asks for the value of an expression (like x + 3, or the sum of solutions), not x itself. Fix: Re-read the question after solving. Use Vieta’s formulas when the question asks for sums or products.
Pitfall 5: Forgetting to set the equation to zero first
WRONG: factoring x² + 5x = 6 as x(x+5) = 6
RIGHT: rewrite as x² + 5x - 6 = 0, then factor
Fix: Always move all terms to one side before factoring or applying the formula.
Related Entries
- Quadratic_Functions_Parabolas
- Systems_Nonlinear_Equations
- Polynomial_Operations_Factoring
- Functions_Concepts_Notation
- Radicals_Rational_Exponents
Quick Reference Card
| Formula / Rule | Expression |
|---|---|
| Standard form | ax² + bx + c = 0 |
| Quadratic formula | x = (-b ± √(b²-4ac)) / 2a |
| Discriminant | D = b² - 4ac |
| D > 0 | 2 real solutions |
| D = 0 | 1 repeated solution |
| D < 0 | No real solutions |
| Sum of roots | -b/a |
| Product of roots | c/a |
| Difference of squares | a² - b² = (a+b)(a-b) |
| Perfect square | (a+b)² = a²+2ab+b² |
| U-substitution trigger | f(u²) = 0 form |