A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. Quadratic equations appear in physics (projectile motion), engineering (structural curves), finance (break-even analysis), and everyday problems like calculating areas. There are three main methods for solving them, and knowing which to use saves time and reduces errors.
Factoring is the fastest method when it works. The idea is to rewrite the quadratic as a product of two binomials, then set each factor equal to zero.
Example: Solve x² + 5x + 6 = 0
Find two numbers that multiply to 6 (the constant term) and add to 5 (the coefficient of x). Those numbers are 2 and 3. So x² + 5x + 6 = (x + 2)(x + 3) = 0. Setting 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.
When to use factoring: Factoring works best when the coefficients are small integers and the equation factors cleanly. If you cannot find integer factors within 30 seconds, switch to the quadratic formula. Use the Quadratic Equation Solver to check your work or solve any quadratic instantly.
The quadratic formula works for every quadratic equation, regardless of whether it factors neatly. For ax² + bx + c = 0:
x = (−b ± √(b² − 4ac)) ÷ 2a
Example: Solve 2x² + 7x − 15 = 0
Here a = 2, b = 7, c = −15. Plugging in: x = (−7 ± √(49 + 120)) ÷ 4 = (−7 ± √169) ÷ 4 = (−7 ± 13) ÷ 4. This gives x = 6/4 = 1.5 and x = −20/4 = −5.
The expression under the square root, b² − 4ac, is called the discriminant. It tells you the nature of the solutions before you finish calculating:
| Discriminant Value | Number of Solutions | What It Means |
|---|---|---|
| b² − 4ac > 0 | 2 real solutions | The parabola crosses the x-axis twice |
| b² − 4ac = 0 | 1 real solution | The parabola touches the x-axis at its vertex |
| b² − 4ac < 0 | No real solutions | The parabola never crosses the x-axis (solutions are complex numbers) |
Completing the square converts any quadratic into vertex form, revealing both the solutions and the shape of the parabola. While less commonly used for raw solving, it is essential for deriving the quadratic formula itself, understanding the vertex of a parabola, and working with conic sections in advanced math.
Example: Solve x² + 6x + 2 = 0
Move the constant: x² + 6x = −2. Take half the coefficient of x (which is 3), square it (9), and add to both sides: x² + 6x + 9 = 7. The left side is now a perfect square: (x + 3)² = 7. Take the square root: x + 3 = ±√7. Solve: x = −3 ± √7, giving x ≈ −0.354 and x ≈ −5.646.
| Method | Best When | Advantages | Limitations |
|---|---|---|---|
| Factoring | Small integer coefficients | Fastest; no calculator needed | Only works if equation factors cleanly |
| Quadratic Formula | Any quadratic equation | Always works; gives exact answers | More arithmetic; easy to make sign errors |
| Completing the Square | Finding vertex; deriving formulas | Reveals parabola structure | More steps than other methods |
When you throw a ball upward, its height follows a quadratic equation: h(t) = −16t² + v₀t + h₀, where v₀ is the initial velocity in feet per second and h₀ is the starting height. If you throw a ball upward at 64 ft/s from ground level, h(t) = −16t² + 64t. Setting h = 0 to find when it lands: −16t(t − 4) = 0, giving t = 0 (launch) and t = 4 seconds (landing). Maximum height occurs at the vertex: t = −64/(2 × −16) = 2 seconds, h(2) = 64 feet.
A company’s profit function is often quadratic: P(x) = −2x² + 100x − 800, where x is units sold in thousands. Setting P(x) = 0 and solving finds the break-even points: x = (−100 ± √(10000 − 6400)) ÷ −4 = (−100 ± 60) ÷ −4, giving x = 10 and x = 40. The company breaks even at 10,000 and 40,000 units, with maximum profit at x = 25 (the vertex).
A farmer has 200 feet of fencing and wants to enclose a rectangular area against an existing wall. If the width is x, the length is 200 − 2x, and the area is A = x(200 − 2x) = −2x² + 200x. Maximum area occurs at the vertex: x = −200/(2 × −2) = 50 feet, giving dimensions of 50 × 100 feet = 5,000 square feet.
Forgetting the ± in the quadratic formula. The ± means there are (usually) two solutions. Students who write only the positive root miss the second answer.
Sign errors with negative coefficients. In −3x² + 12x − 9 = 0, a = −3, not 3. Getting the sign of a wrong flips the entire calculation. It helps to write out a, b, and c explicitly before substituting into the formula.
Not simplifying the discriminant first. Compute b² − 4ac as a separate step before putting it under the square root. This reduces errors and lets you check the discriminant’s sign to predict the number of solutions.
Dividing by 2a incorrectly. The entire numerator (−b ± √discriminant) is divided by 2a, not just part of it. Using parentheses helps: x = (−b ± √D) / (2a).
The Equation Solver can verify your solutions for quadratics and other equation types.
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