What Is a Quadratic Equation?
A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b, and c are constants and a is not equal to zero. The term "quadratic" comes from the Latin word "quadratus," meaning square, because the variable is squared (raised to the power of 2).
Quadratic equations appear everywhere in real life. When you throw a ball in the air, its height over time follows a quadratic path. Engineers use them to design bridges and arches. Economists use them to model profit and cost functions. Understanding how to solve them is one of the most important skills in algebra.
The graph of a quadratic equation is called a parabola — a U-shaped curve that either opens upward (when a > 0) or downward (when a < 0). The solutions to the equation, also called roots or zeros, are the x-values where the parabola crosses the x-axis.
Method 1: Factoring
Factoring is usually the fastest method when it works. The idea is to rewrite the quadratic expression as a product of two binomials, then use the zero product property: if AB = 0, then either A = 0 or B = 0.
To factor x² + 5x + 6 = 0, you need two numbers that multiply to 6 (the constant term) and add to 5 (the coefficient of x). Those numbers are 2 and 3, so the equation factors as (x + 2)(x + 3) = 0. Setting each factor equal to zero gives x = -2 or x = -3.
For equations where the leading coefficient is not 1, like 2x² + 7x + 3 = 0, you can use the AC method. Multiply a and c (2 × 3 = 6), then find two numbers that multiply to 6 and add to 7. Those are 6 and 1. Rewrite the middle term: 2x² + 6x + x + 3 = 0. Factor by grouping: 2x(x + 3) + 1(x + 3) = 0, which gives (2x + 1)(x + 3) = 0. The solutions are x = -1/2 and x = -3.
When Factoring Doesn't Work
Not every quadratic can be factored neatly with integers. For example, x² + 3x + 1 = 0 has irrational roots. When you cannot find integer factors, move on to the quadratic formula or completing the square. A quick check: compute the discriminant b² - 4ac. If it is not a perfect square, the equation will not factor over the integers.
Method 2: The Quadratic Formula
The quadratic formula works for every quadratic equation, no matter what. For ax² + bx + c = 0, the solutions are x = (-b ± √(b² - 4ac)) / (2a). This single formula handles every case — rational roots, irrational roots, and even complex roots.
Let's solve 3x² - 5x - 2 = 0. Here a = 3, b = -5, and c = -2. The discriminant is (-5)² - 4(3)(-2) = 25 + 24 = 49. Since 49 is a perfect square (√49 = 7), we get rational solutions: x = (5 + 7) / 6 = 2 or x = (5 - 7) / 6 = -1/3.
Now try x² + 2x - 7 = 0. The discriminant is 4 + 28 = 32, which is not a perfect square. The solutions are x = (-2 ± √32) / 2 = (-2 ± 4√2) / 2 = -1 ± 2√2. These are irrational numbers, approximately 1.83 and -3.83.
Understanding the Discriminant
The discriminant (b² - 4ac) tells you exactly how many real solutions exist before you even solve the equation. If the discriminant is positive, there are two distinct real roots. If it equals zero, there is exactly one repeated real root (the parabola just touches the x-axis). If it is negative, there are no real roots — only two complex conjugate roots.
This is extremely useful on exams. Before spending time solving, compute the discriminant to know what kind of answer to expect.
Method 3: Completing the Square
Completing the square transforms the equation into the form (x + p)² = q, which can be solved by taking square roots. This method is also how the quadratic formula itself is derived, and it is essential for converting quadratic functions to vertex form.
To solve x² + 6x + 5 = 0 by completing the square: move the constant to the right side to get x² + 6x = -5. Take half the coefficient of x (which is 3), square it (which is 9), and add it to both sides: x² + 6x + 9 = 4. The left side is now a perfect square: (x + 3)² = 4. Take the square root of both sides: x + 3 = ±2. So x = -1 or x = -5.
When a is not 1, divide the entire equation by a first. For example, 2x² + 8x + 6 = 0 becomes x² + 4x + 3 = 0 after dividing by 2. Then proceed as before: x² + 4x = -3, add 4 to both sides to get (x + 2)² = 1, so x = -1 or x = -3.
How to Choose the Right Method
Start by trying to factor — it is the fastest when it works. If the equation has simple coefficients and the discriminant is a perfect square, factoring will likely succeed. If factoring does not work within about 30 seconds, switch to the quadratic formula. Use completing the square when you specifically need the vertex form of a quadratic function, or when a problem explicitly asks for it.
On standardized tests like the SAT and ACT, most quadratic equations are designed to be factorable. In physics and engineering courses, the quadratic formula is more commonly needed because the coefficients come from real measurements and rarely produce clean factors.
Common Mistakes to Avoid
The most common mistake is forgetting that the quadratic formula has a ± sign, meaning there are usually two solutions. Always check both. Another frequent error is sign mistakes when plugging negative values into the formula — be especially careful with (-b) when b is already negative.
Students also often forget to set the equation equal to zero before solving. If you have x² + 3x = 10, you must first rewrite it as x² + 3x - 10 = 0 before factoring or using the formula.
Finally, always verify your solutions by substituting them back into the original equation. This catches arithmetic errors and confirms your answer is correct.
Practice Problems
Try solving these on your own, then check your work: (1) x² - 7x + 12 = 0 [Answer: x = 3 or x = 4]. (2) 2x² + 3x - 5 = 0 [Answer: x = 1 or x = -5/2]. (3) x² + 4x + 1 = 0 [Answer: x = -2 ± √3]. (4) 4x² - 12x + 9 = 0 [Answer: x = 3/2, a repeated root].
If you get stuck on any of these, snap a photo and let ScanSolve walk you through the solution step by step.
