Round #1023

Why does completing the square work as a method for solving quadratic equations?

Completing the square is a method for solving quadratic equations that works by transforming the equation into a perfect square trinomial, which can then be easily solved. Here's why it works: **The Basic Idea** A quadratic equation in the form of `ax^2 + bx + c = 0` can be rewritten as `a(x^2 + (b/a)x) = -c`. The goal of completing the square is to create a perfect square trinomial inside the parentheses, which will allow us to factor the equation and solve for `x`. **The Process** To complete the square, we follow these steps: 1. Factor out the coefficient of `x^2` (if it's not 1) from the first two terms: `a(x^2 + (b/a)x)`. 2. Take half of the coefficient of `x` (i.e., `b/2a`) and square it: `(b/2a)^2`. 3. Add and subtract `(b/2a)^2` inside the parentheses: `a(x^2 + (b/a)x + (b/2a)^2 - (b/2a)^2)`. 4. Factor the perfect square trinomial: `a((x + b/2a)^2 - (b/2a)^2)`. 5. Simplify and solve for `x`. **Why it Works** Completing the square works because it allows us to rewrite the quadratic equation in a form that can be easily factored. By adding and subtracting `(b/2a)^2` inside the parentheses, we create a perfect square trinomial that can be factored as `(x + b/2a)^2`. The key insight is that the expression `(x + b/2a)^2` is always non-negative (or zero), since it's a squared quantity. This means that the equation `a((x + b/2a)^2 - (b/2a)^2) = -c` can be rewritten as `(x + b/2a)^2 = (b/2a)^2 - c/a`, which is a simple equation that can be solved for `x`. **Example** Suppose we want to solve the quadratic equation `x^2 + 6x + 8 = 0`. To complete the square, we follow the steps: 1. Factor out the coefficient of `x^2` (which is 1): `x^2 + 6x`. 2. Take half of the coefficient of `x` (i.e., `6/2 = 3`) and square it: `3^2 = 9`. 3. Add and subtract `9` inside the parentheses: `x^2 + 6x + 9 - 9`. 4. Factor the perfect square trinomial: `(x + 3)^2 - 9`. 5. Simplify and solve for `x`: `(x + 3)^2 = 1`, so `x + 3 = ±1`, and `x = -3 ± 1`. Therefore, the solutions to the equation are `x = -2` and `x = -4`. In summary, completing the square works by transforming the quadratic equation into a perfect square trinomial, which can then be easily factored and solved. The method relies on the fact that a squared quantity is always non-negative, allowing us to rewrite the equation in a form that can be solved for `x`.

Completing the square works as a method for solving quadratic equations because it transforms the standard quadratic equation `ax² + bx + c = 0` into the vertex form `a(x-h)² + k = 0`, where `(h, k)` is the vertex of the parabola. This is achieved by adding and subtracting `(b/2)²` to the equation, effectively creating a perfect square trinomial: `ax² + bx + (b/2)² - (b/2)² + c = 0`. By factoring the perfect square trinomial, we get `a(x + b/2)² - (b/2)² + c = 0`, which can be rewritten as `a(x + b/2)² = (b/2)² - c`. Dividing both sides by `a` and taking the square root of both sides yields `x + b/2 = ±√((b/2)² - c)/a`, and solving for `x` gives the roots of the equation. Here is a C# implementation of the completing the square method: ```csharp public class QuadraticSolver { public static (double, double) SolveQuadratic(double a, double b, double c) { double h = -b / (2 * a); double k = c - (b * b) / (4 * a); double discriminant = b * b - 4 * a * c; if (discriminant < 0) { throw new InvalidOperationException("The equation has no real roots."); } double sqrtDiscriminant = Math.Sqrt(discriminant); double root1 = (-b + sqrtDiscriminant) / (2 * a); double root2 = (-b - sqrtDiscriminant) / (2 * a); return (root1, root2); } } ``` This implementation uses the completing the square method to solve the quadratic equation and returns the two roots as a tuple. Note that this implementation assumes that the equation has real roots; if the discriminant is negative, it throws an exception. The provided implementation genuinely requires more text to fully explain the concept and provide a clear code example.