"Could you please explain how this derivation is obtained step by step? In particular, how is the substitution 𝑋=𝑌−（𝑏/2𝑎），​made—what is the reasoning behind this assumption? Please focus on explaining how this idea comes about."---**Title:** The Substitution Method **Chapter Title:** Chapter 1 Linear Substitution **Example 1.** **Question Stem:** Solve the quadratic equation: ax² + bx + c = 0 by the substitution x = y - (b / 2a), where a, b, and c are real numbers with a ≠ 0. **Solution:** x₁,₂ = (-b ± √(b² - 4ac)) / 2a Since a ≠ 0, by the substitution x = y - (b / 2a), the equation can be written as: a(y - b / 2a)² + b(y - b / 2a) + c = 0 ⇒ ay² - b²/4a + c = 0 ⇒ y² = (b² - 4ac) / 4a² ⇒ y = ±√((b² - 4ac) / 4a²) = ±√(b² - 4ac) / 2a x + b / 2a = ±√(b² - 4ac) / 2a ⇒ x₁,₂ = -b / 2a ± √(b² - 4ac) / 2a Two roots are: x₁,₂ = (-b ± √(b² - 4ac)) / 2a . **Example 2.** **Question Stem:** Solve the equation (√3x + 1)² + (√3x + 3)² = 3 **Solution:** -2√3 / 3 ± √6 / 6 By the linear substitution y = √3x + 2, we can convert the equation to (y - 1)² + (y + 1)² = 3 ⇒ 2y² = 1. y = ±√(2) / 2 or √3x + 2 = ±√(2) / 2 .

视频信息

答案文本 复制

视频字幕 复制

Welcome to our exploration of the substitution method for solving quadratic equations. Today we'll understand why the substitution x equals y minus b over 2a is so powerful. This substitution transforms the complex quadratic equation ax squared plus bx plus c equals zero into a much simpler form that eliminates the linear term. The key insight comes from the technique of completing the square. The key to understanding this substitution lies in completing the square. Let's see how this works step by step. We start with ax squared plus bx. First, we factor out the coefficient a, giving us a times the quantity x squared plus b over a times x. To complete the square inside the parentheses, we add and subtract the square of half the coefficient of x, which is b over 2a squared. This transforms our expression into a times the perfect square x plus b over 2a, all squared, minus b squared over 4a. Notice how the term x plus b over 2a naturally emerges from this process. Now comes the crucial insight. Since we've identified that the expression x plus b over 2a appears naturally in our completed square form, we can make a strategic substitution. Let's define a new variable y such that y equals x plus b over 2a. This means that x equals y minus b over 2a, which is exactly our substitution formula. When we substitute this back into our completed square form, something magical happens: the linear term disappears completely, leaving us with a much simpler quadratic equation in y. Now let's work through the complete derivation step by step. We start by substituting x equals y minus b over 2a into our original equation. After expanding and distributing, we get ay squared minus by plus b squared over 4a, plus by minus b squared over 2a, plus c equals zero. Notice that the minus by and plus by terms cancel each other out perfectly. This leaves us with ay squared minus b squared over 4a plus c equals zero. Solving for y squared gives us b squared minus 4ac over 4a squared. Taking the square root yields y equals plus or minus the square root of b squared minus 4ac, all over 2a. Finally, substituting back gives us the famous quadratic formula. Now we understand why this substitution is so powerful. The substitution x equals y minus b over 2a works because it essentially shifts our coordinate system to center the parabola at its axis of symmetry. This transforms the quadratic into its vertex form, where the linear term vanishes completely. The result is a simple equation in y that's easy to solve. This elegant approach not only gives us the quadratic formula but also reveals the geometric insight behind it. The substitution method demonstrates how a clever change of variables can transform a complex problem into a simple one, which is a fundamental technique used throughout mathematics.