📚 Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式
Quadratic equations are a central topic in IGCSE Mathematics. You will need to solve them by factorising, completing the square, and using the quadratic formula. Understanding these methods also helps you interpret graphs and answer questions about the nature of roots.
二次方程是 IGCSE 数学的核心主题。你需要掌握因式分解法、配方法和求根公式。理解这些方法也有助于你解读图像，并回答有关根的性质的问题。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation of degree 2. This means the highest power of the variable, usually x, is 2. In IGCSE Maths, we typically work with one variable.
二次方程是次数为 2 的多项式方程。这意味着变量的最高次数，通常是 x，为 2。在 IGCSE 数学中，我们通常研究含有一个变量的二次方程。
ere a, b and c are constants and a ≠ 0. The condition a ≠ 0 is essential because otherwise the equation becomes linear.
任何二次方程都可以写成标准形式 ax² + bx + c = 0，其中 a、b、c 是常数，且 a ≠ 0。a ≠ 0 这一条件非常重要，因为如果 a = 0，方程就变成了一次方程。
2. Standard Form and Key Terms | 标准形式与关键术语
is the constant term. For example, in 3x² - 5x + 2 = 0, we have a = 3, b = -5, and c = 2.
系数 a 是 x² 前面的数，b 是 x 前面的数，c 是常数项。例如，在 3x² - 5x + 2 = 0 中，a = 3，b = -5，c = 2。
A root of the equation is a value of x that makes the equation true. A quadratic equation can have two distinct real roots, one repeated real root, or no real roots, depending on the discriminant.
方程的根是使等式成立的 x 值。二次方程可以有两个不相等的实根、一个重根或没有实根，具体取决于判别式。
It is always useful to identify a, b and c before starting a solution. This is especially important for the quadratic formula.
在开始求解之前，先确定 a、b、c 的值总是很有用的。这对使用求根公式尤其重要。
3. Solving by Factorising | 因式分解法
Factorising is often the fastest method when the quadratic expression can be written as a product of two linear factors. If the product of two factors is zero, then at least one factor must be zero.
当二次式可以写成两个一次因式的乘积时，因式分解通常是最快的方法。如果两个因式的乘积为零，那么至少有一个因式必须为零。
rise the left-hand side into two brackets. Finally, set each bracket equal to zero and solve for x.
步骤如下：首先将方程写成 ax² + bx + c = 0 的形式。然后将左边分解成两个括号。最后令每个括号等于零并解出 x。
This method works well when a, b and c are small integers and the roots are rational numbers. If factorising looks difficult, use completing the square or the quadratic formula.
当 a、b、c 是较小的整数且根为有理数时，这种方法非常有效。如果因式分解看起来困难，就使用配方法或求根公式。
4. Worked Example: Factorising | 因式分解例题
s that multiply to 12 and add to 7. These numbers are 3 and 4, so the expression factorises as (x + 3)(x + 4).
解方程 x² + 7x + 12 = 0。我们需要两个数，乘积为 12，和为 7。这两个数是 3 和 4，所以该式可因式分解为 (x + 3)(x + 4)。
equal to zero gives x + 3 = 0 or x + 4 = 0. Therefore the solutions are x = -3 or x = -4.
令每个因式等于零，得到 x + 3 = 0 或 x + 4 = 0。因此解为 x = -3 或 x = -4。
You should always check your solutions by substituting them back into the original equation. Both values make x² + 7x + 12 equal to zero.
你应该始终将解代回原方程进行检验。这两个值都能使 x² + 7x + 12 等于零。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form (x + p)² + q. This form is useful for solving equations, finding the vertex of a parabola, and understanding transformations.
配方法将二次式改写为 (x + p)² + q 的形式。这种形式对于解方程、求抛物线的顶点以及理解图像变换都非常有用。
of the form x² + bx, we add and subtract (b/2)². This gives x² + bx = (x + b/2)² - (b/2)².
对于 x² + bx 形式的表达式，我们加上并减去 (b/2)²，得到 x² + bx = (x + b/2)² - (b/2)²。
3)² - 9 because (6/2)² = 9. If there is a constant term, move it to the other side first.
例如，x² + 6x 可以写成 (x + 3)² - 9，因为 (6/2)² = 9。如果有常数项，应先将它移到等号另一边。
6. Worked Example: Completing the Square | 配方法例题
completing the square. First move the constant term to the right-hand side: x² + 6x = -2.
用配方法解 x² + 6x + 2 = 0。首先将常数项移到右边：x² + 6x = -2。
/2)² = 9, we can write x² + 6x as (x + 3)² - 9. So the equation becomes (x + 3)² - 9 = -2.
在左边配方。因为 (6/2)² = 9，我们可以将 x² + 6x 写成 (x + 3)² - 9。因此方程变为 (x + 3)² - 9 = -2。
Simplify to get (x + 3)² = 7. Taking the square root of both sides gives x + 3 = ±√7. Hence the roots are x = -3 + √7 and x = -3 - √7.
化简得到 (x + 3)² = 7。两边开平方得到 x + 3 = ±√7。因此根为 x = -3 + √7 和 x = -3 - √7。
7. The Quadratic Formula | 求根公式
the form ax² + bx + c = 0. It is derived by completing the square on the general equation.
求根公式可以解任何形如 ax² + bx + c = 0 的二次方程。它是由对一般方程进行配方法推导出来的。
x = (-b ± √b² - 4ac) ÷ 2a
The expression under the square root, b² - 4ac, is called the discriminant. It tells you how many real roots the equation has without needing to solve it completely.
根号下的表达式 b² - 4ac 称为判别式。它可以在不需要完全求解的情况下告诉你方程有多少个实根。
When using the formula, always write down the values of a, b and c before substituting. Use brackets for negative numbers to avoid sign errors.
使用公式时，在代入前一定要先写下 a、b、c 的值。对负数使用括号，以避免符号错误。
8. Worked Example: Quadratic Formula | 求根公式例题
Solve 2x² - 3x - 2 = 0 using the quadratic formula. Here a = 2, b = -3 and c = -2.
用求根公式解 2x² - 3x - 2 = 0。这里 a = 2，b = -3，c = -2。
he formula: x = ( -(-3) ± √((-3)² - 4 × 2 × (-2)) ) ÷ (2 × 2). This simplifies to x = (3 ± √9 + 16) ÷ 4 = (3 ± √25) ÷ 4.
代入公式：x = ( -(-3) ± √((-3)² - 4 × 2 × (-2)) ) ÷ (2 × 2)。化简为 x = (3 ± √9 + 16) ÷ 4 = (3 ± √25) ÷ 4。
Since √25 e x = (3 + 5) ÷ 4 = 2 or x = (3 - 5) ÷ 4 = -0.5. The two solutions are x = 2 and x = -0.5.
因为 √25 = 5，所以 x = (3 + 5) ÷ 4 = 2 或 x = (3 - 5) ÷ 4 = -0.5。两个解为 x = 2 和 x = -0.5。
9. Discriminant and Nature of Roots | 判别式与根的性质
The discriminant is defined as Δ = b²
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