📚 Solving Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 求解一元二次方程：因式分解、配方法与求根公式
A quadratic equation is a central topic in IGCSE Mathematics, appearing in algebra, graphs, mensuration and problem solving. You must be able to recognise the standard form, solve by factorising, complete the square, apply the quadratic formula, and use the discriminant to describe the roots. This revision guide explains each method with worked examples and common exam tips.
一元二次方程是 IGCSE 数学的核心主题，出现在代数、图像、测量和问题解决中。你必须能够识别标准形式，掌握因式分解法、配方法、求根公式，并会用判别式描述根的性质。本复习指南逐一讲解每种方法，并配有例题和常见考试提示。
1. What Is a Quadratic Equation? | 什么是一元二次方程
nts and a ≠ 0. The highest power of the variable x is 2, which gives the equation its name.
二次方程是任何可以整理成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数且 a ≠ 0。变量 x 的最高次数为 2，因此称为二次方程。
ratic equations. The equation x³ + 2x = 0 is not quadratic because the highest power is 3.
例如，x² − 5x + 6 = 0、2x² + 3x − 2 = 0 和 x² − 4 = 0 都是二次方程。方程 x³ + 2x = 0 不是二次方程，因为最高次数是 3。
In IGCSE questions, a quadratic may be hidden inside a word problem, a fraction, or a geometric formula. You should always try to simplify it into standard form before solving.
在 IGCSE 题目中，二次方程可能隐藏在应用题、分式或几何公式中。你应始终先将其化为标准形式，再求解。
2. Standard Form and Rearranging | 标准形式与整理
ually written in descending powers of x. Keeping all terms on one side and zero on the other is essential for factorising and for identifying a, b and c.
二次方程的标准形式是 ax² + bx + c = 0，各项通常按 x 的降幂排列。将所有项移到一边、另一边为零，对于因式分解和确定 a、b、c 至关重要。
example, to solve x² = 5x − 6, first subtract 5x and add 6 to both sides: x² − 5x + 6 = 0.
例如，求解 x² = 5x − 6 时，先将两边减去 5x 并加上 6：x² − 5x + 6 = 0。
If the equation contains fractions, multiply through by the lowest common denominator. If it contains brackets, expand them first. This prevents mistakes in the coefficients.
如果方程含有分式，应两边乘以最小公分母；如果含有括号，应先展开。这样可以避免系数出错。
3. Solving by Factorising When a = 1 | 当 a = 1 时用因式分解求解
form x² + bx + c = 0. To factorise, look for two numbers that multiply to c and add to b.
当 x² 的系数为 1 时，方程形如 x² + bx + c = 0。进行因式分解时，需寻找两个数，它们的乘积为 c，和为 b。
duct 6 and sum −5. The numbers are −2 and −3 because (−2) × (−3) = 6 and (−2) + (−3) = −5.
对于 x² − 5x + 6 = 0，我们需要两个乘积为 6、和为 −5 的数。这两个数是 −2 和 −3，因为 (−2) × (−3) = 6 且 (−2) + (−3) = −5。
) = 0. Set each factor to zero: x − 2 = 0 or x − 3 = 0. The solutions are x = 2 and x = 3.
因此因式分解形式为 (x − 2)(x − 3) = 0。令每个因式等于零：x − 2 = 0 或 x − 3 = 0。解为 x = 2 和 x = 3。
ing: (x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6, which matches the original equation.
始终通过展开进行检验：(x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6，与原方程一致。
4. Solving by Factorising When a ≠ 1 | 当 a ≠ 1 时用因式分解求解
When a ≠ 1, factorising is slightly harder. One reliable method is to split the middle term. First multiply a and c, then find two numbers with that product and sum b.
当 a ≠ 1 时，因式分解稍难。一种可靠的方法是拆分中间项。首先将 a 与 c 相乘，然后找到两个数，使它们的乘积等于 ac，和等于 b。
a × c = 2 × (−2) = −4. We need two numbers with product −4 and sum 3: these are 4 and −1.
对于 2x² + 3x − 2 = 0，a = 2，b = 3，c = −2。a × c = 2 × (−2) = −4。我们需要两个乘积为 −4、和为 3 的数：4 和 −1。
in pairs: 2x(x + 2) − 1(x + 2) = 0. The common bracket is (x + 2), so (2x − 1)(x + 2) = 0.
重写中间项：2x² + 4x − x − 2 = 0。然后分组分解：2x(x + 2) − 1(x + 2) = 0。公共括号为 (x + 2)，所以 (2x − 1)(x + 2) = 0。
x = 1/2 and x = −2. Always multiply your factors back to check the middle term is correct.
根为 x = 1/2 和 x = −2。务必将因式乘回，检查中间项是否正确。
5. The Zero Factor Principle | 零因式原理
st one of the factors must be zero. In symbols, if A × B = 0, then A = 0 or B = 0 or both.
因式分解法之所以有效，是因为零因式原理：如果两个或多个因式的乘积为零，那么至少有一个因式为零。用符号表示，如果 A × B = 0，则 A = 0 或 B = 0，或两者同时为零。
have (x − 2)(x − 3) = 6, you cannot simply set each factor to 6; the product must be zero.
这就是为什么在应用该原理之前，方程一边必须为零。如果你有 (x − 2)(x − 3) = 6，不能简单令每个因式等于 6；乘积必须为零。
Also, never divide both sides of a quadratic equation by an expression containing x, such as x or (x − 1), because that can remove a valid solution. Always factorise instead of dividing.
此外，切勿将二次方程两边同时除以含有 x 的表达式，如 x 或 (x − 1)，因为那样可能会删除一个有效解。应始终通过因式分解而不是除法。
6. Completing the Square | 配方法
ethod is useful when the quadratic does not factorise easily, or when you need to write the equation in vertex form.
配方法将二次方程改写为 (x + p)² + q = 0 的形式。当方程不易因式分解，或需要写成顶点式时，这一方法非常有用。
coefficient of x, which is 6/2 = 3, and square it: 3² = 9. Write x² + 6x as (x + 3)² − 9.
对于 x² + 6x + 2 = 0，从 x² + 6x 开始。取 x 项系数的一半，即 6/2 = 3，并将其平方：3² = 9。将 x² + 6x 写成 (x + 3)² − 9。
Substitute back: (x + 3)² − 9 + 2 = 0, so (x + 3)² = 7. Then x + 3 = ±√7, giving x = −3 + √7 and x = −3 − √7.
代回原式：(x + 3)² − 9 + 2 = 0，所以 (x + 3)² = 7。于是 x + 3 = ±√7，得到 x = −3 + √7 和 x = −3 − √7。
quare. For example, 2x² + 8x + 4 = 0 becomes x² + 4x + 2 = 0 before completing the square.
如果 a ≠ 1，在配方前应先将每一项除以 a。例如，2x² + 8x + 4 = 0 在配方前先化为 x² + 4x + 2 = 0。
7. The Quadratic Formula | 求根公式
ion in standard form ax² + bx + c = 0, even when factorising is difficult. The formula is:
求根公式可以求解任何标准形式 ax² + bx + c = 0 的二次方程，即使在因式分解困难时也可使用。公式为：
x = (−b ± √b² − 4ac) / 2a
from ax² + bx + c = 0. Careful substitution is essential, especially with negative signs.
这里 a、b、c 是 ax² + bx + c = 0 中的系数。代入时必须细心，尤其是负号。
−2. Substituting into the formula gives x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2) = (−3 ± √9 + 16) / 4 = (−3 ± √25) / 4.
对于 2x² + 3x − 2 = 0，a = 2，b = 3，c = −2。代入公式得 x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2) = (−3 ± √9 + 16) / 4 = (−3 ± √25) / 4。
Since √25 5, x = (−3 + 5) / 4 = 1/2 or x = (−3 − 5) / 4 = −2. These match the factorised solutions.
因为 √25 = 5，所以 x = (−3 + 5) / 4 = 1/2 或 x = (−3 − 5) / 4 = −2。这与因式分解得到的解一致。
8. The Discriminant and the Nature of Roots | 判别式与根的性质
The expression inside the square root in the quadratic formula, b² − 4ac, is called the discriminant. It tells you how many real roots the equation has without actually solving it.
求根公式中平方根内的表达式 b² − 4ac 称为判别式。它可以在不实际求解的情况下告诉你方程有多少个实数根。
- If b² − 4ac > 0, the equation has two distinct real roots. | 如果 b² − 4ac > 0，方程有两个不相等的实数根。
- If b² − 4ac = 0, the equation has one repeated real root. | 如果 b² − 4ac = 0，方程有一个重复的实数根。
- If b² − 4ac < 0, the equation has no real roots. | 如果 b² − 4ac < 0，方程没有实数根。
ant (−5)² − 4 × 1 × 6 = 25 − 24 = 1, which is positive, so it has two distinct real roots.
例如，x² − 5x + 6 = 0 的判别式为 (−5)² − 4 × 1 × 6 = 25 − 24 = 1，为正数，因此有两个不相等的实数根。
f k for which a quadratic has equal roots. In that case, set b² − 4ac = 0 and solve for k.
IGCSE 题目可能会要求求使二次方程有两个相等根时 k 的值。此时应令 b² − 4ac = 0 并求解 k。
9. Choosing the Best Method | 选择最佳求解方法
In the exam, choose the most efficient method. Factorising is fastest for simple quadratics with integer roots. Completing the square is useful when you need exact sur
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