📚 Solving Quadratic Equations | 解二次方程
A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. In IGCSE Mathematics, solving quadratic equations appears in both Core and Extended papers, and this skill underpins many other topics such as graphing functions, solving inequalities, and solving simultaneous equations.
二次方程是最高次数为 2 的多项式方程，即变量的最高次幂为 2。在 IGCSE 数学中，解二次方程同时出现在 Core 和 Extended 试卷中，并且这一技能是函数作图、解不等式以及解联立方程等众多主题的基础。
0. If a = 0, the equation becomes linear and the methods in this article no longer apply.
二次方程的一般形式如下所示。其中 a、b、c 为常数，且 a ≠ 0。若 a = 0，方程将退化为一次方程，本文介绍的方法便不再适用。
ax² + bx + c = 0
A quadratic equation has at most two solutions, which are often called its roots. These roots can be two distinct real numbers, one repeated real number, or two non-real complex numbers, depending on the value of the discriminant.
二次方程最多有两个解，通常称为根。根据判别式的取值，这些根可以是两个不同的实数、一个重数实数根，或两个非实数的复数根。
1. Expanding and Factorising | 展开与因式分解
Before solving equations, you must be fluent in expanding brackets and factorising quadratics. Expanding means removing brackets by multiplying terms, while factorising is the reverse process: writing an expression as a product of two linear factors.
在解方程之前，你必须熟练掌握展开括号和因式分解二次式。展开是通过逐项相乘去掉括号，而因式分解是相反的过程：将一个表达式写成两个一次因式相乘的形式。
Consider the product (x + 2)(x + 3). Multiplying each term in the first bracket by each term in the second gives x² + 2x + 3x + 6, which simplifies to x² + 5x + 6.
考虑乘积 (x + 2)(x + 3)。将第一个括号中的每一项乘以第二个括号中的每一项，得到 x² + 2x + 3x + 6，化简为 x² + 5x + 6。
mbers are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3). Always check by expanding your answer.
要对 x² + 5x + 6 进行因式分解，需要找两个数，它们相乘得 6 且相加得 5。这两个数是 2 和 3，因此 x² + 5x + 6 = (x + 2)(x + 3)。务必通过展开来检验答案。
), and perfect squares, a² + 2ab + b² = (a + b)². These appear frequently in IGCSE papers.
两个特殊模式值得记忆：平方差公式 a² - b² = (a - b)(a + b)，以及完全平方公式 a² + 2ab + b² = (a + b)²。它们在 IGCSE 试卷中频繁出现。
2. Solving by Factorisation | 因式分解法求解
ckly. This property states that if A × B = 0, then at least one of A or B must equal zero.
若一个二次方程可以因式分解，利用零乘积性质即可快速求解。该性质指出：若 A × B = 0，则 A 和 B 中至少有一个等于零。
x - 2 = 0 or x - 3 = 0, so x = 2 or x = 3. These two values are the roots of the equation.
解方程 x² - 5x + 6 = 0。因式分解得 (x - 2)(x - 3) = 0。因此 x - 2 = 0 或 x - 3 = 0，所以 x = 2 或 x = 3。这两个值就是方程的根。
Always rearrange the equation into the form 'expression = 0' etting to set the product equal to zero, or dividing both sides by x when x could be zero.
在因式分解之前，务必先把方程整理成 '表达式 = 0' 的形式。常见错误包括：因式分解后忘记令乘积等于零，或在 x 可能为 0 的情况下两边同时除以 x。
3. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This form immediately reveals the turning point of the corresponding parabola and is especially useful when the equation does not factorise easily.
配方法将二次式改写为 a(x + p)² + q 的形式。这种形式可以直接显示对应抛物线的顶点坐标，并且在方程不易因式分解时尤为有用。
quare it: 3² = 9. Then write x² + 6x = (x + 3)² - 9. Therefore x² + 6x + 2 = (x + 3)² - 7.
以 x² + 6x + 2 为例。将 x 的系数除以 2：6 ÷ 2 = 3。再平方：3² = 9。然后写成 x² + 6x = (x + 3)² - 9。因此 x² + 6x + 2 = (x + 3)² - 7。
+ 3)² - 7 = 0, rearrange to (x + 3)² = 7 and take the square root of both sides: x + 3 = ±√7, so x = -3 ± √7. This gives the two exact roots in surd form.
解 (x + 3)² - 7 = 0 时，先整理为 (x + 3)² = 7，然后两边开平方：x + 3 = ±√7，所以 x = -3 ± √7。这样得到的两个根就是根号形式的精确值。
t first. For example, 2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² - 4] + 5 = 2(x + 2)² - 3.
当 x² 的系数不为 1 时，先将其提出。例如，2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² - 4] + 5 = 2(x + 2)² - 3。
4. The Quadratic Formula | 求根公式
The quadratic formula is the most general method for solving quadratic equations. It works for every quadratic equation, whether or not it can be factorised, and it is the safest method to use when you are unsure.
求根公式是解二次方程最通用的方法。它适用于所有二次方程，无论能否因式分解，也是当你没有把握时最稳妥的方法。
oots are given by the formula below. You should memorise this formula for the non-calculator paper.
已知方程 ax² + bx + c = 0，其根由以下公式给出。在非计算器试卷中，你需要记住这个公式。
x = (-b ± √b² - 4ac) / 2a
x² - 4x - 3 = 0 using the formula. Here a = 2, b = -4, c = -3. Substituting gives x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = 1 ± √10/2.
用公式法解 2x² - 4x - 3 = 0。这里 a = 2，b = -4，c = -3。代入得 x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = 1 ± √10/2。
Before substituting, always check that the equation is in standard form. Rearranging terms correctly is essential: any sign errors in a, b or c will change the answer completely.
代入之前，务必检查方程是否已化为标准形式。正确整理各项至关重要：a、b、c 中任何一个符号出错都会完全改变答案。
5. The Discriminant | 判别式
The discriminant is the expression b² - 4ac, found under the square root in the quadratic formula. It tells us the nature of the roots without solving the equation fully.
判别式是求根公式中根号下的表达式 b² - 4ac。它无需完整解方程就能判断根的性质。
The three cases are summarised
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