📚 Mastering Quadratic Equations for IGCSE Mathematics | 掌握 IGCSE 数学二次方程
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, coordinate geometry, and word problems. A strong understanding of how to solve and sketch quadratics will help you achieve high marks across several exam papers.
二次方程是 IGCSE 数学中最重要的主题之一。它们出现在代数、坐标几何和应用题中。扎实掌握二次方程的解法与图像画法，能帮助你在多张试卷中取得高分。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation of degree 2. It involves a variable, usually x, whose highest power is x². The equation can contain x² terms, x terms, and constant terms.
二次方程是次数为 2 的多项式方程。它含有一个变量，通常是 x，最高次幂为 x²。方程中可以包含 x² 项、x 项和常数项。
, the equation becomes linear, not quadratic. In IGCSE exams, you will be asked to solve quadratics, interpret their roots, and sketch their graphs.
二次方程的一般形式为 ax² + bx + c = 0，其中 a ≠ 0。如果 a = 0，方程就变成了一次方程，而不是二次方程。在 IGCSE 考试中，你会被要求解二次方程、解释其根并绘制其图像。
2. Standard Form and Key Terms | 标准形式与关键术语
is the quadratic coefficient, b is the linear coefficient, and c is the constant term. Identifying a, b, and c correctly is essential before using many solution methods.
标准形式为 ax² + bx + c = 0。其中 a 是二次项系数，b 是一次项系数，c 是常数项。在使用许多解法之前，正确识别 a、b 和 c 至关重要。
example, 3x − 2 = 5x² can be rearranged to 5x² − 3x + 2 = 0, so a = 5, b = −3, and c = 2.
在选择解法之前，必须先把方程整理成标准形式。将所有项移到一边并化简。例如，3x − 2 = 5x² 可以整理为 5x² − 3x + 2 = 0，因此 a = 5，b = −3，c = 2。
3. Solving by Factorising | 因式分解法
s that multiply to give ac and add to give b, then split the middle term or use trial and error.
当二次方程有简单的整数根时，因式分解通常是最快的方法。将 ax² + bx + c 写成 (px + q)(rx + s) = 0。你需要找到两个数，它们相乘得到 ac，相加得到 b，然后拆分中间项或使用试错法。
After factorising, use the zero-product rule: if two factors multiply to zero, at least one factor must equal zero. Set each bracket equal to 0 and solve each linear equation.
因式分解后，使用零乘积性质：如果两个因式相乘为零，那么至少有一个因式为零。令每个括号等于 0，然后解每个一元一次方程。
= −3. Always check by expanding your factors to confirm they give the original quadratic.
例：x² + 5x + 6 = 0 因式分解为 (x + 2)(x + 3) = 0，解得 x = −2 或 x = −3。一定要通过展开因式来检验是否得到原方程。
4. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. This is especially useful when factorising is difficult or when you need to find the vertex of a parabola.
配方法将二次方程改写为 (x + p)² + q 的形式。当因式分解困难或需要求抛物线顶点时，这种方法尤其有用。
ect square: x² + bx + (b ÷ 2)² = (x + b ÷ 2)². Rearrange and solve by taking square roots.
对于 x² + bx + c = 0，加上并减去 (b ÷ 2)²。这会产生完全平方：x² + bx + (b ÷ 2)² = (x + b ÷ 2)²。整理后通过取平方根求解。
te x² + 6x + 9 − 9 + 5 = 0, so (x + 3)² − 4 = 0. Then x + 3 = ±2, giving x = −1 or x = −5.
例：x² + 6x + 5 = 0。6 的一半是 3，3² = 9。写为 x² + 6x + 9 − 9 + 5 = 0，因此 (x + 3)² − 4 = 0。则 x + 3 = ±2，解得 x = −1 或 x = −5。
If a ≠ 1, factor out a from the x² and x terms before completing the square. Do not forget to keep the equation balanced.
如果 a ≠ 1，在配方前先把 a 从 x² 项和 x 项中提取出来。不要忘记保持方程平衡。
5. The Quadratic Formula | 求根公式
t possible. It is derived from completing the square on the general form ax² + bx + c = 0.
求根公式可以解任何二次方程，即使无法因式分解。它是由对一般形式 ax² + bx + c = 0 使用配方法推导出来的。
The formula is:
x = (−b ± √b² − 4ac) ÷ 2a
公式为：x = (−b ± √b² − 4ac) ÷ 2a
Identify a, b, and c from the equation in standard form, substitute them into the formula, and simplify carefully. Pay attention to the sign of each coefficient.
从标准形式的方程中确定 a、b 和 c，代入公式并仔细化简。注意每个系数的符号。
Example: For 2x² + 3x − 2 = 0, a = 2, b = 3, c = −2. Then x = (−3 ± √9 + 16) ÷ 4 = (−3 ± 5) ÷ 4, giving x = 1/2 or x = −2.
例：对于 2x² + 3x − 2 = 0，a = 2，b = 3，c = −2。则 x = (−3 ± √9 + 16) ÷ 4 = (−3 ± 5) ÷ 4，解得 x = 1/2 或 x = −2。
6. The Discriminant and Nature of Roots | 判别式与根的性质
the number and type of roots without solving the equation. It comes from the expression under the square root in the quadratic formula.
判别式 D = b² − 4ac 可以在不解方程的情况下告诉你根的数量和类型。它来自求根公式中平方根下的表达式。
ated real root. If D < 0, there are no real roots because you cannot take the square root of a negative number in the real number system.
若 D > 0，方程有两个不同的实数根；若 D = 0，方程有一个重根；若 D < 0，方程没有实数根，因为在实数范围内不能对负数取平方根。
ample: For x² − 4x + 4 = 0, D = (−4)² − 4(1)(4) = 0, so there is one repeated root, x = 2.
这尤其适用于判断图像是与 x 轴相交、相切还是不相交的问题。例：对于 x² − 4x + 4 = 0，D = (−4)² − 4(1)(4) = 0，所以有一个重根 x = 2。
7. Graphs of Quadratic Functions | 二次函数图像
. If a > 0, the parabola opens upward and has a minimum point. If a < 0, it opens downward and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。若 a > 0，抛物线开口向上，有一个最小值点；若 a < 0，抛物线开口向下，有一个最大值点。
the quadratic equation ax² + bx + c = 0. These are also called x-intercepts or solutions.
图像与 x 轴的交点对应于二次方程 ax² + bx + c = 0 的实数根。这些点也称为 x 轴截距或解。
graph usually requires the y-intercept, the x-intercepts if they exist, and the turning point.
y 轴截距为 (0, c)，因为当 x = 0 时，y = c。绘制二次函数图像通常需要 y 轴截距、x 轴截距（若存在）以及顶点。
8. Turning Points and Symmetry | 顶点与对称性
A parabola has a line of symmetry through its turning point, also called the vertex. The vertex is the minimum point when a > 0 and the maximum point when a < 0.
抛物线有一条经过顶点（转折点）的对称轴。当 a > 0 时顶点是最小值点，当 a < 0 时顶点是最大值点。
. Completing the square gives the vertex directly: y = (x + p)² + q has vertex at (−p, q).
顶点的 x 坐标由 x = −b ÷ (2a) 给出。将该值代入方程即可求出 y 坐标。配方法可以直接给出顶点：y = (x + p)² + q 的顶点为 (−p, q)。
+ 3, the vertex has x = 4 ÷ 2 = 2, and y = (2)² − 4(2) + 3 = −1, so the vertex is (2, −1).
例：对于 y = x² − 4x + 3，顶点 x 坐标为 4 ÷ 2 = 2，y = (2)² − 4(2) + 3 = −1，因此顶点为 (2, −1)。
9. Applications and Word Problems | 应用题与文字题
Quadratic equations model many real-life situations, including area, projectile motion, and profit problems. The key is to translate the words into an equation, then solve it.
二次方程可以模拟许多实际情境，包括面积、抛体运动和利润问题。关键是把文字转化为方程，然后求解。
− 40 = 0. Factorising gives (x + 8)(x − 5) = 0, so x = 5 because width cannot be negative.
常见的面积问题：一个矩形的长比宽多 3 cm，面积为 40 cm²。设宽为 x，则 x(x + 3) = 40，即 x² + 3x − 40 = 0。因式分解得 (x + 8)(x − 5) = 0，所以 x = 5，因为宽度不能为负数。
Solve the resulting quadratic, then check that the answers make sense in the original context. Negative lengths, for example, must be rejected.
解出所得二次方程，然后检验答案在实际情境中是否有意义。例如负长度必须舍去。
= −5t² + vt + h₀. Setting h equal to a specific value leads to a quadratic equation in t.
在抛体运动中，高度 h 与时间 t 的关系常为 h = −5t² + vt + h₀。令 h 等于某个具体值就会得到关于 t 的二次方程。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
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