📚 Mastering Quadratic Equations | 掌握二次方程
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, coordinate geometry, and real-world modelling. This revision guide covers the key methods, formulas and exam strategies you need to solve quadratic equations confidently.
二次方程是 IGCSE 数学最重要的主题之一，广泛出现在代数、坐标几何和实际建模中。本复习指南涵盖关键方法、公式与考试策略，帮助你自信地求解二次方程。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation of degree 2. The highest power of the variable x is 2. Its graph is a parabola, which opens upwards if the leading coefficient is positive and downwards if it is negative.
二次方程是一个次数为 2 的多项式方程，变量 x 的最高次数是 2。它的图像是一条抛物线，首项系数为正时开口向上，为负时开口向下。
ax² + bx + c = 0, a ≠ 0
Here a, b and c are constants. The condition a ≠ 0 ensures the equation is genuinely quadratic; if a = 0, the equation becomes linear.
其中 a、b、c 为常数。a ≠ 0 保证方程确实是二次方程；若 a = 0，则方程变为一次方程。
2. Standard Form and Coefficients | 标准形式与系数
Always rearrange a quadratic equation into standard form before solving. Collect all terms on one side so that the other side equals zero. This makes the coefficients clear and ready for any method.
求解前必须先将二次方程化为标准形式，把所有项移到一边，使另一边等于 0。这样系数会变得清晰，也便于使用任何方法。
2x² − 5x = 3 → 2x² − 5x − 3 = 0
For this equation, the coefficients are a = 2, b = −5 and c = −3. In the standard form ax² + bx + c = 0, a is the coefficient of x², b is the coefficient of x, and c is the constant term.
对于这个方程，系数为 a = 2，b = −5，c = −3。在标准形式 ax² + bx + c = 0 中，a 是 x² 的系数，b 是 x 的系数，c 是常数项。
3. Solving by Factorising | 因式分解法
Factorising works when the quadratic can be written as a product of two linear factors. Set each factor equal to zero and solve the resulting linear equations.
当二次方程可以写成两个一次因式的乘积时，可使用因式分解法。令每个因式等于 0，再求解所得的一次方程。
x² + 5x + 6 = 0 → (x + 2)(x + 3) = 0 → x = −2 or x = −3
When a = 1, the key is to find two numbers that multiply to c and add to b. For a ≠ 1, trial and checking is often needed, for example:
当 a = 1 时，关键是找到两个数，它们的乘积为 c，和为 b。当 a ≠ 1 时，通常需要尝试和检验，例如：
2x² − 5x − 3 = 0 → (2x + 1)(x − 3) = 0 → x = −1/2 or x = 3
Another special case is the difference of two squares, such as x² − 9 = 0, which gives (x − 3)(x + 3) = 0 and therefore x = ±3.
另一个特殊情况是平方差，例如 x² − 9 = 0，可化为 (x − 3)(x + 3) = 0，因此 x = ±3。
4. Solving by Completing the Square | 配方法
Completing the square transforms ax² + bx + c into the form a(x + p)² + q. This method is especially useful for finding the vertex of a parabola and for solving equations that do not factorise neatly.
配方法将 ax² + bx + c 转化为 a(x + p)² + q 的形式。这种方法对于求抛物线的顶点以及求解不易因式分解的方程尤其有用。
x² + 6x + 5 = 0 → (x + 3)² − 9 + 5 = 0 → (x + 3)² = 4 → x + 3 = ±2 → x = −1 or x = −5
For a monic quadratic x² + bx, add and subtract (b/2)² to create a perfect square. The term (b/2)² is found by halving the coefficient of x and squaring it.
对于首项系数为 1 的二次式 x² + bx，加上并减去 (b/2)² 来构造完全平方。(b/2)² 由 x 的系数一半再平方得到。
5. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation, including those that do not factorise nicely. It is derived from completing the square on the general form ax² + bx + c = 0.
求根公式可以求解任何二次方程，包括不能整齐因式分解的方程。它由一般形式 ax² + bx + c = 0 配方推导而来。
x = (−b ± √(b² − 4ac)) / 2a
Substitute the values of a, b and c carefully, paying close attention to negative signs. The expression under the square root is the discriminant, which controls the number of real solutions.
代入 a、b、c 时要仔细，尤其注意负号。根号下的表达式是判别式，它决定实根的个数。
3x² − 2x − 4 = 0 → x = (2 ± √(4 + 48)) / 6 = (2 ± √52) / 6 = (1 ± √13) / 3
Unless a question asks for a decimal approximation, leave your answer in exact form using surds or fractions.
除非题目要求取小数近似值，否则应使用根式或分数保留精确形式。
6. Discriminant and Nature of Roots | 判别式与根的性质
The discriminant is defined as Δ = b² − 4ac. It tells you how many real roots a quadratic equation has without solving the equation fully.
判别式定义为 Δ = b² − 4ac。它可以在不完整求解方程的情况下判断二次方程有多少个实根。
| Discriminant | Nature of roots |
|---|---|
| Δ > 0 | Two distinct real roots |
| Δ = 0 | One repeated real root |
| Δ < 0 | No real roots |
Δ > 0 表示有两个不同的实根；Δ = 0 表示有一个重复实根；Δ < 0 表示没有实根。例如 x² − 4x + 4 = 0 的判别式为 16 − 16 = 0，因此只有一个根 x = 2。
Δ > 0 gives two distinct real roots; Δ = 0 gives one repeated real root; Δ < 0 gives no real roots. For example, x² − 4x + 4 = 0 has discriminant 16 − 16 = 0, so it has only one root, x = 2.
7. Graphical Interpretation | 图像解读
The solutions of ax² + bx + c = 0 are exactly the x-intercepts of the graph y = ax² + bx + c. The parabola crosses the x-axis at those solution values.
方程 ax² + bx + c = 0 的解正是图像 y = ax² + bx + c 与 x 轴的交点。抛物线在这些解的位置穿过 x 轴。
y = x² − 4x + 3 → x-intercepts at x = 1 and x = 3; vertex at x = 2
The vertex of a parabola has x-coordinate x = −b/(2a). The axis of symmetry passes through this x-value, so the roots are symmetric about the vertex.
抛物线的顶点横坐标为 x = −b/(2a)。对称轴经过这个 x 值，因此两个根关于顶点对称。
If the discriminant is negative, the parabola does not cross the x-axis at all, which matches the fact that the equation has no real roots.
如果判别式为负，抛物线完全不与 x 轴相交，这与方程没有实根的事实一致。
8. Forming Quadratic Equations from Roots | 由根构造二次方程
If the roots of a quadratic equation are α and β, the equation can be written as x² − (α + β)x + αβ = 0. This is very useful when you are given roots rather than an equation.
若二次方程的根为 α 和 β，则该方程可写作 x² − (α + β)x + αβ = 0。当题目给出根而不是方程时，这非常有用。
Sum of roots S = α + β = −b/a, Product of roots P = αβ = c/a
Example: roots 3 and −2 give x² − (3 + (−2))x + (3)(−2) = 0, which simplifies to x² − x − 6 = 0.
例子：根为 3 和 −2，得到 x² − (3 + (−2))x + (3)(−2) = 0，化简为 x² − x − 6 = 0。
You can also use these relationships to find unknown coefficients when one root is given, by substituting the known root into the equation.
当已知一个根时，你也可以利用这些关系，将已知根代入方程来求未知系数。
9. Applications and Word Problems | 应用与文字题
Many IGCSE problems ask you to form a quadratic equation from a real-life situation, such as area, projectile motion or product relationships. The steps are always the same: define variables, form an equation, solve it, and check answers in context.
许多 IGCSE 题目要求你从实际情境中建立二次方程，例如面积、抛体运动或乘积关系。步骤总是相同：设变量、建立方程、求解、并在情境中检验答案。
Consecutive integers: n(n + 1) = 72 → n² + n − 72 = 0 → (n + 9)(n − 8) = 0 → n = 8 (positive)
So the two consecutive positive integers are 8 and 9. The negative solution n = −9 is rejected because the question asks for positive integers.
因此这两个连续正数为 8 和 9。负解 n = −9 被舍去，因为题目要求正数。
Rectangle: width x, length x + 3, area 40 → x(x + 3) = 40 → x² + 3x − 40 = 0 → x = 5 or x = −8
Since a width cannot be negative, x = 5 and the length is 8. Always reject solutions that are impossible in the real-world situation.
由于宽度不能为负，所以 x = 5，长度为 8。始终舍去在实际情境中不可能的解。
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