📚 Quadratic Equations: From Factoring to the Formula | 二次方程：从因式分解到求根公式
Quadratic equations are among the most frequently tested topics in IGCSE Mathematics. Whether you are solving for roots, sketching a parabola, or interpreting a word problem, mastering quadratics is essential for exam success.
二次方程是 IGCSE 数学中考查频率最高的知识点之一。无论是求根、画抛物线图像，还是解答应用题，掌握二次方程都是取得考试成功的关键。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. The highest power of the variable x is 2, which gives the equation its name “quadratic.”
二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。变量的最高次幂为 2，这正是其名称”二次”的由来。
Examples include 2x² − 5x + 3 = 0 and 3x² = 12. In the second example, the equation can be rearranged to 3x² − 12 = 0, which still fits the general form with b = 0.
例如 2x² − 5x + 3 = 0 和 3x² = 12。第二个例子整理后可得 3x² − 12 = 0，仍符合一般形式，其中 b = 0。
Notice that a linear equation has only one solution, but a quadratic equation can have two solutions, one solution, or no real solutions. This depends on the value of the discriminant, which we will examine later.
注意，一次方程只有一个解，而二次方程可以有两个解、一个解，或没有实数解。这取决于判别式的值，我们稍后会详细讨论。
2. Solving by Factoring | 因式分解法
Factoring is often the fastest method when the quadratic expression can be written as a product of two linear binomials. We rely on the zero product property: if AB = 0, then either A = 0 or B = 0.
当二次表达式可以写成两个一次二项式的乘积时，因式分解通常是最快捷的方法。我们依据零积性质：若 AB = 0，则 A = 0 或 B = 0。
For example, solve x² − 5x + 6 = 0. We look for two numbers that multiply to 6 and add to −5. The numbers are −2 and −3, so we factor to obtain (x − 2)(x − 3) = 0. Setting each factor to zero gives x = 2 or x = 3.
例如，解方程 x² − 5x + 6 = 0。我们需要寻找两个数，乘积为 6 且和为 −5。这两个数是 −2 和 −3，因此因式分解为 (x − 2)(x − 3) = 0。令每个因子为零，可得 x = 2 或 x = 3。
Another example: 2x² + 7x + 3 = 0. We find two numbers that multiply to 2 × 3 = 6 and add to 7, which are 6 and 1. Rewriting the middle term, we have 2x² + 6x + x + 3 = 0, then factor by grouping: 2x(x + 3) + 1(x + 3) = 0, giving (2x + 1)(x + 3) = 0. Therefore x = −½ or x = −3.
再看一个例子：2x² + 7x + 3 = 0。我们寻找两个数，乘积为 2 × 3 = 6 且和为 7，即 6 和 1。将中间项拆分，得到 2x² + 6x + x + 3 = 0，然后分组因式分解：2x(x + 3) + 1(x + 3) = 0，即 (2x + 1)(x + 3) = 0。因此 x = −½ 或 x = −3。
This technique requires careful attention to signs. Negative coefficients often mislead students, so always check your factors by expanding them back to the original expression.
这种方法需要特别注意符号。负系数常常会误导学生，因此务必通过展开来验证因子是否正确还原为原表达式。
3. Solving by Completing the Square | 配方法
Completing the square is a powerful algebraic technique that transforms a quadratic into a perfect square plus a constant. It is especially useful when factoring is not straightforward.
配方法是一种强大的代数技巧，它把二次式转化为一个完全平方加上一个常数。当因式分解不容易时，这种方法尤其有用。
Consider x² + 6x + 2 = 0. First, halve the coefficient of x, which is 6 ÷ 2 = 3. Then we construct (x + 3)² − 9 + 2 = 0, which simplifies to (x + 3)² − 7 = 0. Hence (x + 3)² = 7, so x + 3 = ±√7, giving x = −3 ± √7.
考虑 x² + 6x + 2 = 0。首先，取 x 系数的一半，即 6 ÷ 2 = 3。然后构造 (x + 3)² − 9 + 2 = 0，化简得 (x + 3)² − 7 = 0。因此 (x + 3)² = 7，所以 x + 3 = ±√7，即 x = −3 ± √7。
For a quadratic where a ≠ 1, first divide the entire equation by a. For instance, 2x² − 4x − 3 = 0 becomes x² − 2x − 1.5 = 0. Then halve −2 to get −1, so we write (x − 1)² − 1 − 1.5 = 0, leading to (x − 1)² = 2.5, and therefore x = 1 ± √2.5 = 1 ± √10⁄2.
对于 a ≠ 1 的二次方程，先将整个方程除以 a。例如，2x² − 4x − 3 = 0 变为 x² − 2x − 1.5 = 0。然后取 −2 的一半得 −1，于是写成 (x − 1)² − 1 − 1.5 = 0，得到 (x − 1)² = 2.5，因此 x = 1 ± √2.5 = 1 ± √10⁄2。
Completing the square also reveals the vertex form of a parabola, y = a(x − h)² + k, where (h, k) is the turning point. This makes it doubly valuable in graph sketching.
配方法还可以揭示抛物线的顶点式 y = a(x − h)² + k，其中 (h, k) 是顶点坐标。这使得它在画图时具有双重价值。
4. The Quadratic Formula | 求根公式
The quadratic formula is the universal solver: given ax² + bx + c = 0, the solutions are given by x = (−b ± √(b² − 4ac)) / (2a). You should memorise this formula and understand what each part means.
求根公式是通用的解法：对于 ax² + bx + c = 0，解为 x = (−b ± √(b² − 4ac)) / (2a)。你应该牢记这个公式，并理解每个部分的含义。
Let us solve 3x² − 5x − 2 = 0 using the formula. Here a = 3, b = −5, c = −2. Substitute to obtain the expression:
让我们用公式求解 3x² − 5x − 2 = 0。这里 a = 3，b = −5，c = −2。代入得到表达式：
x = (5 ± √(25 + 24)) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6
Thus x = (5 + 7)/6 = 2 or x = (5 − 7)/6 = −⅓. The quadratic formula always works, provided the discriminant b² − 4ac is not negative when only real roots are required.
因此 x = (5 + 7)/6 = 2 或 x = (5 − 7)/6 = −⅓。求根公式总是有效的，前提是当只需要实数根时，判别式 b² − 4ac 不为负。
When substituting into the formula, be extremely careful with negative values of b. A common error is writing −(−5) as −5. Always rewrite the equation in standard form before identifying a, b, and c.
代入公式时，要特别小心 b 为负数的情况。常见错误是把 −(−5) 写成 −5。务必先将方程整理为标准形式，再确定 a、b、c 的值。
5. The Discriminant and the Nature of Roots | 判别式与根的性质
The discriminant is the expression Δ = b² − 4ac. Its value tells us how many real roots a quadratic equation has, without solving it completely.
判别式是表达式 Δ = b² − 4ac。它的值告诉我们二次方程有多少个实数根，无需完全求解。
When Δ > 0, the equation has two distinct real roots. When Δ = 0, there is exactly one repeated real root, and the parabola touches the x-axis at a single point. When Δ < 0, there are no real roots, and the parabola does not intersect the x-axis.
当 Δ > 0 时，方程有两个不相等的实数根。当 Δ = 0 时，恰好有一个重根，抛物线在 x 轴上的一个点处相切。当 Δ < 0 时，没有实数根，抛物线不与 x 轴相交。
| Discriminant Δ = b² − 4ac | Nature of Roots | Graph Interpretation |
| Δ > 0 | Two distinct real roots | Intersects x-axis twice |
| Δ = 0 | One repeated root | Touches x-axis once |
| Δ < 0 | No real roots | Does not meet x-axis |
| 判别式 Δ = b² − 4ac | 根的性质 | 图像意义 |
| Δ > 0 | 两个不等实数根 | 与 x 轴有两个交点 |
| Δ = 0 | 一个重根 | 与 x 轴有一个切点 |
| Δ < 0 | 无实数根 | 不与 x 轴相交 |
Consider x² + 2x + 5 = 0. The discriminant is 2² − 4(1)(5) = 4 − 20 = −16, which is negative. Therefore there are no real solutions, and the graph lies entirely above the x-axis.
考虑 x² + 2x + 5 = 0。判别式为 2² − 4(1)(5) = 4 − 20 = −16，是负数。因此没有实数解，图像完全位于 x 轴上方。
6. Solving Word Problems | 解应用题
Many IGCSE questions present quadratic equations through real-world scenarios. A typical example involves a rectangular garden whose area is given, with length expressed in terms of width.
许多 IGCSE 考题通过实际场景呈现二次方程。一个典型例子是：已知矩形花园的面积，长用宽来表示。
Suppose a rectangle has length (x + 3) cm and width x cm, and its area is 40 cm². Then x(x + 3) = 40, which expands to x² + 3x − 40 = 0. Factoring gives (x + 8)(x − 5) = 0, so x = 5 or x = −8. Since width cannot be negative, we reject x = −8 and conclude the width is 5 cm and the length is 8 cm.
假设一个矩形的长为 (x + 3) cm，宽为 x cm，面积为 40 cm²。那么 x(x + 3) = 40，展开得 x² + 3x − 40 = 0。因式分解得 (x + 8)(x − 5) = 0，因此 x = 5 或 x = −8。由于宽不能为负，我们舍去 x = −8，得出宽为 5 cm，长为 8 cm。
Another classic application is projectile motion: h(t) = −5t² + 20t + 1 describes the height of a ball above the ground. To find when the ball hits the ground, set h(t) = 0 and solve using the quadratic formula.
另一个经典应用是抛体运动：h(t) = −5t² + 20t + 1 描述球离地面的高度。要求球何时落地，令 h(t) = 0，然后用求根公式求解。
Word problems require careful translation into equations. Always define your variable clearly, state the quadratic equation, solve it, and check whether each solution makes sense in the context.
应用题需要仔细地将文字转化为方程。务必明确设定未知数，写出二次方程，求解之后，还要检查每个解在实际情况中是否合理。
7. Sketching Quadratic Graphs | 二次函数图像
A quadratic function y = ax² + bx + c always produces a parabola. When a > 0, the parabola opens upward; when a < 0, it opens downward. The coefficient a also affects the width of the graph.
二次函数 y = ax² + bx + c 的图像总是一条抛物线。当 a > 0 时，抛物线开口向上；当 a < 0 时，开口向下。系数 a 还影响图像的宽窄。
The roots of the equation ax² + bx + c = 0 are the x-intercepts of the graph. The vertex or turning point can be found at x = −b/(2a), and its y-coordinate is obtained by substituting this value back into the function.
方程 ax² + bx + c = 0 的根就是图像与 x 轴的交点。顶点坐标可通过 x = −b/(2a) 求得，再将此值代回函数得到 y 坐标。
When sketching, four features suffice: the direction of opening, the y-intercept, the x-intercepts, and the turning point. For practice, sketch y = x² − 4x + 3. The factorised form is y = (x − 1)(x − 3), so the x-intercepts are 1 and 3. The turning point has x = 2, and substituting gives y = −1.
画图时，四个特征就足够了：开口方向、y 截距、x 截距和顶点。例如练习画 y = x² − 4x + 3。因式分解形式为 y = (x − 1)(x − 3)，所以 x 截距为 1 和 3。顶点的 x 坐标为 2，代回得 y = −1。
8. Common Mistakes and Exam Tips | 常见错误与考试技巧
Students commonly forget to set the equation to zero before factoring or applying the formula. Always rearrange to ax² + bx + c = 0 first. Another frequent error is losing the ± sign when taking square roots.
学生常犯的错误包括：在因式分解或使用公式前忘记将方程化为零。务必先将方程整理为 ax² + bx + c = 0。另一个常见错误是取平方根时遗漏 ± 号。
When given a quadratic in non-standard form, such as x² = 5x + 6, do not divide both sides by x. This would lose the root x = 0. Instead, rearrange to x² − 5x − 6 = 0, then factor.
当二次方程不是标准形式时，比如 x² = 5x + 6，不要两边同时除以 x，否则会丢失 x = 0 这个根。应整理为 x² − 5x − 6 = 0，再因式分解。
For the graph of y = ax² + bx + c, ensure you identify the sign of a carefully. A negative a means the parabola opens downwards, and the vertex is at the maximum point, not the minimum.
对于 y = ax² + bx + c 的图像，务必仔细判断 a 的符号。a 为负时抛物线开口向下，顶点是最大值点，而不是最小值点。
Finally, always check your solutions by substituting them back into the original equation. This takes only a few seconds and can save you from losing valuable marks.
最后，务必把解代回原方程检验。这只需几秒钟，却可以帮你避免丢失宝贵的分数。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导