📚 Mastering Quadratic Equations for IGCSE Mathematics | 精通 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, graph sketching, and in word problems involving area, motion, and profit. Understanding how to solve and sketch quadratics is essential for success in both Core and Extended papers.
二次方程是 IGCSE 数学中最重要的主题之一。它出现在代数、坐标几何、函数图像以及涉及面积、运动和利润的应用题中。掌握二次方程的求解与作图方法，对于 Core 和 Extended 考试都至关重要。
1. What Is a Quadratic Equation? | 什么是二次方程
A quadratic equation is an equation that can be written in the general form ax² + bx + c = 0, where a, b and c are constants, and a is not equal to zero. The highest power of the variable x is 2, which is why it is called quadratic.
二次方程是可以写成一般形式 ax² + bx + c = 0 的方程，其中 a、b、c 是常数，且 a 不等于零。变量 x 的最高次数是 2，因此称为二次方程。
The condition a ≠ 0 is important: if a = 0, the equation becomes linear, bx + c = 0. A quadratic equation must always contain an x² term.
条件 a ≠ 0 非常重要：如果 a = 0，方程就变为一次方程 bx + c = 0。二次方程必须始终含有 x² 项。
ax² + bx + c = 0, a ≠ 0
Common examples include x² + 5x + 6 = 0, 2x² − 3x + 1 = 0, and x² − 9 = 0. The third example is a special case where b = 0.
常见例子包括 x² + 5x + 6 = 0、2x² − 3x + 1 = 0 和 x² − 9 = 0。第三个例子是 b = 0 的特殊情况。
2. Standard Form and Coefficients | 标准形式与系数
Before solving a quadratic equation, you should rearrange it into standard form: ax² + bx + c = 0. This makes it easy to identify the coefficients a, b and c, which are needed for factorising, completing the square, and the quadratic formula.
在求解二次方程之前，应将其化为标准形式：ax² + bx + c = 0。这样可以轻松确定系数 a、b 和 c，便于因式分解、配方法和使用求根公式。
For example, in the equation 3x² − 7x + 2 = 0, the coefficient a = 3, b = −7, and c = 2. Always include the sign in front of the coefficient.
例如，在方程 3x² − 7x + 2 = 0 中，系数 a = 3、b = −7、c = 2。确定系数时始终要带上它前面的符号。
If an equation is given as 4x − x² = 5, rearrange it to −x² + 4x − 5 = 0 or multiply both sides by −1 to get x² − 4x + 5 = 0. Then a = 1, b = −4, and c = 5.
如果给出的方程是 4x − x² = 5，应将其整理为 −x² + 4x − 5 = 0，或将两边同时乘以 −1 得到 x² − 4x + 5 = 0。此时 a = 1、b = −4、c = 5。
3. Solving by Factorising | 因式分解法
Factorising is usually the fastest method when the quadratic has simple integer roots. The idea is to write ax² + bx + c as a product of two brackets, such as (px + q)(rx + s) = 0.
当二次方程具有简单的整数根时，因式分解通常是最快的方法。其思路是将 ax² + bx + c 写成两个括号的乘积，例如 (px + q)(rx + s) = 0。
For x² + 5x + 6 = 0, we look for two numbers that multiply to 6 and add to 5. These are 2 and 3, so we write (x + 2)(x + 3) = 0. Then either x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3.
对于 x² + 5x + 6 = 0，我们要找两个数，它们的乘积是 6，和是 5。这两个数是 2 和 3，因此写成 (x + 2)(x + 3) = 0。于是 x + 2 = 0 或 x + 3 = 0，解得 x = −2 或 x = −3。
When a ≠ 1, more care is needed. For 2x² + 7x + 3, multiply a and c to get 2 × 3 = 6. We need two numbers that multiply to 6 and add to 7: these are 6 and 1. Split the middle term and factor by grouping: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
当 a ≠ 1 时需要更加仔细。对于 2x² + 7x + 3，先将 a 和 c 相乘得到 2 × 3 = 6。我们需要两个数，乘积为 6，且和为 7：它们是 6 和 1。拆开中间项并分组分解：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
Then set each bracket equal to zero: 2x + 1 = 0 gives x = −1/2, and x + 3 = 0 gives x = −3. Always substitute your answers back into the original equation to check.
然后令每个括号等于零：2x + 1 = 0 得 x = −1/2，x + 3 = 0 得 x = −3。最后始终要把答案代回原方程进行检验。
4. Solving by the Quadratic Formula | 求根公式法
The quadratic formula works for every quadratic equation, even when factorising is difficult or impossible. It is derived from completing the square and is given in the IGCSE formula sheet, but you must know how to use it correctly.
求根公式适用于所有二次方程，即使因式分解困难或无法进行时也能使用。它由配方法推导而来，IGCSE 公式表中会给出，但你必须知道如何正确使用。
x = (−b ± √(b² − 4ac)) / 2a
For the equation 2x² − 3x − 2 = 0, identify a = 2, b = −3, c = −2. Substitute into the formula: x = (3 ± √((−3)² − 4 × 2 × (−2))) / (2 × 2).
对于方程 2x² − 3x − 2 = 0，确定 a = 2、b = −3、c = −2。代入公式：x = (3 ± √((−3)² − 4 × 2 × (−2))) / (2 × 2)。
Simplify the discriminant: (−3)² = 9, and −4 × 2 × (−2) = 16, so the expression under the square root is 25. Thus x = (3 ± √25) / 4 = (3 ± 5) / 4. This gives x = 8/4 = 2 or x = −2/4 = −1/2.
化简判别式：(−3)² = 9，且 −4 × 2 × (−2) = 16，因此根号下的表达式为 25。于是 x = (3 ± √25) / 4 = (3 ± 5) / 4。解得 x = 8/4 = 2 或 x = −2/4 = −1/2。
Remember to write the formula accurately and to be careful with negative signs. A common error is writing −b incorrectly when b is already negative.
请准确书写公式并注意负号。常见错误是当 b 本身为负数时错误地处理 −b。
5. Completing the Square | 配方法
Completing the square is a method that rewrites a quadratic expression in the form a(x − h)² + k. It is useful for finding the turning point of a graph and for solving equations when factorising is not straightforward.
配方法是一种将二次表达式改写为 a(x − h)² + k 形式的方法。它有助于确定图像的顶点，并在因式分解不方便时求解方程。
To complete the square for x² + 6x + 1, take half of the coefficient of x, which is 6 ÷ 2 = 3, square it to get 9, then write (x + 3)² − 9 + 1. This simplifies to (x + 3)² − 8.
对 x² + 6x + 1 进行配方时，取 x 系数的一半，即 6 ÷ 2 = 3，将其平方得到 9，然后写成 (x + 3)² − 9 + 1。化简得 (x + 3)² − 8。
If a ≠ 1, factor the coefficient of x² out of the first two terms first. For example, 2x² + 8x + 5 becomes 2(x² + 4x) + 5. Since half of 4 is 2 and 2² = 4, write 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3.
如果 a ≠ 1，应先将 x² 的系数从前两项中提出。例如，2x² + 8x + 5 变为 2(x² + 4x) + 5。因为 4 的一半是 2，且 2² = 4，所以写成 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。
When solving x² + 6x + 1 = 0, set (x + 3)² − 8 = 0. Then (x + 3)² = 8, so x + 3 = ±√8 and x = −3 ± √8. You can simplify √8 to 2√2 if required.
解方程 x² + 6x + 1 = 0 时，令 (x + 3)² − 8 = 0。于是 (x + 3)² = 8，所以 x + 3 = ±√8，x = −3 ± √8。如果题目要求，可进一步将 √8 化简为 2√2。
6. The Discriminant | 判别式
The discriminant is the expression b² − 4ac found under the square root in the quadratic formula. It tells you the number and type of real roots without solving the equation.
判别式是求根公式中根号下的表达式 b² − 4ac。它可以在不解方程的情况下判断实根的个数和类型。
| Discriminant Δ = b² − 4ac | Real roots | Graph behaviour |
| Δ > 0 | Two distinct real roots | Graph crosses x-axis twice |
| Δ = 0 | One repeated real root | Graph touches x-axis once |
| Δ < 0 | No real roots | Graph does not cross x-axis |
For example, x² − 4x + 5 has Δ = (−4)² − 4 × 1 × 5 = 16 − 20 = −4. Since Δ < 0, the equation has no real solutions, and its graph lies entirely above the x-axis.
例如，x² − 4x + 5 的判别式 Δ = (−4)² − 4 × 1 × 5 = 16 − 20 = −4。由于 Δ < 0，该方程没有实数解，其图像完全位于 x 轴上方。
If Δ = 0, the quadratic is a perfect square trinomial. For instance, x² − 6x + 9 has Δ = 36 − 36 = 0, so the only root is x = 3, and the factorised form is (x − 3)².
如果 Δ = 0，则二次式是完全平方式。例如，x² − 6x + 9 的 Δ = 36 − 36 = 0，因此唯一的根是 x = 3，因式分解形式为 (x − 3)²。
7. Sketching Quadratic Graphs | 二次函数图像
The graph of y = ax² + bx + c is a parabola. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上，且有一个最低点。如果 a < 0，抛物线开口向下，且有一个最高点。
To sketch a quadratic graph, find the y-intercept by setting x = 0. This gives y = c. Then find the x-intercepts by solving ax² + bx + c = 0, either by factorising or using the quadratic formula.
绘制二次函数图像时，令 x = 0 可求出 y 轴截距，得到 y = c。再通过因式分解或求根公式解 ax² + bx + c = 0 来求出 x 轴截距。
The axis of symmetry is the vertical line x = −b / 2a. The turning point lies on this line. Substitute x = −b / 2a into the equation to find its y-coordinate.
对称轴是竖直直线 x = −b / 2a。顶点就位于这条直线上。将 x = −b / 2a 代入方程即可求出顶点的 y 坐标。
For y = x² − 4x + 3, we have a = 1, b = −4, c = 3. The y-intercept is 3. Factorising gives (x − 1)(x − 3) = 0, so the x-intercepts are 1 and 3. The axis of symmetry is x = 2, and the turning point is (2, −1).
对于 y = x² − 4x + 3，有 a = 1、b = −4、c = 3。y 轴截距为 3。因式分解得 (x − 1)(x − 3) = 0，因此 x 轴截距为 1 和 3。对称轴为 x = 2，顶点为 (2, −1)。
8. Roots, Intercepts and Turning Point | 根、截距与顶点
The roots of a quadratic equation ax² + bx + c = 0 are the values of x that make y = 0. On a graph, they are the x-coordinates where the parabola crosses or touches the x-axis.
二次方程 ax² + bx + c = 0 的根是使 y = 0 的 x 值。在图像上，它们是抛物线与 x 轴相交或相切处的横坐标。
The y-intercept is the point where the graph crosses the y-axis, found by setting x = 0. This coordinate is always (0, c).
y 轴截距是图像与 y 轴相交的点，令 x = 0 即可求出。该坐标始终是 (0, c)。
The turning point is also called the vertex. Its x-coordinate is x = −b / 2a, and its y-coordinate is found by substitution. The vertex is a minimum if a > 0 and a maximum if a < 0.
顶点也称 turning point。其 x 坐标为 x = −b / 2a，y 坐标通过代入求得。当 a > 0 时该点为最小值，当 a < 0 时为最大值。
In completing-the-square form y = a(x − h)² + k, the vertex is simply (h, k). Be careful with signs: y = (x + 3)² − 8 has vertex (−3, −8), not (3, −8).
在配方式 y = a(x − h)² + k 中，顶点就是 (h, k)。注意符号：y = (x + 3)² − 8 的顶点是 (−3, −8)，而不是 (3, −8)。
9. Real-Life Applications | 实际应用
IGCSE word problems often use quadratic equations to model area, projectile motion, or profit. Set up the equation from the given information, then solve it using a suitable method, and check that your answers make sense in context.
IGCSE 应用题经常使用二次方程来建模面积、抛体运动或利润问题。根据已知信息建立方程，然后选择合适的方法求解，并检查答案是否符合实际情况。
For example, a rectangular garden has length 3 m more than its width, and its area is 40 m². Let the width be x, so the length is x + 3. The area equation is x(x + 3) = 40, which expands to x² + 3x − 40 = 0.
例如，一个长方形花园的长比宽多 3 m，面积为 40 m²。设宽为 x，则长为 x + 3。面积方程为 x(x + 3) = 40，展开得 x² + 3x − 40 = 0。
Factorising gives (x + 8)(x − 5) = 0, so x = −8 or x = 5. Since width cannot be negative, x = 5 m, and the length is 8 m.
因式分解得 (x + 8)(x − 5) = 0，因此 x = −8 或 x = 5。由于宽度不能为负，所以 x = 5 m，长为 8 m。
In projectile problems, the height h of an object might be given by h = −5t² + 20t + 10. Setting h = 0 gives a quadratic equation for the time when the object lands.
在抛体运动问题中，物体的高度 h 可能由 h = −5t² + 20t + 10 给出。令 h = 0 可得到物体落地时间对应的二次方程。
10. Common IGCSE Exam Tips | 常见考试技巧
Always show your method clearly, even if you use a calculator. Many IGCSE questions award marks for correct substitution into the quadratic formula or for setting up the equation, not just for the final answer.
即使使用计算器，也要清晰地写出解题过程。IGCSE 许多题目会给正确代入求根公式或正确建立方程的步骤分，而不仅仅是最终答案分。
After solving, substitute each root back into the original equation to check that it gives zero. This catches sign errors and careless mistakes.
求解后，将每个根代回原方程检验是否等于零。这可以发现符号错误和粗心造成的失误。
If a question asks for exact answers, leave roots in surd form, such as x = −3 ± √8, rather than rounding them. If it asks for decimal answers, round only at the final step.
如果题目要求精确答案，根应保留为无理数形式，如 x = −3 ± √8，而不是取近似值。如果要求小数答案，只应在最后一步进行四舍五入。
When sketching, label the y-intercept, x-intercepts, and turning point clearly. The sketch should show the correct shape, with the parabola opening upwards for a > 0 and downwards for a < 0.
作图时，清楚标注 y 轴截距、x 轴截距和顶点。草图应显示正确形状，即 a > 0 时抛物线开口向上，a < 0 时开口向下。
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