📚 Quadratic Equations: Solving and Graphing | IGCSE 数学：二次方程求解与图像
Quadratic equations appear throughout the IGCSE syllabus, from pure algebraic manipulation to real-world modelling. A confident command of factorisation, completing the square, the quadratic formula, and graph interpretation is essential for high marks in both Core and Extended papers.
二次方程贯穿 IGCSE 数学大纲，从纯代数运算到实际问题建模都有考查。熟练掌握因式分解、配方法、二次公式以及图像解读，是 Core 和 Extended 试卷取得高分的关键。
1. What is a Quadratic Equation? | 什么是二次方程？
t common form is ax² + bx + c = 0, where a, b, and c are constants and a must not be zero.
二次方程是指未知数最高次数为 2 的多项式方程。它最常见的形式是 ax² + bx + c = 0，其中 a、b、c 是常数，且 a 不能为零。
If a were zero, the x² term would disappear and the equation would become linear rather than quadratic. The graph of a quadratic expression is always a parabola, which may open upwards or downwards.
如果 a 等于零，x² 项就会消失，方程会变成一次方程而不是二次方程。二次表达式的图像总是一条抛物线，可能开口向上或向下。
ax² + bx + c = 0, a ≠ 0
2. Standard Form and Coefficients | 标准式与系数
s form, a is called the quadratic coefficient, b is the linear coefficient, and c is the constant term.
在求解二次方程之前，通常最好先将其整理为标准式：ax² + bx + c = 0。在这个形式中，a 称为二次项系数，b 称为一次项系数，c 称为常数项。
0, we have a = 3, b = −7, and c = 2. In x² − 5x + 6 = 0, we have a = 1, b = −5, and c = 6.
例如，在方程 3x² − 7x + 2 = 0 中，a = 3，b = −7，c = 2。在方程 x² − 5x + 6 = 0 中，a = 1，b = −5，c = 6。
Getting the right values of a, b, and c is especially important when using the quadratic formula, because sign errors are a common cause of lost marks.
正确确定 a、b、c 的值在使用二次公式时尤其重要，因为符号错误是常见的失分原因。
3. Solving by Factorisation | 因式分解法求解
Factorisation is often the fastest method when the quadratic can be written as a product of two linear factors. The key idea is the zero product property: if a product equals zero, then at least one of its factors must be zero.
当二次式可以写成两个一次因式的乘积时，因式分解通常是最快的方法。核心思想是零积性质：如果乘积等于零，那么至少有一个因式等于零。
6 and add to −5. These numbers are −2 and −3, so the equation becomes (x − 2)(x − 3) = 0.
对于类似 x² − 5x + 6 = 0 的简单二次方程，我们需要找到两个数，它们的乘积为 6，和为 −5。这两个数是 −2 和 −3，因此方程变为 (x − 2)(x − 3) = 0。
(x − 2)(x − 3) = 0 ⇒ x = 2 or x = 3
When a is not 1, you may need to split the middle term or use trial and error to find the correct factors. Always expand your factors at the end to check that they match the original expression.
当 a 不等于 1 时，可能需要拆分中间项或通过试错找到正确的因式。最后一定要将因式展开，检查是否与原式一致。
4. Solving by Completing the Square | 配方法求解
Completing the square rewrites a quadratic expression in the form a(x − h)² + k. This is useful for solving equations, finding turning points, and proving the quadratic formula.
配方法将二次表达式改写为 a(x − h)² + k 的形式。这种方法对解方程、求顶点以及证明二次公式都很有用。
use the identity x² + bx = (x + b/2)² − (b/2)². For example, x² + 6x becomes (x + 3)² − 9.
对于 x² + bx，配方时使用恒等式 x² + bx = (x + b/2)² − (b/2)²。例如，x² + 6x 可写成 (x + 3)² − 9。
x² + 6x + 2 = 0 ⇒ (x + 3)² − 9 + 2 = 0 ⇒ (x + 3)² = 7
Taking square roots then gives x + 3 = ±√7, so x = −3 ± √7. Completing the square is particularly reliable when factorisation is not obvious.
接着开平方得到 x + 3 = ±√7，所以 x = −3 ± √7。当因式分解不明显时，配方法尤其可靠。
5. The Quadratic Formula | 二次公式法
ard form. It is derived by completing the square on the general equation ax² + bx + c = 0.
二次公式可以求解任何标准形式的二次方程。它是由一般方程 ax² + bx + c = 0 通过配方法推导出来的。
x = (−b ± √b² − 4ac) ÷ (2a)
x − 3 = 0, we have a = 2, b = −4, and c = −3. Substituting into the formula gives x = (4 ± √16 + 24) ÷ 4 = (4 ± √40) ÷ 4.
例如，在 2x² − 4x − 3 = 0 中，a = 2，b = −4，c = −3。代入公式得到 x = (4 ± √16 + 24) ÷ 4 = (4 ± √40) ÷ 4。
This simplifies to x = 1 ± ½√10. The formula is especially useful when the roots are irrational or when factorisation is difficult.
化简后得到 x = 1 ± ½√10。当根是无理数或因式分解困难时，二次公式尤其有用。
6. The Discriminant | 判别式
lls us how many real roots a quadratic equation has without actually solving the equation.
判别式是二次公式中根号下的部分：Δ = b² − 4ac。它可以在不解方程的情况下告诉我们二次方程有多少个实根。
- Δ > 0: two distinct real roots | 两个不同的实根
- Δ = 0: one repeated real root | 一个重根
- Δ < 0: no real roots | 没有实根
x² − 4x + 4 = 0, Δ = (−4)² − 4 × 1 × 4 = 0, so the equation has one repeated root: x = 2.
例如，在 x² − 4x + 4 = 0 中，Δ = (−4)² − 4 × 1 × 4 = 0，所以方程有一个重根：x = 2。
| Discriminant | Number of real roots |
| b² − 4ac > 0 | 2 distinct roots |
| b² − 4ac = 0 | 1 repeated root |
| b² − 4ac < 0 | 0 real roots |
7. Graphs of Quadratic Functions | 二次函数图像
a graph called a parabola. The sign of a determines whether the parabola opens upwards or downwards.
二次函数 y = ax² + bx + c 的图像是一条抛物线。a 的正负决定了抛物线开口向上还是向下。
- If a > 0, the parabola opens upwards and has a minimum point.
- If a < 0, the parabola opens downwards and has a maximum point.
0, y = c. The graph is always symmetric about a vertical line called the axis of symmetry.
常数 c 是 y 轴截距，因为当 x = 0 时，y = c。图像总是关于一条称为对称轴的竖直线对称。
Axis of symmetry: x = −b ÷ (2a)
8. Turning Point and Axis of Symmetry | 顶点与对称轴
The turning point is the point where a parabola changes direction. For a positive quadratic it is the minimum, and for a negative quadratic it is the maximum.
顶点是抛物线改变方向的点。对于 a 为正的二次函数，它是最小值点；对于 a 为负的二次函数，它是最大值点。
−b ÷ (2a). The y-coordinate is found by substituting this x-value back into the function.
从标准式出发，顶点的 x 坐标是 x = −b ÷ (2a)。将这个 x 值代回函数即可求出 y 坐标。
mple, y = x² − 4x + 1 can be written as y = (x − 2)² − 3, so the turning point is (2, −3).
将二次函数写成顶点式 y = a(x − h)² + k 可以直接看出顶点为 (h, k)。例如，y = x² − 4x + 1 可以写成 y = (x − 2)² − 3，因此顶点是 (2, −3)。
9. Roots and x-Intercepts | 根与 x 轴交点
aph crosses or touches the x-axis. To find these intercepts, we set y = 0 and solve for x.
二次方程 ax² + bx + c = 0 的根是图像与 x 轴相交或相切处的 x 值。为了求这些交点，我们令 y = 0 并解出 x。
If the discriminant is positive, the graph crosses the x-axis at two distinct points. If it is zero, the graph touches the x-axis at exactly one point, which is the turning point. If it is negative, the graph does not meet the x-axis at all.
如果判别式为正，图像与 x 轴在两个不同点相交。如果判别式为零，图像恰好在顶点处与 x 轴相切。如果判别式为负，图像完全不与 x 轴相交。
This connection between algebraic roots and graphical intercepts is frequently tested in IGCSE questions, especially when sketching curves.
代数根与图像交点之间的联系在 IGCSE 题目中经常考查，特别是在绘制曲线草图时。
10. Applications and Word Problems | 应用题与文字题
Quadratic equations are often used to model area, projectile motion, profit, and consecutive-number problems. In word problems, the solution must usually be interpreted in the context of the question.
二次方程通常用于建立面积、抛体运动、利润和连续整数等问题的模型。在文字题中，解通常需要结合题目情境进行解释。
the width be x, so the length is x + 3. Then x(x + 3) = 54, which gives x² + 3x − 54 = 0.
例题：一个长方形花园的长比宽多 3 m，面积为 54 m²。设宽为 x，则长为 x + 3。于是 x(x + 3) = 54，得到 x² + 3x − 54 = 0。
x² + 3x − 54 = 0 ⇒ (x + 9)(x − 6) = 0 ⇒ x = 6 or x = −9
s x = 6 m. Always check that your answers make sense in the original real-world situation.
由于宽度不能为负数，有效解是 x = 6 m。最终一定要检查答案在原始实际情境中是否合理。
11. Common Mistakes and Exam Tips | 常见错误与考试技巧
Many marks are lost through avoidable errors such as forgetting to rearrange the equation to equal zero, omitting signs, or failing to consider both solutions from a square root.
许多失分都来自可以避免的错误，例如忘记将方程整理为等于零、遗漏符号，或者开平方后没有同时考虑正负两个解。
- Always write the equation in standard form before choosing a method.
- Check factorisations by expanding them back.
- When using the formula, be very careful with negative b values.
- Use the discriminant to check the number of real roots before solving.
- Leave answers in exact form unless the question asks for decimals.
中文要点：
- 选择解法前，先把方程写成标准式。
- 通过展开检验因式分解是否正确。
- 使用公式时，要特别注意 b 为负数的情况。
- 求解前可用判别式判断实根个数。
- 除非题目要求小数，否则答案保留精确形式。
Finally, substitute your roots back into the original equation to verify that they satisfy it. This habit will help you catch sign errors and avoid unnecessary marks being lost in the exam.
最后，将根代回原方程进行验证。这个习惯可以帮助你发现符号错误，避免在考试中不必要的失分。
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