📚 IGCSE Mathematics: Quadratic Equations and Their Graphs | IGCSE 数学：二次方程及其图像
A quadratic equation is one of the most important topics in the IGCSE Mathematics syllabus. It connects algebra, graphing, inequalities, and real-life modelling. Mastering quadratic equations not only helps you solve standard problems but also builds a strong foundation for advanced algebra.
二次方程是 IGCSE 数学大纲中最重要的主题之一。它把代数、图像、不等式和现实建模联系在一起。掌握二次方程不仅能帮助你解决标准问题，还能为更高级的代数打下扎实基础。
1. What is a quadratic equation? | 什么是二次方程？
A quadratic equation in one variable is an equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of x is 2, so the corresponding function y = ax² + bx + c produces a parabola when graphed.
一元二次方程是可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。因为 x 的最高次数是 2，所以对应的函数 y = ax² + bx + c 在作图时会形成一条抛物线。
ax² + bx + c = 0, a ≠ 0
In IGCSE questions, you will be asked to solve quadratic equations, sketch their graphs, interpret the discriminant, and apply them to word problems such as projectile motion or area optimisation.
在 IGCSE 题目中，你会被要求解二次方程、绘制其图像、解释判别式，并将其应用于抛体运动或面积优化等文字题。
2. Solving by factorising | 因式分解法
When the quadratic expression can be factorised, this is usually the fastest method. For example, to solve x² − 5x + 6 = 0, find two numbers that multiply to give 6 and add to give −5. These are −2 and −3, so the equation becomes (x − 2)(x − 3) = 0.
当二次式可以因式分解时，这通常是最快的方法。例如，解 x² − 5x + 6 = 0，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。
Using the zero product property, each bracket can be set to zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. Always expand the brackets as a quick check that the factorised form matches the original quadratic.
利用零乘积性质，每个括号可以设为零：x − 2 = 0 得 x = 2，x − 3 = 0 得 x = 3。始终展开括号作为快速检查，确认因式分解形式与原二次式一致。
x² − 5x + 6 = (x − 2)(x − 3) = 0 ⇒ x = 2 or x = 3
3. Completing the square | 配方法
Completing the square rewrites x² + bx + c in the form (x + p)² + q. This is especially useful for finding the turning point of a graph and for solving quadratics that do not factorise neatly.
配方法将 x² + bx + c 写成 (x + p)² + q 的形式。这对于求图像的顶点以及解不易因式分解的二次方程特别有用。
To complete the square for x² + 6x + 2, take half of the coefficient of x, which is 3, and square it to get 9. Add and subtract 9: x² + 6x + 9 − 9 + 2 = (x + 3)² − 7. So the expression x² + 6x + 2 is equivalent to (x + 3)² − 7.
例如对 x² + 6x + 2 配方，取 x 系数的一半为 3，平方得 9。加上再减去 9：x² + 6x + 9 − 9 + 2 = (x + 3)² − 7。因此 x² + 6x + 2 等价于 (x + 3)² − 7。
x² + bx + c = (x + b/2)² − (b/2)² + c
This method also reveals the minimum or maximum value of the quadratic function, which is q when p = −b/2.
该方法还揭示了二次函数的最小值或最大值，即当 p = −b/2 时，值为 q。
4. The quadratic formula | 求根公式
The quadratic formula gives the roots of any quadratic equation ax² + bx + c = 0. It is derived by completing the square and works even when factorisation is difficult or impossible.
求根公式给出任何二次方程 ax² + bx + c = 0 的根。它由配方法推导而来，即使因式分解困难或不可行时也总是有效。
x = ( −b ± √(b² − 4ac) ) ÷ (2a)
For example, solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2. Substitute into the formula: x = ( −3 ± √(9 + 16) ) ÷ 4 = ( −3 ± √25 ) ÷ 4 = ( −3 ± 5 ) ÷ 4. This gives x = 1/2 or x = −2.
例如，解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入公式：x = ( −3 ± √(9 + 16) ) ÷ 4 = ( −3 ± √25 ) ÷ 4 = ( −3 ± 5 ) ÷ 4。得到 x = 1/2 或 x = −2。
Always keep the ± sign until the final step. Many exam errors come from dropping one root too early or forgetting that the entire numerator must be divided by 2a.
始终保留 ± 号直到最后一步。许多考试错误来自过早丢掉一个根，或忘记整个分子都要除以 2a。
5. The discriminant and nature of roots | 判别式与根的性质
The discriminant is the part under the square root in the quadratic formula: Δ = b² − 4ac. It tells you how many real roots the equation has without solving it.
判别式是求根公式中平方根下的部分：Δ = b² − 4ac。它让你无需解方程就能知道方程有多少个实数根。
- If Δ > 0, there are two distinct real roots.
- 若 Δ > 0，则有两个不同的实数根。
- If Δ = 0, there is one repeated real root, meaning the graph touches the x-axis at only one point.
- 若 Δ = 0，则有一个重复的实数根，意味着图像仅在一个点与 x 轴相切。
- If Δ < 0, there are no real roots; the parabola does not cross the x-axis.
- 若 Δ < 0，则没有实数根；抛物线不与 x 轴相交。
For example, for x² + 4x + 4 = 0, Δ = 4² − 4 × 1 × 4 = 0, so there is one repeated root. For x² + x + 1 = 0, Δ = 1 − 4 = −3, so there are no real roots.
例如，对于 x² + 4x + 4 = 0，Δ = 4² − 4 × 1 × 4 = 0，因此有一个重复根。对于 x² + x + 1 = 0，Δ = 1 − 4 = −3，因此没有实数根。
6. Sketching quadratic graphs | 二次函数图像草图
The graph of y = ax² + bx + c is a parabola. The sign of a determines its direction: if a > 0, it 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。a 的符号决定其开口方向：如果 a > 0，开口向上且有最低点；如果 a < 0，开口向下且有最高点。
The y-intercept is simply c, because when x = 0, y = c. The x-intercepts are the real roots of ax² + bx + c = 0, if they exist. You can find them by factorising, the formula, or completing the square.
y 轴截距就是 c，因为当 x = 0 时，y = c。x 轴截距是 ax² + bx + c = 0 的实数根（如果存在）。你可以通过因式分解、公式或配方法找到它们。
For example, the graph of y = x² − 4x + 3 has a y-intercept at 3 and x-intercepts at x = 1 and x = 3 because x² − 4x + 3 = (x − 1)(x − 3).
例如，y = x² − 4x + 3 的图像 y 轴截距为 3，x 轴截距为 x = 1 和 x = 3，因为 x² − 4x + 3 = (x − 1)(x − 3)。
7. Turning points and axis of symmetry | 顶点与对称轴
Every parabola has a turning point, also called the vertex. Its x-coordinate is given by x = −b ÷ (2a). The corresponding y-coordinate can be found by substituting this x value into the original function, or by using y = −Δ ÷ (4a).
每条抛物线都有一个顶点，也称为 turning point。其 x 坐标由 x = −b ÷ (2a) 给出。对应的 y 坐标可以将此 x 值代入原函数求出，或使用 y = −Δ ÷ (4a)。
Vertex = ( −b ÷ (2a), −Δ ÷ (4a) )
The axis of symmetry is the vertical line that passes through the vertex: x = −b ÷ (2a). This line divides the parabola into two mirror-image halves and is essential for accurate graph sketching.
对称轴是经过顶点的竖直线：x = −b ÷ (2a)。这条线将抛物线分成两个镜像对称的部分，对于准确绘制图像非常重要。
For y = x² − 4x + 3, the vertex is at x = 4 ÷ 2 = 2, and y = 2² − 8 + 3 = −1, so the vertex is (2, −1). The minimum value of the function is therefore −1.
对于 y = x² − 4x + 3，顶点在 x = 4 ÷ 2 = 2 处，y = 2² − 8 + 3 = −1，因此顶点为 (2, −1)。所以该函数的最小值是 −1。
8. Solving quadratic inequalities | 解二次不等式
A quadratic inequality such as x² − 4x + 3 < 0 is solved by first finding the roots of the related equation x² − 4x + 3 = 0. The roots are x = 1 and x = 3, so these values split the number line into three intervals.
形如 x² − 4x + 3 < 0 的二次不等式，首先要求出相关方程 x² − 4x + 3 = 0 的根。根为 x = 1 和 x = 3，因此这些值将数轴分成三个区间。
Test a value from each interval: for x < 1, choose x = 0, the expression is positive; for 1 < x < 3, choose x = 2, it is negative; for x > 3, choose x = 4, it is positive. Since we want the expression to be less than zero, the solution is 1 < x < 3.
从每个区间取一个值检验：对于 x < 1，取 x = 0，表达式为正；对于 1 < x < 3，取 x = 2，表达式为负；对于 x > 3，取 x = 4，表达式为正。因为我们需要表达式小于零，所以解为 1 < x < 3。
If the inequality had been x² − 4x + 3 ≥ 0, the solution would have been x ≤ 1 or x ≥ 3. Always pay attention to whether the inequality is strict or includes equality.
如果不等式是 x² − 4x + 3 ≥ 0，解将是 x ≤ 1 或 x ≥ 3。始终注意不等式是严格的还是包含等号。
9. Applications in real-life problems | 实际应用问题
Quadratic equations model many real-life situations, including projectile motion, area problems, and profit maximisation. A typical example is the height of a ball thrown upwards: h = −5t² + 20t + 1, where h is the height in metres and t is time in seconds.
二次方程可以建模许多现实情况，包括抛体运动、面积问题和利润最大化。一个典型例子是向上抛出的球的高度：h = −5t² + 20t + 1，其中 h 为高度（米），t 为时间（秒）。
To find when the ball hits the ground, set h = 0: −5t² + 20t + 1 = 0. Using the quadratic formula gives t ≈ −0.05 or t ≈ 4.05 seconds. Only the positive value makes sense in this context, so the ball lands after about 4.05 seconds.
要求球何时落地，令 h = 0：−5t² + 20t + 1 = 0。使用求根公式得到 t ≈ −0.05 或 t ≈ 4.05 秒。在这个情境中只有正值有意义，因此球大约在 4.05 秒后落地。
Area problems often lead to quadratics as well. For instance, if the length of a rectangle is 3 cm longer than its width and its area is 40 cm², then x(x + 3) = 40, giving x² + 3x − 40 = 0, which solves to x = 5 or x = −8. The width must be positive, so x = 5 cm.
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