Quadratic Equations and Parabolas | 二次方程与抛物线

📚 Quadratic Equations and Parabolas | 二次方程与抛物线

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, and countless real-world problems. This revision article will guide you through the essential skills: solving quadratic equations, analysing the discriminant, sketching parabolas, and applying these ideas to word problems.

二次方程是IGCSE数学中最重要的课题之一。它出现在代数、图像以及无数实际问题中。本复习文章将引导你掌握核心技能：解二次方程、分析判别式、绘制抛物线草图，以及将这些思想应用于文字题。

1. Standard Form of a Quadratic Equation | 二次方程的标准形式

A quadratic equation in one variable can be written as ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The highest power of the variable is 2. If a = 0, the equation becomes linear, not quadratic.

一元二次方程可以写成 ax² + bx + c = 0，其中 a、b、c 是实数且 a ≠ 0。变量的最高次数是2。如果 a = 0，方程就变成一次方程，而不是二次方程。

ax² + bx + c = 0 (a ≠ 0)

For example, 2x² − 5x + 3 = 0 is quadratic, but x² − 4 = 0 is also quadratic because it can be written as x² + 0x − 4 = 0.

例如，2x² − 5x + 3 = 0 是二次方程，而 x² − 4 = 0 也是二次方程，因为它可以写成 x² + 0x − 4 = 0。

2. Solving by Factorisation | 因式分解法

The simplest way to solve a quadratic equation is to factorise the left-hand side. If the product of two linear expressions is zero, then at least one of them must equal zero. This is called the zero product property.

解二次方程最简单的方式是对左边进行因式分解。如果两个一次表达式的乘积为零，那么其中至少一个必须等于零。这称为零乘积性质。

Example: Solve x² − 5x + 6 = 0. First factorise: (x − 2)(x − 3) = 0. Then set each factor equal to zero: x − 2 = 0 or x − 3 = 0. Hence x = 2 or x = 3.

例：解 x² − 5x + 6 = 0。首先因式分解：(x − 2)(x − 3) = 0。然后令每个因子等于零：x − 2 = 0 或 x − 3 = 0。因此 x = 2 或 x = 3。

Always expand your factorisation to check it is correct. If the coefficient of x² is not 1, you may need to use the “ac method” or trial and error.

一定要展开因式来检查是否正确。如果 x² 的系数不是 1，你可能需要采用“ac 法”或试错法。

3. Solving by Completing the Square | 配方法

Completing the square rewrites ax² + bx + c in the form a(x − h)² + k. This is useful for solving equations that do not factorise easily, and it also reveals the vertex of the parabola.

配方法将 ax² + bx + c 改写为 a(x − h)² + k 的形式。这可用于解不易因式分解的方程，也能直接揭示抛物线的顶点。

For a monic quadratic x² + bx, add and subtract (b/2)². For example, x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7.

对于首项系数为1的二次式 x² + bx，加上并减去 (b/2)²。例如，x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7。

To solve x² + 6x + 2 = 0, write (x + 3)² = 7, then x + 3 = ±√7, so x = −3 ± √7.

要解 x² + 6x + 2 = 0，写成 (x + 3)² = 7，然后 x + 3 = ±√7，所以 x = −3 ± √7。

4. The Quadratic Formula | 二次公式

The quadratic formula works for every quadratic equation. It is derived by completing the square on the general form. You are usually allowed to use this formula in the IGCSE exam, but you must memorise it carefully.

二次公式适用于所有二次方程。它通过对一般形式配方推导而来。在IGCSE考试中通常允许使用该公式，但你必须准确记住它。

x = (−b ± √(b² − 4ac)) / (2a)

Example: Solve 2x² + 3x − 5 = 0. Here a = 2, b = 3, c = −5. Substitute into the formula:

例：解 2x² + 3x − 5 = 0。这里 a = 2, b = 3, c = −5。代入公式：

x = (−3 ± √(9 + 40)) / 4 = (−3 ± 7) / 4

Therefore x = 1 or x = −5/2.

因此 x = 1 或 x = −5/2。

5. The Discriminant | 判别式

The expression b² − 4ac under the square root in the quadratic formula is called the discriminant. It tells us how many real roots the equation has.

二次公式根号下的表达式 b² − 4ac 称为判别式。它告诉我们方程有多少个实数根。

If b² − 4ac > 0, there are two distinct real roots.
If b² − 4ac = 0, there is exactly one repeated real root.
If b² − 4ac < 0, there are no real roots.

如果 b² − 4ac > 0，方程有两个不同的实数根。
如果 b² − 4ac = 0，方程有一个重根。
如果 b² − 4ac < 0，方程没有实数根。

For example, the equation x² + 2x + 5 = 0 has discriminant 2² − 4×1×5 = 4 − 20 = −16, which is negative, so it has no real roots.

例如，方程 x² + 2x + 5 = 0 的判别式为 2² − 4×1×5 = 4 − 20 = −16，是负数，所以它没有实数根。

6. Graph of a Quadratic Function | 二次函数的图像

The graph of y = ax² + bx + c is always a parabola. When a > 0, the parabola opens upwards; when a < 0, it opens downwards. The larger the absolute value of a, the narrower the parabola.

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。|a| 越大，抛物线越窄。

To sketch a parabola, you need the y-intercept (0, c), the x-intercepts (solutions of ax² + bx + c = 0), and the vertex. The vertex can be found using x = −b/(2a).

画抛物线草图时，你需要 y 轴截距 (0, c)、x 轴截距（即方程 ax² + bx + c = 0 的解）以及顶点。顶点可用 x = −b/(2a) 求得。

For instance, y = x² − 4x + 3 has y-intercept 3, x-intercepts 1 and 3, and vertex x = 2, y = −1. Plot these points to make an accurate sketch.

例如，y = x² − 4x + 3 的 y 截距为 3，x 截距为 1 和 3，顶点为 x = 2, y = −1。标出这些点即可画出准确的草图。

7. Vertex and Axis of Symmetry | 顶点与对称轴

Every parabola has a vertical axis of symmetry that passes through the vertex. The equation of the axis is x = −b/(2a). The y-coordinate of the vertex is obtained by substituting this x-value back into the original function.

每条抛物线都有一条穿过顶点的竖直对称轴。对称轴的方程是 x = −b/(2a)。顶点的 y 坐标是将该 x 值代回原函数得到的。

Alternatively, if the quadratic is written as y = a(x − h)² + k, then the vertex is simply (h, k) and the axis is x = h. This form is called the vertex form.

另一种方式，如果二次函数写成 y = a(x − h)² + k，那么顶点就是 (h, k)，对称轴是 x = h。这种形式称为顶点式。

Example: y = 2(x − 3)² + 4 has vertex (3, 4) and axis x = 3.

例：y = 2(x − 3)² + 4 的顶点为 (3, 4)，对称轴为 x = 3。

8. Quadratic Inequalities | 二次不等式

Quadratic inequalities are solved by first finding the roots of the corresponding quadratic equation, then testing intervals on a number line.

解二次不等式时，首先求出对应二次方程的根，然后在数轴上测试区间。

For example, solve x² − x − 6 > 0. Factorise: (x − 3)(x + 2) > 0. The critical values are x = −2 and x = 3. The inequality is true when x < −2 or x > 3.

例如，解 x² − x − 6 > 0。因式分解：(x − 3)(x + 2) > 0。临界值是 x = −2 和 x = 3。当 x < −2 或 x > 3 时不等式成立。

If the inequality is ≥ or ≤, include the critical values. If the parabola opens upward, the graph is above the x-axis outside the roots and below it between the roots.

如果不等式是 ≥ 或 ≤，则包含临界值。若抛物线开口向上，则在两根之外图像在 x 轴上方，在两根之间在 x 轴下方。

9. Applications of Quadratic Equations | 二次方程的应用

Quadratic equations can model projectile motion, area problems, and revenue optimisation. For instance, the height h(t) of an object thrown upward can be h(t) = −5t² + 20t + 2, where t is time.

二次方程可以模拟抛体运动、面积问题和收益优化。例如，一个物体抛出的高度可表示为 h(t) = −5t² + 20t + 2，其中 t 是时间。

To find when the object hits the ground, set h(t) = 0 and solve for t. To find the maximum height, use the vertex formula t = −b/(2a).

要计算物体何时落地，令 h(t) = 0 并解出 t。要计算最大高度，使用顶点公式 t = −b/(2a)。

In geometry, if a rectangle has area 48 m² and its length is 4 m longer than its width, let w be the width. Then w(w + 4) = 48, so w² + 4w − 48 = 0. Factorising gives (w + 8)(w − 6) = 0, so w = 6 m because width cannot be negative.

在几何中，如果一个矩形面积为 48 平方米，且长比宽多 4 米，设宽为 w。则 w(w + 4) = 48，所以 w² + 4w − 48 = 0。因式分解得 (w + 8)(w − 6) = 0，因此 w = 6 米，因为宽不能为负。

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Quadratic Equations and Parabolas | 二次方程与抛物线