Quadratic Equations for IGCSE Mathematics: Solving ax² + bx + c = 0 | IGCSE 数学：求解二次方程 ax² + bx + c = 0

📚 Quadratic Equations for IGCSE Mathematics: Solving ax² + bx + c = 0 | IGCSE 数学：求解二次方程 ax² + bx + c = 0

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, problem solving and even in geometry questions. In this revision guide you will learn how to recognise a quadratic equation, solve it by factorising, by the quadratic formula and by completing the square, and how to use the discriminant and graphs to understand its roots.

二次方程是 IGCSE 数学中最重要的主题之一。它们出现在代数、图像、应用题甚至几何题中。在本复习指南中，你将学会如何识别二次方程，如何通过因式分解、二次公式和配方法求解，以及如何利用判别式和图像理解方程的根。

1. From Linear to Quadratic | 从一次方程到二次方程

A linear equation such as 2x + 3 = 11 has the unknown x only to the first power. A quadratic equation introduces a squared term, usually x². Its standard form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. If a were zero, the equation would become linear rather than quadratic.

一次方程如 2x + 3 = 11 中未知数 x 只出现一次方。二次方程引入了平方项，通常是 x²。它的标准形式为 ax² + bx + c = 0，其中 a、b、c 为常数且 a ≠ 0。如果 a 为零，该方程就变成一次方程而不是二次方程。

In IGCSE questions, you may be asked to solve the equation, find the values of x for which a function equals zero, sketch the related parabola, or interpret the roots in a real-life context. Understanding the structure ax² + bx + c = 0 is therefore essential.

在 IGCSE 题目中，你可能需要解方程、求出函数值为零时的 x 值、画出相关抛物线，或在现实情境中解释根的意义。因此理解 ax² + bx + c = 0 的结构非常重要。

2. Standard Form and Key Vocabulary | 标准形式与关键术语

In the equation ax² + bx + c = 0, the number a is called the leading coefficient, b is the linear coefficient, and c is the constant term. For example, in 3x² − 5x + 2 = 0, we have a = 3, b = −5 and c = 2.

在方程 ax² + bx + c = 0 中，数字 a 称为二次项系数，b 称为一次项系数，c 称为常数项。例如在 3x² − 5x + 2 = 0 中，a = 3，b = −5，c = 2。

The solutions of a quadratic equation are called roots, zeros or x-intercepts. A root is a value of x that makes the equation true. On a graph, the roots are the points where the parabola crosses the x-axis.

二次方程的解称为根、零点或 x 截距。根是使方程成立的 x 值。在图像上，根是抛物线与 x 轴相交的点。

Quadratic expression: an expression of degree 2, such as x² + 4x − 7.

二次式：次数为 2 的表达式，例如 x² + 4x − 7。
Quadratic equation: a statement that a quadratic expression equals zero, such as x² + 4x − 7 = 0.

二次方程：二次式等于零的等式，例如 x² + 4x − 7 = 0。
Parabola: the U-shaped graph of a quadratic function y = ax² + bx + c.

抛物线：二次函数 y = ax² + bx + c 的 U 形图像。

3. Solving by Factorising | 因式分解法求解

Factorising is usually the fastest method when the quadratic expression can be written as a product of two linear brackets. The key idea is to find two numbers that multiply to give the constant term and add to give the linear coefficient.

当二次式可以写成两个一次括号的乘积时，因式分解通常是最快的方法。关键在于找到两个数，使它们相乘等于常数项，相加等于一次项系数。

For a simple example, solve x² − 5x + 6 = 0. We need two numbers whose product is +6 and sum is −5. The numbers are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3). Setting each bracket to zero gives x = 2 or x = 3.

举一个简单的例子：解 x² − 5x + 6 = 0。我们需要两个数，乘积为 +6，和为 −5。这两个数是 −2 和 −3，因此 x² − 5x + 6 = (x − 2)(x − 3)。令每个括号为零，得到 x = 2 或 x = 3。

When a ≠ 1, try to factor by inspection or by splitting the middle term. For example, solve 2x² + 7x + 3 = 0. We look for two numbers that multiply to 2 × 3 = 6 and add to 7. The numbers are 6 and 1, so we split the middle term: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Therefore x = −½ or x = −3.

当 a ≠ 1 时，可以尝试通过观察或裂项来因式分解。例如解 2x² + 7x + 3 = 0。我们寻找两个数，乘积为 2 × 3 = 6，和为 7。这两个数是 6 和 1，于是我们裂项：2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。因此 x = −½ 或 x = −3。

Always check your factors by expanding them. If the original quadratic does not factorise using integers, use the quadratic formula or completing the square instead.

一定要通过展开来检验因式。如果原二次式不能使用整数因式分解，就改用二次公式或配方法。

4. Solving by the Quadratic Formula | 二次公式法求解

The quadratic formula solves any quadratic equation ax² + bx + c = 0. It is derived from completing the square and is given by:

二次公式可以求解任何二次方程 ax² + bx + c = 0。它由配方法推导而来，公式如下：

x = (−b ± √(b² − 4ac)) / (2a)

You must remember this formula exactly. The expression b² − 4ac under the square root is called the discriminant, and it controls how many real roots the equation has.

你必须准确记住这个公式。平方根下的表达式 b² − 4ac 称为判别式，它决定了方程有多少个实数根。

For example, solve x² + 6x + 4 = 0. Here a = 1, b = 6 and c = 4. Substitute into the formula: x = (−6 ± √(6² − 4 × 1 × 4)) / (2 × 1) = (−6 ± √(36 − 16)) / 2 = (−6 ± √20) / 2 = (−6 ± 2√5) / 2 = −3 ± √5.

例如，解 x² + 6x + 4 = 0。这里 a = 1，b = 6，c = 4。代入公式：x = (−6 ± √(6² − 4 × 1 × 4)) / (2 × 1) = (−6 ± √(36 − 16)) / 2 = (−6 ± √20) / 2 = (−6 ± 2√5) / 2 = −3 ± √5。

The formula is especially useful when the quadratic cannot be factorised. Always simplify the square root and the fraction if possible, and be careful with negative signs when substituting.

当二次式无法因式分解时，该公式尤其有用。尽可能化简平方根和分数，代入时注意负号。

5. Solving by Completing the Square | 配方法求解

Completing the square rewrites a quadratic expression x² + bx + c in the form (x + p)² + q. This method is tested in IGCSE because it also helps find the vertex of a parabola.

配方法将二次式 x² + bx + c 改写为 (x + p)² + q 的形式。IGCSE 会考查这一方法，因为它还有助于求出抛物线的顶点。

To complete the square for x² + bx, add and subtract (b/2)². For example, x² + 6x becomes x² + 6x + (6/2)² − (6/2)² = (x + 3)² − 9. Therefore x² + 6x + 4 = (x + 3)² − 9 + 4 = (x + 3)² − 5.

对 x² + bx 配方时，要加上并减去 (b/2)²。例如 x² + 6x 变为 x² + 6x + (6/2)² − (6/2)² = (x + 3)² − 9。因此 x² + 6x + 4 = (x + 3)² − 9 + 4 = (x + 3)² − 5。

To solve x² + 6x + 4 = 0, set (x + 3)² − 5 = 0. Then (x + 3)² = 5, so x + 3 = ±√5 and x = −3 ± √5. This matches the quadratic formula answer.

解 x² + 6x + 4 = 0 时，令 (x + 3)² − 5 = 0。于是 (x + 3)² = 5，所以 x + 3 = ±√5，x = −3 ± √5。这与二次公式的答案一致。

If the leading coefficient is not 1, first take it out as a factor before completing the square inside the bracket.

如果二次项系数不是 1，先把该系数提取出来，再在括号内完成配方。

6. The Discriminant and Nature of Roots | 判别式与根的性质

The discriminant is the part of the quadratic formula under the square root: Δ = b² − 4ac. It tells you the nature of the roots without actually solving the equation.

判别式是二次公式中平方根下的部分：Δ = b² − 4ac。它无需真正解方程就能告诉你根的性质。

Discriminant Δ = b² − 4ac	Nature of roots	Graph behaviour
Δ > 0	Two distinct real roots	Parabola crosses x-axis twice
Δ = 0	One repeated real root	Parabola touches x-axis at vertex
Δ < 0	No real roots	Parabola does not cross x-axis

For example, in x² − 4x + 4 = 0, Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0. The equation has one repeated root, x = 2, and the graph of y = x² − 4x + 4 touches the x-axis at x = 2.

例如，在 x² − 4x + 4 = 0 中，Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0。该方程有一个重根 x = 2，y = x² − 4x + 4 的图像在 x = 2 处与 x 轴相切。

In x² + x + 1 = 0, Δ = 1² − 4 × 1 × 1 = −3. Since Δ < 0, the equation has no real roots, and the parabola lies entirely above the x-axis because a > 0.

在 x² + x + 1 = 0 中，Δ = 1² − 4 × 1 × 1 = −3。因为 Δ < 0，该方程没有实数根，且由于 a > 0，抛物线完全位于 x 轴上方。

7. Graphs of Quadratic Functions | 二次函数的图像

A quadratic function y = ax² + bx + c has a parabola graph. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.

二次函数 y = ax² + bx + c 的图像是抛物线。若 a > 0，抛物线开口向上并有最低点；若 a < 0，开口向下并有最高点。

The x-intercepts of the graph are the solutions of ax² + bx + c = 0. The y-intercept is c, because when x = 0, y = c. The axis of symmetry is the vertical line x = −b / (2a), and the vertex lies on this line.

图像与 x 轴的交点就是 ax² + bx + c = 0 的解。y 截距为 c，因为当 x = 0 时 y = c。对称轴是竖直线 x = −b / (2a)，顶点就在这条线上。

For example, y = x² − 4x + 3 has a = 1, b = −4 and c = 3. Its roots are x = 1 and x = 3, its y-intercept is 3, its axis of symmetry is x = 2, and its vertex is (2, −1).

例如，y = x² − 4x + 3 中 a = 1，b = −4，c = 3。它的根为 x = 1 和 x = 3，y 截距为 3，对称轴为 x = 2，顶点为 (2, −1)。

Sketching a quadratic graph requires you to find the roots, the y-intercept, and the vertex. Mark these points clearly and draw a smooth U-shaped curve through them.

画二次函数草图时，需要求出根、y 截距和顶点。清楚地标出这些点，并通过它们画一条光滑的 U 形曲线。

8. Solving Quadratic Inequalities | 解二次不等式

IGCSE extended papers may ask you to solve inequalities such as x² − 5x + 6 > 0. First solve the related equation x² − 5x + 6 = 0 to get x = 2 and x = 3. These values divide the number line into three intervals: x < 2, 2 < x < 3, and x > 3.

IGCSE 扩展卷可能会要求解诸如 x² − 5x + 6 > 0 的不等式。首先解相关方程 x² − 5x + 6 = 0，得到 x = 2 和 x = 3。这些值将数轴分成三个区间：x < 2、2 < x < 3 和 x > 3。

Test a value from each interval in the original inequality. For x = 0, 0² − 5 × 0 + 6 = 6 > 0, so x < 2 works. For x = 2.5, 2.5² − 5 × 2.5 + 6 = −0.25, so the interval 2 < x < 3 does not work. For x = 4, 4² − 5 × 4 + 6 = 2 > 0, so x > 3 works. The solution is x < 2 or x > 3.

从每个区间中取一个值代入原不等式检验。x = 0

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Quadratic Equations for IGCSE Mathematics: Solving ax² + bx + c = 0 | IGCSE 数学：求解二次方程 ax² + bx + c = 0