Mastering Quadratic Equations for IGCSE Mathematics | 掌握 IGCSE 数学二次方程

📚 Mastering Quadratic Equations for IGCSE Mathematics | 掌握 IGCSE 数学二次方程

Quadratic equations form one of the most important topics in IGCSE Mathematics. They appear across algebra, graphs, and real-world problem solving, so understanding how to solve and interpret them is essential for success in both Core and Extended papers. This revision article covers the standard form, four main solution methods, the discriminant, graph features, and common exam applications.

二次方程是 IGCSE 数学中最重要的主题之一。它广泛出现在代数、图像以及实际问题中，因此掌握二次方程的求解与图像解释对 Core 和 Extended 考试都至关重要。本篇复习文章将涵盖标准形式、四种主要解法、判别式、图像特征以及常见考试应用。

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1. What Is a Quadratic Equation? | 什么是二次方程？

A quadratic equation is a polynomial equation in which the highest power of the variable is 2. In IGCSE Mathematics, the variable is usually x, and the equation can contain an x² term, an x term, and a constant term. For example, x² – 5x + 6 = 0 and 3x² + 2x – 1 = 0 are both quadratic equations.

二次方程是未知数的最高次数为 2 的多项式方程。在 IGCSE 数学中，未知数通常为 x，方程中可以包含 x² 项、x 项和常数项。例如 x² – 5x + 6 = 0 和 3x² + 2x – 1 = 0 都是二次方程。

A linear equation has degree 1, such as 2x + 3 = 7, while a quadratic equation has degree 2. The word ‘quadratic’ comes from the Latin word ‘quadratus’, meaning square, because the variable is squared. Quadratic equations can have two solutions, one solution, or no real solution, depending on the values in the equation.

一次方程的次数为 1，例如 2x + 3 = 7；而二次方程的次数为 2。’quadratic’ 一词来自拉丁语 ‘quadratus’，意为“平方”，因为未知数被平方。二次方程可能有两个解、一个解或没有实数解，具体取决于方程中的系数值。

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

The standard form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. The condition a ≠ 0 is essential because if a = 0, the x² term disappears and the equation becomes linear, not quadratic.

二次方程的标准形式为 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。条件 a ≠ 0 非常关键，因为如果 a = 0，x² 项消失，方程就变成一次方程，而不再是二次方程。

The coefficient a is called the leading coefficient, the coefficient b is the coefficient of x, and c is the constant term. For example, in 4x² – 7x + 2 = 0, we have a = 4, b = -7, and c = 2. Identifying these values correctly is vital when using the quadratic formula or the discriminant.

系数 a 称为首项系数，b 是 x 的系数，c 是常数项。例如，在 4x² – 7x + 2 = 0 中，a = 4、b = -7、c = 2。正确识别这些值对于使用公式法或判别式至关重要。

Always rearrange an equation into standard form before solving. For example, 3x² = 5x – 2 should be rewritten as 3x² – 5x + 2 = 0, so that the right-hand side equals zero.

求解前必须先将方程整理为标准形式。例如，3x² = 5x – 2 应改写为 3x² – 5x + 2 = 0，使等号右边为 0。

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3. Solving by Factorising | 因式分解法

Factorising is usually the fastest method when the quadratic expression can be written as a product of two linear factors. The key idea is the zero product property: if (x – p)(x – q) = 0, then either x – p = 0 or x – q = 0, giving the solutions x = p or x = q.

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。其关键思想是零乘积性质：如果 (x – p)(x – q) = 0，那么 x – p = 0 或 x – q = 0，因此解为 x = p 或 x = q。

For example, to solve x² – 5x + 6 = 0, first factorise the left side as (x – 2)(x – 3) = 0. Then set each factor equal to zero: x – 2 = 0 gives x = 2, and x – 3 = 0 gives x = 3. The two solutions are x = 2 and x = 3.

例如，解 x² – 5x + 6 = 0 时，先把左边因式分解为 (x – 2)(x – 3) = 0，然后令每个因式等于 0：x – 2 = 0 得 x = 2，x – 3 = 0 得 x = 3。两个解为 x = 2 和 x = 3。

Special patterns such as the difference of two squares can also be solved by factorising. For example, x² – 9 = 0 becomes (x + 3)(x – 3) = 0, so x = -3 or x = 3. When the coefficient of x² is not 1, you may need to factorise by grouping or by trial and error.

特殊形式如平方差也可以通过因式分解求解。例如，x² – 9 = 0 可化为 (x + 3)(x – 3) = 0，因此 x = -3 或 x = 3。当 x² 的系数不为 1 时，可能需要分组分解或试错法进行因式分解。

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4. Solving by Completing the Square | 配方法

Completing the square is a powerful method that works for any quadratic equation. The goal is to rewrite the quadratic expression in the form (x + p)² + q, from which the solutions can be found by taking square roots. The method is especially useful when the quadratic does not factorise easily.

配方法是一种适用于任何二次方程的有效方法。其目标是将二次式改写为 (x + p)² + q 的形式，然后通过开平方求出解。当二次式不易因式分解时，配方法尤其有用。

To complete the square for x² + bx, add and subtract (b/2)². For example, to solve x² + 6x + 5 = 0, first move the constant to the right: x² + 6x = -5. Then add (6/2)² = 9 to both sides to get x² + 6x + 9 = 4. This becomes (x + 3)² = 4, so x + 3 = ±2, giving x = -1 or x = -5.

为 x² + bx 配方时，需要加上并减去 (b/2)²。例如，解 x² + 6x + 5 = 0 时，先把常数项移到右边：x² + 6x = -5，然后两边同时加上 (6/2)² = 9，得到 x² + 6x + 9 = 4。左边可写成 (x + 3)² = 4，因此 x + 3 = ±2，解得 x = -1 或 x = -5。

If the leading coefficient is not 1, divide the whole equation by a before completing the square. For example, in 2x² + 8x + 6 = 0, first divide by 2 to get x² + 4x + 3 = 0, then proceed as above.

如果首项系数不是 1，应先在整个方程中除以 a，再进行配方。例如，对于 2x² + 8x + 6 = 0，先除以 2 得到 x² + 4x + 3 = 0，然后按上述步骤继续。

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5. The Quadratic Formula | 公式法

The quadratic formula gives the solutions of ax² + bx + c = 0 directly. It states that x = (-b ± √(b² – 4ac)) / (2a). This formula is extremely reliable because it works for every quadratic equation, even when factorising is difficult or impossible.

公式法可以直接给出 ax² + bx + c = 0 的解。公式为 x = (-b ± √(b² – 4ac)) / (2a)。该公式非常可靠，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。

For example, to solve 2x² + 3x – 2 = 0, identify a = 2, b = 3, and c = -2. Substituting into the formula gives x = (-3 ± √(3² – 4 × 2 × (-2))) / (2 × 2) = (-3 ± √(9 + 16)) / 4 = (-3 ± √25) / 4 = (-3 ± 5) / 4. The two solutions are x = (-3 + 5)/4 = 1/2 and x = (-3 – 5)/4 = -2.

例如，解 2x² + 3x – 2 = 0，先确定 a = 2、b = 3、c = -2。代入公式得 x = (-3 ± √(3² – 4 × 2 × (-2))) / (2 × 2) = (-3 ± √(9 + 16)) / 4 = (-3 ± √25) / 4 = (-3 ± 5) / 4。两个解为 x = (-3 + 5)/4 = 1/2 和 x = (-3 – 5)/4 = -2。

Remember to use brackets carefully when substituting negative values. The term -b changes sign if b is negative, and the discriminant b² – 4ac must be calculated before taking the square root. Simplifying surds may be required if the discriminant is not a perfect square.

代入负值时务必小心使用括号。如果 b 为负数，-b 会变号；判别式 b² – 4ac 必须在开平方之前计算。如果判别式不是完全平方数，可能需要化简根式。

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

The expression under the square root in the quadratic formula, D = b² – 4ac, is called the discriminant. It tells us the nature of the roots without actually solving the equation. This is a common IGCSE exam topic, especially in questions that ask for the number of real solutions.

公式法中平方根号下的式子 D = b² – 4ac 称为判别式。它可以在不实际求解方程的情况下判断根的性质。这是 IGCSE 考试中的常见考点，尤其在询问实数解数量的题目中经常出现。

Discriminant | 判别式	Nature of roots | 根的性质
D > 0	Two distinct real roots | 两个不同的实数根
D = 0	One repeated real root | 一个重根（两个相等的实数根）
D < 0	No real roots | 无实数根

For example, in x² – 4x + 4 = 0, the discriminant is b² – 4ac = (-4)² – 4 × 1 × 4 = 16 – 16 = 0, so there is exactly one repeated real root. In x² + x + 1 = 0, the discriminant is 1² – 4 × 1 × 1 = -3, which is negative, so there are no real roots.

例如，在 x² – 4x + 4 = 0 中，判别式为 b² – 4ac = (-4)² – 4 × 1 × 4 = 16 – 16 = 0，因此恰好有一个重根。在 x² + x + 1 = 0 中，判别式为 1² – 4 × 1 × 1 = -3，为负数，所以没有实数根。

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7. Graphing Quadratic Functions | 二次函数图像

A quadratic function can be written as y = ax² + bx + c, and its graph is a parabola. If a > 0, the parabola opens upward and has a minimum point. If a < 0, it opens downward and has a maximum point. The shape is always symmetric, and the line of symmetry passes through the vertex.

二次函数可以写成 y = ax² + bx + c，其图像是一条抛物线。如果 a > 0，抛物线开口向上，具有最低点；如果 a < 0，开口向下，具有最高点。图像始终是对称的，对称轴经过顶点。

The x-coordinate of the vertex is given by x = -b / (2a), and the y-coordinate is found by substituting this value back into the function. The y-intercept is simply the constant term c, because when x = 0, y = c. The x-intercepts, if any, are the real roots of the equation ax² + bx + c = 0.

顶点的 x 坐标由 x = -b / (2a) 给出，y 坐标可将该值代回函数求得。y 轴截距就是常数项 c，因为当 x = 0 时，y = c。如果存在 x 轴截距，它们就是方程 ax² + bx + c = 0 的实数根。

For example, for y = x² – 4x + 3, the vertex has x-coordinate x = -(-4) / (2 × 1) = 2. Substituting into the function gives y = (2)² – 4 × 2 + 3 = -1, so the vertex is (2, -1). The y-intercept is 3, and since x² – 4x + 3 = (x – 1)(x – 3), the x-intercepts are x = 1 and x = 3.

例如，对于 y = x² – 4x + 3，顶点的 x 坐标为 x = -(-4) / (2 × 1) = 2。代入函数得 y = (2)² – 4 × 2 + 3 = -1，所以顶点为 (2, -1)。y 轴截距为 3；因为 x² – 4x + 3 = (x – 1)(x – 3)，所以 x 轴截距为 x = 1 和 x = 3。

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8. Applications and Word Problems | 应用与文字题

IGCSE exam questions often embed quadratic equations in real-life contexts, such as area problems, projectile motion, number

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