Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. A solid understanding of how to solve them is essential for success in algebra, coordinate geometry, and problem-solving questions across both Paper 1 and Paper 2.

二次方程是 IGCSE 数学中考查频率最高的内容之一。扎实掌握其解法，对于在 Paper 1 和 Paper 2 中解答代数、坐标几何以及应用题都至关重要。

---

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

A quadratic equation is an equation in which the highest power of the variable is 2. Its general form is shown below.

二次方程是变量的最高次数为 2 的方程，它的一般形式如下所示。

ax² + bx + c = 0

are constants, and a ≠ 0. If a = 0, the equation reduces to a linear equation bx + c = 0.

其中，a、b、c 为常数，且 a ≠ 0。若 a = 0，方程就退化为线性方程 bx + c = 0。

The following examples help clarify what is and is not a quadratic equation.

以下例子有助于辨别哪些是二次方程，哪些不是。

• x² − 5x + 6 = 0 is a quadratic equation.
• x² − 5x + 6 = 0 是二次方程。
• 2x² + 3x − 7 = 0 is also quadratic because the highest power of x is 2.
• 2x² + 3x − 7 = 0 也是二次方程，因为 x 的最高次数为 2。
• x³ + 2x² − 1 = 0 is NOT quadratic, since the highest power is 3.
• x³ + 2x² − 1 = 0 不是二次方程，因为最高次数为 3。

In the IGCSE syllabus, you must be able to solve quadratic equations using three main methods: factorisation, the quadratic formula, and completing the square. Each method is covered in detail below.

在 IGCSE 考纲中，你需要掌握三种解二次方程的主要方法：因式分解法、二次公式法和配方法。下面将逐一详细讲解。

---

2. Solving by Factorisation | 因式分解法

When a quadratic expression can be written as the product of two linear factors, we can solve the equation using the zero product property.

当一个二次表达式可以写成两个一次因式的乘积时，我们就可以利用零乘积性质来解方程。

If AB = 0, then A = 0 or B = 0.

This property states that if the product of two factors is zero, at least one of the factors must be zero.

该性质说明：若两个因式之积为零，则至少有一个因式为零。

Example 1: Solve x² − 5x + 6 = 0.

例 1：解方程 x² − 5x + 6 = 0。

Step 1: Factorise the quadratic. We look for two numbers that multiply to give 6 and add to give −5. These numbers are −2 and −3.

步骤 1：对二次式进行因式分解。我们需要找到两个数，它们相乘得 6，相加得 −5。这两个数是 −2 和 −3。

x² − 5x + 6 = (x − 2)(x − 3)

Step 2: Apply the zero product property.

步骤 2：应用零乘积性质。

(x − 2)(x − 3) = 0 ⇒ x − 2 = 0 or x − 3 = 0

Step 3: Solve each linear equation.

步骤 3：分别解每个一次方程。

x = 2 or x = 3

and x = 3. Always check your answers by substituting them back into the original equation.

因此，方程的解为 x = 2 和 x = 3。务必把答案代回原方程进行检验。

Example 2: Solve 2x² − 7x + 3 = 0.

例 2：解方程 2x² − 7x + 3 = 0。

a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to −7: they are −1 and −6.

这里使用拆项法。先计算 a × c = 2 × 3 = 6，然后找两个数相乘得 6、相加得 −7：它们是 −1 和 −6。

2x² − x − 6x + 3 = 0

x(2x − 1) − 3(2x − 1) = 0

(2x − 1)(x − 3) = 0 ⇒ x = ½ or x = 3

Factorisation is quick and elegant, but it only works when the quadratic has simple rational roots.

因式分解法快捷而简洁，但仅当二次方程具有简单的有理数根时才适用。

---

3. Solving by the Quadratic Formula | 二次公式法

When factorisation is difficult or impossible, we use the quadratic formula. This formula works for all quadratic equations, including those with irrational or complex roots.

当因式分解困难或无法进行时，我们使用二次公式。该公式适用于所有二次方程，包括含有无理根或复数根的情形。

x = (−b ± √b² − 4ac) / (2a)

In the IGCSE examination, this formula is provided on the formula sheet, but you must know how to substitute values correctly.

在 IGCSE 考试中，公式表会提供此公式，但你必须掌握正确的代入方法。

Example: Solve 2x² + 3x − 5 = 0 using the quadratic formula.

示例：用二次公式解方程 2x² + 3x − 5 = 0。

Step 1: Identify a, b and c. Here a = 2, b = 3, c = −5.

步骤 1：确定 a、b、c 的值。这里 a = 2，b = 3，c = −5。

Step 2: Substitute these values into the formula.

步骤 2：将这些值代入公式。

x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2)

Step 3: Simplify the discriminant and the denominator.

步骤 3：化简判别式和分母。

x = (−3 ± √9 + 40) / 4 = (−3 ± √49) / 4

Step 4: Since √49 = 7, we write the two solutions separately.

步骤 4：由于 √49 = 7，我们分别写出两个解。

x = (−3 + 7) / 4 = 1 or x = (−3 − 7) / 4 = −5/2

5. Notice that these same roots could also be found by factorisation: (2x + 5)(x − 1) = 0.

因此解为 x = 1 和 x = −2.5。注意这两个根也可以通过因式分解得到：(2x + 5)(x − 1) = 0。

When using the quadratic formula, be extremely careful with negative signs. A common error is to forget that c is negative, which changes the value of the discriminant.

使用二次公式时，要特别注意负号。一个常见错误是忘记 c 是负数，这会改变判别式的值。

---

广告 / Sponsored

4. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic expression in the form a(x + p)² + q. This method is particularly useful for finding the turning point of a parabola and for solving equations when factorisation is not obvious.

配方法将二次表达式改写为 a(x + p)² + q 的形式。这种方法特别适用于求抛物线的顶点，以及在因式分解不明显时解方程。

The basic technique for x² + bx: take half of b, square it, then add and subtract this value.

x² + bx 的基本配方技巧：取 b 的一半，平方，然后加上再减去这个值。

Example: Solve x² + 6x + 5 = 0 by completing the square.

示例：用配方法解方程 x² + 6x + 5 = 0。

Step 1: Focus on x² + 6x. Half of 6 is 3, so (x + 3)² = x² + 6x + 9.

步骤 1：关注 x² + 6x。6 的一半是 3，所以 (x + 3)² = x² + 6x + 9。

Step 2: Rewrite the original expression.

步骤 2：改写原表达式。

x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4

Step 3: Set the expression equal to 0 and solve.

步骤 3：令表达式等于 0 并求解。

(x + 3)² − 4 = 0 ⇒ (x + 3)² = 4

Step 4: Take the square root of both sides, remembering the ± sign.

步骤 4：两边开平方根，注意 ± 号。

x + 3 = ±2 ⇒ x = −3 + 2 = −1 or x = −3 − 2 = −5

When a ≠ 1, factor out a first before completing the square.

当 a ≠ 1 时，先提取 a 再进行配方。

Example: Solve 2x² + 8x − 10 = 0 by completing the square.

示例：用配方法解 2x² + 8x − 10 = 0。

2(x² + 4x − 5) = 0 ⇒ 2[(x + 2)² − 4 − 5] = 0

2[(x + 2)² − 9] = 0 ⇒ (x + 2)² = 9 ⇒ x + 2 = ±3 ⇒ x = 1 or x = −5

o reveals the turning point of the graph: for y = a(x + p)² + q, the vertex is at (−p, q).

配方同时也能揭示图像的顶点：对于 y = a(x + p)² + q，顶点坐标为 (−p, q)。

---

5. The Discriminant | 判别式

The discriminant is the expression b² − 4ac inside the square root of the quadratic formula. It tells us about the nature of the roots without actually solving the equation.

判别式是二次公式根号内的表达式 b² − 4ac。它无需真正解方程就能判断根的性质。

The following table summarises the three cases.

下表总结了三种情况。

Value of b² − 4ac	Nature of Roots	Graph Interpretation
b² − 4ac > 0	Two distinct real roots	Parabola crosses the x-axis at two points
b² − 4ac = 0	One repeated real root	Parabola touches the x-axis at one point
b² − 4ac < 0	No real roots	Parabola does not intersect the x-axis

判别式 b² − 4ac 的值决定了根的性质：大于 0 时有两个不等实根，等于 0 时有一个重根，小于 0 时没有实根。图像上分别对应抛物线与 x 轴相交于两点、相切于一点、或完全不相交。

Example: Determine the nature of the roots of x² − 4x + 4 = 0.

示例：判断 x² − 4x + 4 = 0 的根的性质。

b² − 4ac = (−4)² − 4 × 1 × 4 = 16 − 16 = 0

n has one repeated real root. Indeed, x² − 4x + 4 = (x − 2)², so x = 2 is a repeated root.

因为判别式为 0，方程有一个重根。事实上 x² − 4x + 4 = (x − 2)²，所以 x = 2 是重根。

Discriminant questions often appear in the form: 'Find the range of k such that the equation has two distinct real roots.' x + 9 = 0 to have two distinct roots, we require k² − 36 > 0, so k > 6 or k < −6.

判别式题目常以"求 k 的范围使得方程有两个不等实根"的形式出现。例如，x² + kx + 9 = 0 有两个不等实根时，要求 k² − 36 > 0，即 k > 6 或 k < −6。

---

6. Graphs of Quadratic Functions | 二次函数图像

The graph of a quadratic function y = ax² + bx + c is a smooth curve called a parabola.

二次函数 y = ax² + bx + c 的图像是一条平滑曲线，称为抛物线。

If a > 0, the parabola opens upward and has a minimum point. If a < 0, it opens downward and has a maximum point.

若 a > 0，抛物线开口向上，有最小值点；若 a < 0，抛物线开口向下，有最大值点。

orrespond to the real roots of the equation ax² + bx + c = 0. The y-intercept is simply c.

图像与 x 轴的交点对应方程 ax² + bx + c = 0 的实根。与 y 轴的交点则是 c。

The vertex (turning point) can be found using the following formula.

顶点（转向点）可通过以下公式求得。

x_vertex = −b / (2a)

Substitute this x-value into the original equation to find the y-coordinate of the vertex.

将此 x 值代入原方程即可求出顶点的 y 坐标。

te the function in completed square form y = a(x + p)² + q; the vertex is then at (−p, q).

或者，将函数改写为配方形式 y = a(x + p)² +

Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）