Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear frequently in the IGCSE Mathematics syllabus. Understanding how to solve them is essential for success in exams and for building a strong foundation in algebra.

二次方程在 IGCSE 数学课程中频繁出现。理解如何求解它们，对于考试取得好成绩以及建立扎实的代数基础都至关重要。

1. Introduction to Quadratic Equations | 二次方程简介

A quadratic equation is a polynomial equation of degree 2. Its general form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable x is 2.

二次方程是次数为 2 的多项式方程。它的一般形式是 ax² + bx + c = 0，其中 a、b 和 c 是常数，且 a ≠ 0。变量 x 的最高次幂是 2。

The solutions to a quadratic equation are called its roots. A quadratic equation can have two distinct real roots, one repeated real root, or two complex roots. In the IGCSE course, we usually focus on real roots.

二次方程的解称为它的根。一个二次方程可以有两个不同的实数根、一个重实数根或两个复数根。在 IGCSE 课程中，我们通常关注实数根。

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

Before solving, it is important to rewrite the equation in the standard form ax² + bx + c = 0. The coefficient a is the quadratic coefficient, b is the linear coefficient, and c is the constant term.

在求解之前，重要的是将方程改写为标准形式 ax² + bx + c = 0。系数 a 是二次项系数，b 是一次项系数，c 是常数项。

For example, the equation 2x² – 3x = 5 should be rearranged to 2x² – 3x – 5 = 0. This makes it ready for factorisation or the quadratic formula.

例如，方程 2x² – 3x = 5 应重新排列为 2x² – 3x – 5 = 0。这样它就可以进行因式分解或使用二次公式求解。

Some equations may need expansion first, such as (x + 2)² = 9. Expand and simplify before using any solving method.

有些方程可能需要先展开，例如 (x + 2)² = 9。在使用任何求解方法之前，先展开并化简。

3. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method when the quadratic expression can be written as a product of two linear factors. The idea is to find two numbers whose product equals ac and whose sum equals b.

当二次表达式可以写成两个一次因式的乘积时，因式分解通常是最快捷的方法。思路是找到两个数，使得它们的乘积等于 ac，且它们的和等于 b。

Consider the equation x² + 5x + 6 = 0. We look for two numbers that multiply to 6 and add to 5. The numbers 2 and 3 satisfy this, so the equation factors as (x + 2)(x + 3) = 0.

考虑方程 x² + 5x + 6 = 0。我们寻找两个数，它们相乘得 6，相加得 5。数字 2 和 3 满足这个条件，因此方程因式分解为 (x + 2)(x + 3) = 0。

Using the zero product property, if the product of two factors is zero, then at least one factor must be zero. Hence x + 2 = 0 or x + 3 = 0, giving x = -2 or x = -3.

利用零乘积性质，如果两个因式的乘积为零，那么至少有一个因式必须为零。因此 x + 2 = 0 或 x + 3 = 0，得到 x = -2 或 x = -3。

If (px + q)(rx + s) = 0, then px + q = 0 or rx + s = 0.

如果 (px + q)(rx + s) = 0，则 px + q = 0 或 rx + s = 0。

When the coefficient a is not 1, you may need to factor by grouping. For example, 2x² + 7x + 3 = 0. Multiply a and c: 2 × 3 = 6. Find two numbers whose product is 6 and sum is 7: 1 and 6. Rewrite the middle term: 2x² + x + 6x + 3 = 0. Then group: x(2x + 1) + 3(2x + 1) = 0, so (x + 3)(2x + 1) = 0.

当系数 a 不为 1 时，可能需要使用分组分解。例如，2x² + 7x + 3 = 0。将 a 和 c 相乘：2 × 3 = 6。找到两个数，它们的乘积是 6，和是 7：1 和 6。重新写出中间项：2x² + x + 6x + 3 = 0。然后分组：x(2x + 1) + 3(2x + 1) = 0，所以 (x + 3)(2x + 1) = 0。

4. Solving by Completing the Square | 配方法

Completing the square is a method that rewrites a quadratic expression as a perfect square plus a constant. This technique is useful when factorisation is difficult and also helps in finding the vertex of a parabola.

配方法是将二次表达式改写为一个完全平方加上一个常数的方法。当因式分解困难时，这种技术很有用，也有助于找到抛物线的顶点。

For a quadratic x² + bx, add and subtract (b/2)². The expression becomes (x + b/2)² – (b/2)². For example, x² + 6x + 2 can be written as (x + 3)² – 9 + 2 = (x + 3)² – 7.

对于二次式 x² + bx，加上并减去 (b/2)²。表达式变成 (x + b/2)² – (b/2)²。例如，x² + 6x + 2 可以写成 (x + 3)² – 9 + 2 = (x + 3)² – 7。

To solve x² + 6x + 2 = 0 by completing the square first write (x + 3)² – 7 = 0. Then add 7 to both sides: (x + 3)² = 7. Take the square root of both sides, remembering the ± sign:

要通过配方法求解 x² + 6x + 2 = 0，首先写出 (x + 3)² – 7 = 0。然后两边加 7：(x + 3)² = 7。两边开平方，记住 ± 符号：

x + 3 = ±√7, so x = -3 ± √7.

x + 3 = ±√7，所以 x = -3 ± √7。

Always write the answer as two separate roots: x = -3 + √7 or x = -3 – √7.

始终将答案写成两个单独的根：x = -3 + √7 或 x = -3 – √7。

5. The Quadratic Formula | 二次公式

The quadratic formula is a universal method for solving any quadratic equation. It is derived from completing the square and is especially useful when factorisation is not obvious.

二次公式是求解任何二次方程的通用方法。它由配方法推导而来，尤其适用于因式分解不明显的情况。

For ax² + bx + c = 0, x = (-b ± √(b² – 4ac)) / (2a).

对于 ax² + bx + c = 0，x = (-b ± √(b² – 4ac)) / (2a)。

Example: Solve 2x² – 4x – 3 = 0. Here a = 2, b = -4, c = -3. Substitute into the formula:

示例：求解 2x² – 4x – 3 = 0。这里 a = 2，b = -4，c = -3。代入公式：

x = (4 ± √((-4)² – 4 × 2 × (-3))) / (2 × 2) = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4.

x = (4 ± √((-4)² – 4 × 2 × (-3))) / (2 × 2) = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4。

Simplify √40 = 2√10, so x = (4 ± 2√10) / 4 = 1 ± (√10)/2. Always simplify radicals and cancel fractions where possible.

化简 √40 = 2√10，所以 x = (4 ± 2√10) / 4 = 1 ± (√10)/2。始终化简根式并尽可能约分。

6. The Discriminant | 判别式

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

判别式是二次公式中根号内的表达式 b² – 4ac。它让我们无需完全解方程即可了解根的性质。

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

如果 b² – 4ac > 0，则有两个不同的实数根。
如果 b² – 4ac = 0，则有一个重实数根。
如果 b² – 4ac < 0，则没有实数根。

For example, consider x² – 6x + 9 = 0. The discriminant is (-6)² – 4 × 1 × 9 = 36 – 36 = 0, so there is exactly one root, x = 3.

例如，考虑 x² – 6x + 9 = 0。判别式为 (-6)² – 4 × 1 × 9 = 36 – 36 = 0，因此恰好有一个根，x = 3。

The discriminant also tells us whether the roots are rational. If the discriminant is a perfect square and a, b, c are rational, the equation can be solved by factorisation.

判别式还告诉我们根是否为有理数。如果判别式是一个完全平方数，且 a、b、c 是有理数，则该方程可以通过因式分解求解。

7. Solving Equations with Fractions | 含分数方程

Some quadratic equations involve algebraic fractions. To solve them, multiply every term by the common denominator to clear the fractions.

有些二次方程包含代数分数。求解时，将每一项乘以公分母以消去分数。

Example: Solve x/(x-1) + 2 = 3x/(x+1). First note that x ≠ 1 and x ≠ -1. Multiply both sides by (x – 1)(x + 1): x(x + 1) + 2(x – 1)(x + 1) = 3x(x – 1).

示例：求解 x/(x-1) + 2 = 3x/(x+1)。首先注意 x ≠ 1 且 x ≠ -1。两边乘以 (x – 1)(x + 1)：x(x + 1) + 2(x – 1)(x + 1) = 3x(x – 1)。

Expand and simplify: x² + x + 2(x² – 1) = 3x² – 3x. This gives x² + x + 2x² – 2 = 3x² – 3x, so 3x² + x – 2 = 3x² – 3x. Cancel 3x² from both sides: x – 2 = -3x, giving 4x = 2, so x = 1/2.

展开并化简：x² + x + 2(x² – 1) = 3x² – 3x。得到 x² + x + 2x² – 2 = 3x² – 3x，所以 3x² + x – 2 = 3x² – 3x。两边消去 3x²：x – 2 = -3x，得到 4x = 2，所以 x = 1/2。

Always check your solutions against the original equation to ensure no denominator becomes zero.

始终检查你的解是否满足原方程，确保分母不为零。

8. Quadratic Graphs and Roots | 二次函数图像与根

The roots of a quadratic equation correspond to the x-intercepts of the graph of the quadratic function y = ax² + bx + c. The graph is a parabola that opens upward when a > 0 and downward when a < 0.

二次方程的根对应于二次函数 y = ax² + bx + c 的图像与 x 轴的交点。图像是一条抛物线，当 a > 0 时开口向上，当 a < 0 时开口向下。

If the equation has two distinct real roots, the parabola crosses the x-axis at two points. If there is one repeated root, the parabola touches the x-axis at one point, called the vertex. If there are no real roots, the parabola does not cross the x-axis.

如果方程有两个不同的实数根，抛物线与 x 轴相交于两点。如果有一个重根，抛物线在一点与 x 轴相切，该点称为顶点。如果没有实数根，抛物线不与 x 轴相交。

The x-coordinate of the vertex is given by x = -b/(2a). You can find the y-coordinate by substituting this x-value into the equation.

顶点的 x 坐标由 x = -b/(2a) 给出。你可以将此 x 值代入方程求得 y 坐标。

9. Problem Solving with Quadratics | 二次方程应用题

Many word problems can be modelled by quadratic equations. The key is to translate the information into an equation, solve it, and then interpret the solutions in the context of the problem.

许多文字题可以用二次方程建立模型。关键是将信息转化为方程，求解，然后根据问题的背景解释解。

For example: The length of a rectangle is 3 cm more than twice its width. Its area is 20 cm². Find the width. Let the width be w cm. Then the length is (2w + 3) cm. The area equation is w(2w + 3) = 20.

例如：一个长方形的长比宽的 2 倍多 3 厘米。它的面积是 20 平方厘米。求宽。设宽为 w 厘米。则长为 (2w + 3) 厘米。面积方程为 w(2w + 3) = 20。

Expanding gives 2w² + 3w = 20, or 2w² + 3w – 20 = 0. Factorise: (2w – 5)(w + 4) = 0. Thus w = 5/2 or w = -4. Since width cannot be negative, w = 2.5 cm.

展开得 2w² + 3w = 20，即 2w² + 3w – 20 = 0。因式分解：(2w – 5)(w + 4) = 0。因此 w = 5/2 或 w = -4。因为宽不能为负，所以 w = 2.5 厘米。

Always reject solutions that do not make sense in the given context, such as negative lengths or times.

始终舍弃在给定情境中没有意义的解，例如负的长度或时间。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

One common mistake is forgetting to rearrange the equation into standard form before applying the formula or factorisation. Always set the equation equal to zero first.

一个常见错误是在应用公式或因式分解之前忘记将方程整理为标准形式。始终先将方程等于零。

Another mistake is losing the ± sign when taking square roots. Remember that x² = k has two solutions: x = √k and x = -√k.

另一个错误是在开平方时丢失 ± 符号。记住 x² = k 有两个解：x = √k 和 x = -√k。

When using the quadratic formula, double-check your signs, especially when b is negative. Substituting carefully can prevent many arithmetic errors.

使用二次公式时，请仔细检查符号，尤其是当 b 为负数时。仔细代入可以避免许多算术错误。

In the exam, if a quadratic cannot be factorised easily, the quadratic formula is always safe. However, if the question asks you to factorise, you must show the factorisation method.

在考试中，如果二次式不容易因式分解，二次公式总是稳妥的。但如果题目要求你因式分解，你必须展示因式分解的方法。

Finally, always check your answers by substituting them back into the original equation. This helps you catch any mistakes and gives you confidence in your solutions.

最后，始终将答案代回原方程进行验证。这有助于你发现任何错误，并对你的解充满信心。

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Solving Quadratic Equations | 解二次方程