Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in algebra, graphs, geometry, and real-world problem solving. This article explains the standard methods for solving quadratic equations, with careful attention to the techniques and common pitfalls that students encounter in exams.

二次方程是 IGCSE 数学中最重要的主题之一。它出现在代数、图像、几何以及实际应用题中。本文将讲解解二次方程的标准方法，并特别关注学生在考试中遇到的技巧和常见错误。

1. Definition and Standard Form | 定义与标准形式

A quadratic equation in one variable is an equation that can be written in the standard form:

ax² + bx + c = 0

where a, b, and c are constants, and a ≠ 0. The highest power of the variable x is 2, which makes the equation quadratic. If a = 0, the equation becomes linear, not quadratic.

一元二次方程是可以写成标准形式的方程：

ax² + bx + c = 0

其中 a、b、c 是常数，且 a ≠ 0。变量 x 的最高次数是 2，因此方程是二次的。如果 a = 0，方程就变成一次方程，而不是二次方程。

For example, 2x² – 5x + 3 = 0 is a quadratic equation with a = 2, b = -5, and c = 3. The values of x that satisfy the equation are called roots or solutions.

例如，2x² – 5x + 3 = 0 是一个二次方程，其中 a = 2，b = -5，c = 3。满足方程的 x 值称为根或解。

2. Solving by Factorisation | 因式分解法

If a quadratic expression can be factorised into two linear factors, then the equation can be solved by setting each factor equal to zero. This method is often the fastest and is preferred when the coefficients are simple.

如果二次表达式可以分解为两个一次因式，那么可以通过令每个因式等于零来解方程。这种方法通常最快，在系数较简单时优先使用。

For example, solve x² – 5x + 6 = 0. We look for two numbers that multiply to 6 and add to -5. These numbers are -2 and -3. So:

例如，解 x² – 5x + 6 = 0。我们寻找两个数，它们相乘得 6，相加得 -5。这两个数是 -2 和 -3。因此：

(x – 2)(x – 3) = 0

x – 2 = 0 或 x – 3 = 0

x = 2 或 x = 3

The solutions are x = 2 and x = 3. Always expand your factors to check that the original equation is recovered.

解为 x = 2 或 x = 3。务必展开因式检查是否能还原原方程。

For equations where the coefficient of x² is not 1, such as 2x² + 5x – 3 = 0, you can use the product-sum method or factor by grouping. The factors are (2x – 1)(x + 3) = 0, giving x = ½ or x = -3.

对于 x² 系数不为 1 的方程，例如 2x² + 5x – 3 = 0，可以使用乘积-求和法或分组分解。因式为 (2x – 1)(x + 3) = 0，于是 x = ½ 或 x = -3。

3. 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. It is useful for solving equations that cannot be factorised easily and for deriving the quadratic formula.

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。它适用于不容易因式分解的方程，也可用于推导二次公式。

For the equation x² + 6x + 2 = 0, first move the constant term to the other side:

对于方程 x² + 6x + 2 = 0，首先将常数项移到另一边：

x² + 6x = -2

Take half of the coefficient of x, which is 3, and square it to get 9. Add 9 to both sides:

取 x 系数的一半是 3，平方得 9。两边同时加 9：

x² + 6x + 9 = 7

Now the left side is a perfect square:

现在左边是一个完全平方：

(x + 3)² = 7

Take the square root of both sides:

两边开平方：

x + 3 = ±√7

x = -3 ± √7

If the coefficient of x² is not 1, divide the whole equation by that coefficient before completing the square.

如果 x² 的系数不是 1，则先在整个方程两边除以该系数，再进行配方。

4. The Quadratic Formula | 二次公式

The quadratic formula solves any quadratic equation of the form ax² + bx + c = 0. It is derived by completing the square on the general equation and gives:

二次公式可以解任何形式为 ax² + bx + c = 0 的二次方程。它是通过对一般方程配方推导得出的公式：

x = (-b ± √(b² – 4ac)) / (2a)

This formula works for every quadratic equation, including those with irrational or complex roots. You should substitute the values of a, b, and c carefully, paying attention to signs.

这个公式适用于所有二次方程，包括具有无理根或复数根的方程。代入 a、b、c 的值时要仔细，特别注意符号。

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

例：解 2x² – 4x – 3 = 0。这里 a = 2，b = -4，c = -3。代入公式得到：

x = (4 ± √((-4)² – 4×2×(-3))) / (2×2)

x = (4 ± √(16 + 24)) / 4

x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2

Always simplify the square root and cancel common factors in the final answer.

最终答案中要化简根式并约去公因数。

5. The Discriminant | 判别式

The expression b² – 4ac inside the quadratic formula is called the discriminant. It determines the nature of the roots without solving the equation.

二次公式中的表达式 b² – 4ac 称为判别式。它可以在不解方程的情况下判断根的性质。

If b² – 4ac > 0, there are two distinct real roots.

如果 b² – 4ac > 0，方程有两个不同的实数根。
If b² – 4ac = 0, there is exactly one repeated real root.

如果 b² – 4ac = 0，方程有一个重实数根。
If b² – 4ac < 0, there are no real roots; the roots are complex.

如果 b² – 4ac < 0，方程没有实数根；根为复数。

For example, the equation x² – 6x + 9 = 0 has discriminant (-6)² – 4×1×9 = 0, so it has one repeated root x = 3.

例如，方程 x² – 6x + 9 = 0 的判别式为 (-6)² – 4×1×9 = 0，因此有一个重根 x = 3。

The discriminant also tells us whether the graph of the quadratic function crosses the x-axis, touches it, or does not meet it at all.

判别式还告诉我们二次函数的图像是与 x 轴相交、相切还是完全不相交。

6. Solving Quadratic Inequalities | 二次不等式

Quadratic inequalities involve expressions like ax² + bx + c > 0 or ≤ 0. To solve them, first solve the corresponding quadratic equation to find the critical points, then test intervals on a number line.

二次不等式涉及形如 ax² + bx + c > 0 或 ≤ 0 的表达式。解这类不等式时，先解对应的二次方程找到临界点，然后在数轴上测试区间。

For example, solve x² – 3x – 4 < 0. First factorise:

例如，解 x² – 3x – 4 < 0。首先因式分解：

(x – 4)(x + 1) < 0

The critical points are x = 4 and x = -1. The sign of the quadratic changes at these points. Testing the intervals gives the solution -1 < x < 4.

临界点是 x = 4 和 x = -1。二次表达式在这些点改变符号。测试各区间后得到解为 -1 < x < 4。

Remember that for > 0, the solution would be x < -1 or x > 4. The graph of the quadratic is a useful visual aid for choosing the correct region.

记住，如果是 > 0，解为 x < -1 或 x > 4。二次函数图像是选择正确区域的有用辅助工具。

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

The graph of a quadratic function y = ax² + bx + c is a parabola. When a > 0, the parabola opens upward and has a minimum point. When a < 0, it opens downward and has a maximum point.

二次函数 y = ax² + bx + c 的图像是抛物线。当 a > 0 时，抛物线开口向上，有最小值点；当 a < 0 时，开口向下，有最大值点。

The x-intercepts of the graph are the roots of the equation ax² + bx + c = 0. The y-intercept is simply c, because when x = 0, y = c.

图像与 x 轴的交点是方程 ax² + bx + c = 0 的根。与 y 轴的交点就是 c，因为当 x = 0 时，y = c。

The axis of symmetry is the vertical line x = -b/(2a). It passes through the vertex of the parabola.

对称轴是竖直线 x = -b/(2a)。它经过抛物线的顶点。

x_vertex = -b / (2a)

To find the y-coordinate of the vertex, substitute x_vertex back into the original equation.

要求顶点的 y 坐标，将 x_vertex 代回原方程即可。

8. Roots and Vertex | 根与顶点

There is a direct connection between the roots of a quadratic equation and the vertex of its graph. The x-coordinate of the vertex is exactly halfway between the two roots, if they exist.

二次方程的根与其图像顶点之间有直接联系。如果根存在，顶点的 x 坐标正好位于两个根的中间。

If the roots are r₁ and r₂, then:

如果根为 r₁ 和 r₂，则：

x_vertex = (r₁ + r₂) / 2

This is also equal to -b/(2a). The minimum or maximum value of the quadratic function is the y-coordinate of the vertex.

这也等于 -b/(2a)。二次函数的最小值或最大值就是顶点的 y 坐标。

For example, the equation x² – 4x + 3 = 0 has roots x = 1 and x = 3. The vertex has x-coordinate (1+3)/2 = 2, and substituting x = 2 gives y = 4 – 8 + 3 = -1. Thus the vertex is (2, -1).

例如，方程 x² – 4x + 3 = 0 的根为 x = 1 和 x = 3。顶点的 x 坐标为 (1+3)/2 = 2，代入 x = 2 得 y = 4 – 8 + 3 = -1。因此顶点为 (2, -1)。

9. Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0, if the roots are α and β, then the sum and product of the roots can be written directly from the coefficients:

对于二次方程 ax² + bx + c = 0，若根为 α 和 β，则根的和与积可以直接用系数表示：

α + β = -b / a

αβ = c / a

These relationships are useful for checking solutions and for constructing a quadratic equation with given roots.

这些关系可用于检验解的正确性，也可用于构造具有给定根的二次方程。

If you know the roots are α and β, the quadratic equation can be written as:

如果已知根为 α 和 β，则二次方程可写为：

x² – (α + β)x + αβ = 0

For example, if the roots are 5 and -2, then the equation is x² – (5 + (-2))x + 5×(-2) = 0, which simplifies to x² – 3x – 10 = 0.

例如，若根为 5 和 -2，则方程为 x² – (5 + (-2))x + 5×(-2) = 0，化简得 x² – 3x – 10 = 0。

10. Word Problems | 应用题

Quadratic equations often arise from geometry, motion, and area problems. The key is to translate the given information into a quadratic equation and then solve it using one of the methods above.

二次方程常常出现在几何、运动和面积问题中。关键是将其中的信息转化为二次方程，然后用上述任意一种方法求解。

Example: A rectangle has length 3 cm more than its width. Its area is 40 cm². Let the width be x cm. Then the length is x + 3 cm, and the area gives:

例：一个长方形的长比宽多 3 cm，面积为 40 cm²。设宽为 x cm，则长为 x + 3 cm，面积给出：

x(x + 3) = 40

x² + 3x – 40 = 0

Factorise: (x + 8)(x – 5) = 0, so x = -8 or x = 5. Since the width cannot be negative, x = 5. Thus the rectangle is 5 cm by 8 cm.

因式分解：(x + 8)(x – 5) = 0，所以 x = -8 或 x = 5。因为宽度不能为负，所以 x = 5。因此长方形为 5 cm 和 8 cm。

Always check whether your final answers make sense in the context of the problem. Reject any solution that does not satisfy the physical or geometric constraints.

始终检查最终答案在问题情境中是否有意义。拒绝任何不满足物理或几何约束的解。

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Solving Quadratic Equations | 解二次方程