Solving Quadratic Equations: Methods and Applications | 一元二次方程的解法与应用

📚 Solving Quadratic Equations: Methods and Applications | 一元二次方程的解法与应用

A quadratic equation is one of the most fundamental topics in algebra. It appears in nearly every mathematics examination, from GCSE to A-level, and serves as the building block for many advanced concepts such as functions, inequalities, and calculus. Mastering the methods of solving quadratic equations is therefore not only a requirement for exams but also a vital skill for further mathematical study.

一元二次方程是代数学中最基础的内容之一。从 GCSE 到 A-Level，几乎所有数学考试都会涉及这一考点，同时它也是函数、不等式和微积分等更高阶概念的基石。因此，熟练掌握一元二次方程的解法不仅是应试的要求，更是深入学习数学的关键技能。

1. The Standard Form | 标准形式

A quadratic equation is any equation that can be written in the standard form:

ax² + bx + c = 0, where a ≠ 0

Here, ‘a’, ‘b’, and ‘c’ are constants, and ‘x’ is the unknown variable. The coefficient ‘a’ cannot be zero, otherwise the equation would become linear. The expression ax² + bx + c is called a quadratic expression, and the highest power of x is 2, which determines the degree of the equation.

一元二次方程是指可以写成标准形式的方程：

ax² + bx + c = 0，其中 a ≠ 0

其中 a、b、c 是常数，x 是未知数。系数 a 不能为零，否则方程将变为一次方程。我们把 ax² + bx + c 称为二次表达式，其中 x 的最高次数为 2，这一次数决定了方程的次数。

2. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method when the quadratic expression can be factored into two linear factors. The principle relies on the zero product property: if the product of two expressions is zero, then at least one of them must be zero.

因式分解法通常是解决一元二次方程最快的方法，其前提是二次表达式可以被分解为两个一次因式的乘积。这一方法的核心依据是零乘积性质：若两个表达式的乘积为零，则其中至少有一个为零。

For example, consider the equation:

x² − 5x + 6 = 0

We look for two numbers that multiply to give 6 and add to give −5. These numbers are −2 and −3. Hence, the equation can be factored as:

(x − 2)(x − 3) = 0

Applying the zero product property gives x − 2 = 0 or x − 3 = 0, so the solutions are x = 2 and x = 3.

例如，考虑方程：

x² − 5x + 6 = 0

我们寻找两个数，它们的乘积为 6 且和为 −5。这两个数是 −2 和 −3。因此，该方程可以分解为：

(x − 2)(x − 3) = 0

根据零乘积性质，可得 x − 2 = 0 或 x − 3 = 0，因此方程的解为 x = 2 和 x = 3。

When the coefficient of x² is not 1, factorisation requires more care. Consider 2x² + 5x − 3 = 0. We look for two numbers whose product is 2 × (−3) = −6 and whose sum is 5. These numbers are 6 and −1. We then rewrite the middle term and factor by grouping.

当 x² 的系数不为 1 时，因式分解需要更加谨慎。例如，考虑 2x² + 5x − 3 = 0。我们需要找到两个数，使其乘积为 2 × (−3) = −6，且和为 5。这两个数是 6 和 −1。然后我们重写中间项并使用分组分解法。

2x² + 6x − x − 3 = 0

2x(x + 3) − 1(x + 3) = 0

(2x − 1)(x + 3) = 0

Therefore, the solutions are x = ½ and x = −3.

因此，方程的解为 x = ½ 和 x = −3。

3. Solving by Completing the Square | 配方法

Completing the square is a powerful algebraic technique that rewrites a quadratic expression in the form a(x + p)² + q. This method is particularly useful when factorisation is not straightforward, and it also provides the foundation for deriving the quadratic formula.

配方法是一种强大的代数技巧，它将二次表达式改写为 a(x + p)² + q 的形式。当因式分解不易实现时，这种方法尤为实用，同时它也是推导求根公式的基础。

The general procedure for completing the square on x² + bx + c is to take half of b, square it, and then adjust the constant term accordingly. Consider the equation:

x² + 6x + 2 = 0

First, we focus on x² + 6x. Half of 6 is 3, and 3² = 9. We can therefore write:

x² + 6x = (x + 3)² − 9

Substituting back into the original equation:

(x + 3)² − 9 + 2 = 0

(x + 3)² − 7 = 0

对 x² + bx + c 进行配方的一般步骤是：取 b 的一半，将其平方，然后相应调整常数项。考虑方程：

x² + 6x + 2 = 0

首先，我们关注 x² + 6x。6 的一半是 3，而 3² = 9。因此我们可以写出：

x² + 6x = (x + 3)² − 9

将其代入原方程：

(x + 3)² − 9 + 2 = 0

(x + 3)² − 7 = 0

Now we can solve for x:

(x + 3)² = 7

x + 3 = ±√7

x = −3 ± √7

Thus, the two solutions are x = −3 + √7 and x = −3 − √7. The notation ± indicates that both the positive and negative square roots must be considered.

现在我们可以求解 x：

(x + 3)² = 7

x + 3 = ±√7

x = −3 ± √7

因此，两个解为 x = −3 + √7 和 x = −3 − √7。符号 ± 表示正平方根和负平方根都必须考虑。

4. The Quadratic Formula | 求根公式

The quadratic formula is the most general method for solving quadratic equations. It works for every quadratic equation, regardless of whether the expression can be factored neatly or not. The formula is derived by completing the square on the general form ax² + bx + c = 0.

求根公式是解一元二次方程最通用的方法。无论二次表达式能否被简洁地因式分解，该公式对一切一元二次方程都适用。这一公式是通过对一般形式 ax² + bx + c = 0 配方推导而来的。

x = (−b ± √(b² − 4ac)) / (2a)

To use this formula, simply identify the values of a, b, and c from the standard form, substitute them into the formula, and simplify. Consider the equation 2x² − 4x − 3 = 0. Here, a = 2, b = −4, and c = −3.

使用该公式时，只需从标准形式中识别出 a、b、c 的值，将其代入公式并化简。例如，考虑方程 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。

x = (4 ± √(16 + 24)) / 4

x = (4 ± √40) / 4

x = (4 ± 2√10) / 4

x = (2 ± √10) / 2

The two solutions are therefore x = (2 + √10)/2 and x = (2 − √10)/2.

因此，两个解为 x = (2 + √10)/2 和 x = (2 − √10)/2。

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

In the quadratic formula, the expression under the square root sign, b² − 4ac, is called the discriminant. Its value determines the number and nature of the roots of a quadratic equation without the need to solve it fully. The discriminant is typically denoted by the symbol Δ (delta).

在求根公式中，根号下的表达式 b² − 4ac 称为判别式。它的值决定了二次方程根的个数和性质，无需完整求解即可判断。判别式通常用符号 Δ（德尔塔）表示。

If Δ > 0, the equation has two distinct real roots.
If Δ = 0, the equation has exactly one real root (a repeated root).
If Δ < 0, the equation has no real roots; instead, it has two complex conjugate roots.

若 Δ > 0，方程有两个不相等的实数根。
若 Δ = 0，方程恰好有一个实数根（重根）。
若 Δ < 0，方程没有实数根，而是有两个共轭复数根。

Consider the equation x² − 4x + 4 = 0. The discriminant is Δ = (−4)² − 4(1)(4) = 16 − 16 = 0. Since Δ = 0, there is exactly one real root, which is x = 2. On the other hand, for x² + 2x + 5 = 0, the discriminant is Δ = 4 − 20 = −16 < 0, so this equation has no real roots.

考虑方程 x² − 4x + 4 = 0。其判别式为 Δ = (−4)² − 4(1)(4) = 16 − 16 = 0。因为 Δ = 0，所以方程恰好有一个实数根，即 x = 2。另一方面，对于 x² + 2x + 5 = 0，判别式为 Δ = 4 − 20 = −16 < 0，所以该方程没有实数根。

When the discriminant is a perfect square (and a, b, c are rational), the equation can also be solved by factorisation. Conversely, a non-perfect-square discriminant indicates that the roots are irrational and cannot be obtained by simple factorisation.

当判别式是完全平方数（且 a、b、c 为有理数）时，方程也可通过因式分解求解。反之，若判别式不是完全平方数，则说明根是无理数，无法通过简单因式分解求得。

6. Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0 with roots α and β, there are important relationships between the roots and the coefficients. These relationships allow us to construct quadratic equations from given roots and to analyse properties of roots without solving the equation explicitly.

对于根为 α 和 β 的一元二次方程 ax² + bx + c = 0 而言，根与系数之间存在着重要的关系。利用这些关系，我们可以根据已知根来构造二次方程，并无需直接求解即可分析根的性质。

Sum of roots: α + β = −b/a

Product of roots: α × β = c/a

根的和：α + β = −b/a

根的积：α × β = c/a

These formulas are derived from the factorised form a(x − α)(x − β) = 0, which expands to ax² − a(α + β)x + aαβ = 0. Comparing coefficients with ax² + bx + c = 0 yields the relationships above.

这些公式由分解形式 a(x − α)(x − β) = 0 推导而来，展开后得到 ax² − a(α + β)x + aαβ = 0。将系数与 ax² + bx + c = 0 进行比较即可得出上述关系。

For example, if a quadratic equation has roots α = 3 and β = −2, then the sum is α + β = 1 and the product is αβ = −6. The corresponding equation is x² − (sum)x + (product) = 0, giving x² − x − 6 = 0.

例如，若某个二次方程的根为 α = 3 和 β = −2，则根的和为 α + β = 1，根的积为 αβ = −6。对应的方程为 x² −（和）x +（积）= 0，即 x² − x − 6 = 0。

7. Applications in Physics and Kinematics | 在物理与运动学中的应用

Quadratic equations arise naturally in physics, particularly in problems involving projectile motion. When an object is thrown upwards with an initial velocity, the height of the object at a given time can be modelled by a quadratic equation.

在物理学中，一元二次方程自然而然地出现，尤其是在涉及抛体运动的问题中。当一个物体以初速度向上抛出时，物体在某一时刻的高度可以用一元二次方程来建模。

Consider a ball thrown upwards with an initial velocity of 20 m/s from a height of 5 metres. The height h after t seconds is given by the equation:

h = −5t² + 20t + 5

This equation uses a simplified gravitational constant of 10 m/s² for ease of calculation. To find when the ball hits the ground, we set h = 0:

−5t² + 20t + 5 = 0

考虑一个球以 20 m/s 的初速度从 5 米的高度向上抛出。t 秒后球的高度 h 由以下方程给出：

h = −5t² + 20t + 5

为便于计算，此方程使用了简化的重力加速度 10 m/s²。为求球何时落地，我们令 h = 0：

−5t² + 20t + 5 = 0

Dividing through by −5 gives t² − 4t − 1 = 0. Using the quadratic formula:

t = (4 ± √(16 + 4)) / 2 = (4 ± √20) / 2

t = 2 ± √5

Since time cannot be negative, we take t = 2 + √5 ≈ 4.24 seconds as the physically meaningful solution.

两边同时除以 −5 得到 t² − 4t − 1 = 0。使用求根公式：

t = (4 ± √(16 + 4)) / 2 = (4 ± √20) / 2

t = 2 ± √5

由于时间不能为负，我们取 t = 2 + √5 ≈ 4.24 秒作为具有物理意义的解。

8. Applications in Geometry | 在几何中的应用

Quadratic equations frequently appear in geometric problems, particularly those involving area and the Pythagorean theorem. Consider a rectangle whose length is 3 metres longer than its width, and whose area is 40 square metres. Let the width be x metres. Then the length is x + 3, and the area equation is:

x(x + 3) = 40

一元二次方程经常出现在几何问题中，尤其是涉及面积和勾股定理的问题。考虑一个长方形，其长比宽多 3 米，面积为 40 平方米。设宽为 x 米，则长为 x + 3，面积方程为：

x(x + 3) = 40

Expanding and rearranging:

x² + 3x − 40 = 0

Factorising, we obtain (x + 8)(x − 5) = 0, which gives x = −8 or x = 5. Since a length cannot be negative, the width is 5 metres and the length is 8 metres.

展开并整理：

x² + 3x − 40 = 0

因式分解得 (x + 8)(x − 5) = 0，即 x = −8 或 x = 5。由于长度不能为负数，所以宽为 5 米、长为 8 米。

Another common geometric application involves right-angled triangles. If the two legs of a right triangle are x and x + 2, and the hypotenuse is 10, then by the Pythagorean theorem:

x² + (x + 2)² = 10²

另一个常见的几何应用涉及直角三角形。若直角三角形的两条直角边分别为 x 和 x + 2，斜边为 10，则根据勾股定理：

x² + (x + 2)² = 10²

Expanding gives x² + x² + 4x + 4 = 100, which simplifies to 2x² + 4x − 96 = 0, and further to x² + 2x − 48 = 0. Factorising yields (x + 8)(x − 6) = 0, so the valid positive solution is x = 6. The two legs are therefore 6 and 8 units long, confirming the well-known 6-8-10 Pythagorean triple.

展开得到 x² + x² + 4x + 4 = 100，化简为 2x² + 4x − 96 = 0，进一步化简为 x² + 2x − 48 = 0。因式分解得 (x + 8)(x − 6) = 0，因此有效的正数解为 x = 6。两条直角边分别为 6 和 8，验证了著名的 6-8-10 勾股数。

9. The Discriminant and Graphical Interpretation | 判别式与图像的解读

The graph of a quadratic function y = ax² + bx + c is a parabola. The number of real roots of the equation ax² + bx + c = 0 corresponds exactly to the number of intersections between the parabola and the x-axis. The discriminant therefore provides crucial information about the position of the parabola relative to the x-axis.

二次函数 y = ax² + bx + c 的图像是一条抛物线。方程 ax² + bx + c = 0 的实数根个数恰好对应抛物线与 x 轴的交点个数。因此，判别式为抛物线与 x 轴的相对位置提供了关键信息。

If Δ > 0, the parabola intersects the x-axis at two distinct points.
If Δ = 0, the parabola touches the x-axis at exactly one point (the vertex lies on the x-axis).
If Δ < 0, the parabola does not intersect the x-axis at all.

若 Δ > 0，抛物线与 x 轴相交于两个不同的点。
若 Δ = 0，抛物线与 x 轴恰好相切于一点（顶点位于 x 轴上）。
若 Δ < 0，抛物线与 x 轴没有交点。

Furthermore, the sign of the coefficient a determines whether the parabola opens upward (a > 0) or downward (a < 0). Combining this with the discriminant allows us to sketch the graph of any quadratic function quickly and accurately.

此外，系数 a 的符号决定了抛物线的开口方向：当 a > 0 时开口向上，当 a < 0 时开口向下。将这一信息与判别式结合使用，我们可以快速准确地画出任意二次函数的图像。

10. Choosing the Best Method | 选择最优解法

In examinations, students are often faced with a quadratic equation and must decide which method to use. The table below summarises the recommended approaches based on the characteristics of the equation.

在考试中，学生经常会面临一道一元二次方程并需要决定使用哪种方法。下表根据方程的特征总结了推荐的做法。

Method	Best Used When	Advantage
Factorisation	The expression is easily factored	Fastest; no formula needed
Completing the Square	The coefficient a = 1 and b is even	Also gives the vertex of the parabola
Quadratic Formula	Factorisation is difficult or impossible	Works universally for all quadratics

解法	适用情形	优势
因式分解法	表达式容易被分解	速度最快；无需套用公式
配方法	系数 a = 1 且 b 为偶数	同时可得到抛物线的顶点
求根公式	因式分解困难或无法分解	对一切一元二次方程通用

A good rule of thumb is to try factorisation first. If the equation factorises within a few seconds, use it. If not, apply the quadratic formula directly, especially when the discriminant is not a perfect square.

一个实用的经验法则是先尝试因式分解。如果方程能在几秒钟内分解，就直接使用这种方法。如果不能，那么直接套用求根公式，尤其是当判别式不是完全平方数时。

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Solving Quadratic Equations: Methods and Applications | 一元二次方程的解法与应用