Complete Guide to Solving Quadratic Equations | 一元二次方程的求解方法全归纳

📚 Complete Guide to Solving Quadratic Equations | 一元二次方程的求解方法全归纳

Quadratic equations are among the most fundamental topics in mathematics, appearing in everything from projectile motion to optimisation problems. In this article, we systematically explore every major method for solving a quadratic equation, from factoring to the quadratic formula, and discuss when each method is most effective.

一元二次方程是数学中最基础也最重要的课题之一，从抛体运动到优化问题，无处不在。本文将系统梳理求解一元二次方程的各种主要方法，从因式分解到求根公式，并讨论每种方法的最佳适用场景。

1. Standard Form and Key Concepts | 标准形式与核心概念

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

一元二次方程是指可以写成以下标准形式的方程：

ax² + bx + c = 0, where a ≠ 0

Here, a is the coefficient of x², b is the coefficient of x, and c is the constant term. The condition a ≠ 0 is essential: if a = 0, the equation becomes linear, not quadratic.

其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。条件 a ≠ 0 至关重要：若 a = 0，方程退化为一次方程，而不是二次方程。

A solution to the equation is called a root or zero of the quadratic. Because a quadratic is a polynomial of degree 2, it always has exactly two roots in the complex number system, though they may be equal (a repeated root) or non-real.

方程的解称为二次方程的根或零点。由于二次多项式是二次的，它在复数范围内总是恰好有两个根，尽管它们可能相等（重根）或不是实数。

2. Method 1: Solving by Factoring | 方法一：因式分解法

Factoring is often the fastest method when the quadratic has simple integer or rational coefficients. The idea is to rewrite ax² + bx + c as a product of two linear factors, then set each factor equal to zero.

因式分解法通常是在二次方程具有简单整数或有理系数时的最快方法。其核心思想是将 ax² + bx + c 重写为两个一次因式的乘积，然后令每个因式等于零。

Example: Solve x² − 5x + 6 = 0

示例：求解 x² − 5x + 6 = 0

Find two numbers whose product is 6 and sum is −5. These are −2 and −3.

寻找两个数，乘积为 6，和为 −5。这两个数是 −2 和 −3。
Rewrite: (x − 2)(x − 3) = 0

重写为：(x − 2)(x − 3) = 0
Set each factor to zero: x − 2 = 0 ⇒ x = 2; x − 3 = 0 ⇒ x = 3.

令每个因式为零：x − 2 = 0 ⇒ x = 2；x − 3 = 0 ⇒ x = 3。
Roots are x = 2 and x = 3.

根为 x = 2 和 x = 3。

When the leading coefficient a ≠ 1, factoring requires a bit more care. For example, solve 2x² − 7x + 3 = 0. Look for factors of the form (2x + m)(x + n) where mn = 3 and 2n + m = −7. In this case, (2x − 1)(x − 3) = 0 gives x = ½ and x = 3.

当首项系数 a ≠ 1 时，因式分解需要更细致的处理。例如，求解 2x² − 7x + 3 = 0。寻找形如 (2x + m)(x + n) 的因式，其中 mn = 3 且 2n + m = −7。在本例中，(2x − 1)(x − 3) = 0 给出 x = ½ 和 x = 3。

When to use factoring	何时使用因式分解
Coefficients are small integers or simple fractions 系数为较小的整数或简单分数	Roots are rational numbers 根为有理数

3. Method 2: Completing the Square | 方法二：配方法

Completing the square transforms the quadratic into the form a(x − h)² + k, from which the roots can be found by isolating the squared term. This method also reveals the vertex of the parabola, making it invaluable for graphing.

配方法将二次式变形为 a(x − h)² + k 的形式，从而可以通过隔离平方项来求根。该方法还能揭示抛物线的顶点，因此对作图非常有价值。

Starting from ax² + bx + c = 0, divide by a (if a ≠ 1), then add and subtract the square of half the coefficient of x.

从 ax² + bx + c = 0 出发，先除以 a（如果 a ≠ 1），然后加上并减去 x 系数一半的平方。

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

示例：用配方法求解 2x² + 8x − 10 = 0。

Divide by 2: x² + 4x − 5 = 0

除以 2：x² + 4x − 5 = 0
Move constant: x² + 4x = 5

移常数项：x² + 4x = 5
Add (4/2)² = 4 to both sides: x² + 4x + 4 = 9

两边加上 (4/2)² = 4：x² + 4x + 4 = 9
Factor the left side: (x + 2)² = 9

左边因式分解：(x + 2)² = 9
Take square roots: x + 2 = ±3 ⇒ x = 1 or x = −5.

开平方：x + 2 = ±3 ⇒ x = 1 或 x = −5。

For a general quadratic, completing the square gives:

对于一般二次方程，配方得到：

ax² + bx + c = a(x + b/2a)² + (4ac − b²)/4a

Setting this to zero and solving for x directly produces the quadratic formula, which we examine next.

令其为零并解出 x，即可直接导出求根公式，我们接下来讨论这一方法。

4. Method 3: The Quadratic Formula | 方法三：求根公式法

The quadratic formula is the most general and reliable method. It works for every quadratic equation, regardless of whether the roots are rational, irrational, or complex.

求根公式是最通用、最可靠的方法。它适用于所有一元二次方程，无论根是有理数、无理数还是复数。

x = (−b ± √(b² − 4ac)) / 2a

The expression under the square root, Δ = b² − 4ac, is called the discriminant.

根号下的表达式 Δ = b² − 4ac 称为判别式。

Example: Solve 3x² − 4x − 2 = 0.

示例：求解 3x² − 4x − 2 = 0。

Here a = 3, b = −4, c = −2. The discriminant is:

这里 a = 3，b = −4，c = −2。判别式为：

Δ = (−4)² − 4 × 3 × (−2) = 16 + 24 = 40

Since Δ = 40 > 0, there are two distinct real roots:

因为 Δ = 40 > 0，有两个不同的实根：

x = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3

Thus the roots are (2 + √10)/3 ≈ 1.72 and (2 − √10)/3 ≈ −1.39.

因此根为 (2 + √10)/3 ≈ 1.72 和 (2 − √10)/3 ≈ −1.39。

Discriminant Δ = b² − 4ac	判别式 Δ = b² − 4ac	Nature of roots	根的性质
Δ > 0	Δ > 0	Two distinct real roots	两个不相等实根
Δ = 0	Δ = 0	One repeated real root	一个重根（两个相等实根）
Δ < 0	Δ < 0	No real roots, two complex roots	无实根，有两个复数根

5. Method 4: Graphical Method | 方法四：图像法

The roots of a quadratic equation ax² + bx + c = 0 correspond to the x-intercepts of the parabola y = ax² + bx + c. By sketching or plotting the graph, we can estimate the solutions where the curve crosses the x-axis.

一元二次方程 ax² + bx + c = 0 的根对应抛物线 y = ax² + bx + c 与 x 轴的交点。通过绘制图像，我们可以估计曲线与 x 轴交点处的解。

The sign of the discriminant tells us the number of intercepts:

判别式的符号决定了交点的个数：

Δ > 0: the parabola crosses the x-axis at two distinct points.

Δ > 0：抛物线与 x 轴有两个不同的交点。
Δ = 0: the parabola touches the x-axis at exactly one point (the vertex lies on the axis).

Δ = 0：抛物线与 x 轴恰好相切于一点（顶点在轴上）。
Δ < 0: the parabola does not cross the x-axis at all; the roots are complex.

Δ < 0：抛物线完全不与 x 轴相交；根为复数。

The graphical method is rarely used for exact answers, but it provides strong intuition and confirms the plausibility of algebraic solutions.

图像法很少用于求精确解，但它能提供直观认识，并帮助确认代数解的合理性。

6. Method 5: Vieta’s Formulas | 方法五：韦达定理

Vieta’s formulas relate the roots of a quadratic to its coefficients without solving the equation explicitly. If the roots are r₁ and r₂, then:

韦达定理无需解方程，直接将方程根与其系数联系起来。若根为 r₁ 和 r₂，则：

r₁ + r₂ = −b/a

r₁ × r₂ = c/a

These formulas are extremely useful in problems that ask for the sum or product of roots, or when one root is known and the other needs to be found.

这些公式在处理求根的和或积、或已知一根求另一根的问题时极为有用。

Example: If one root of x² − 7x + 12 = 0 is 3, find the other root.

示例：若 x² − 7x + 12 = 0 的一个根为 3，求另一个根。

Using sum: r₁ + r₂ = 7, so 3 + r₂ = 7 ⇒ r₂ = 4. Checking product: 3 × 4 = 12 ✓.

利用和的关系：r₁ + r₂ = 7，因此 3 + r₂ = 7 ⇒ r₂ = 4。检验积：3 × 4 = 12 ✓。

Vieta’s formulas also help when constructing a quadratic equation with given roots: x² − (sum)x + (product) = 0.

韦达定理还用于由给定根构造二次方程：x² − (和)x + (积) = 0。

7. Special Cases and Shortcuts | 特殊情形与简化技巧

Certain forms of quadratic equations admit shortcuts.

某些特殊形式的一元二次方程可以使用简便方法。

Missing constant term (c = 0): ax² + bx = x(ax + b) = 0, so x = 0 or x = −b/a.

缺少常数项（c = 0）：ax² + bx = x(ax + b) = 0，因此 x = 0 或 x = −b/a。
Missing linear term (b = 0): ax² + c = 0 ⇒ x² = −c/a, giving x = ±√(−c/a). If −c/a < 0, the roots are purely imaginary.

缺少一次项（b = 0）：ax² + c = 0 ⇒ x² = −c/a，从而 x = ±√(−c/a)。若 −c/a < 0，则根为纯虚数。
Perfect square trinomial: x² + 2px + p² = (x + p)² = 0 gives a repeated root x = −p.

完全平方三项式：x² + 2px + p² = (x + p)² = 0，给出重根 x = −p。
Difference of squares: x² − p² = (x − p)(x + p) = 0 gives x = ±p.

平方差：x² − p² = (x − p)(x + p) = 0，给出 x = ±p。

Recognising these patterns instantly saves time in exams.

在考试中快速识别这些模式可以节省大量时间。

8. Worked Mixed Examples | 综合例题精讲

Let us apply all methods to a single equation to see how they compare.

让我们用同一方程演示各种方法的对比。

Equation: x² + 6x + 9 = 0

方程：x² + 6x + 9 = 0

Factoring: (x + 3)(x + 3) = 0 ⇒ x = −3 (repeated).

因式分解：(x + 3)(x + 3) = 0 ⇒ x = −3（重根）。
Completing the square: (x + 3)² = 0 ⇒ x = −3.

配方法：(x + 3)² = 0 ⇒ x = −3。
Formula: Δ = 36 − 36 = 0, x = (−6 ± 0)/2 = −3.

求根公式：Δ = 36 − 36 = 0，x = (−6 ± 0)/2 = −3。
Vieta: r₁ + r₂ = −6 and r₁r₂ = 9; the only possible value is r₁ = r₂ = −3.

韦达定理：r₁ + r₂ = −6 且 r₁r₂ = 9；唯一可能为 r₁ = r₂ = −3。

Equation: x² − 4x + 13 = 0 (complex roots)

方程：x² − 4x + 13 = 0（复数根）

Complete the square:

配方：

(x − 2)² + 9 = 0 ⇒ (x − 2)² = −9 ⇒ x − 2 = ±3i ⇒ x = 2 ± 3i

The discriminant is Δ = 16 − 52 = −36 < 0, confirming a pair of complex conjugates.

判别式 Δ = 16 − 52 = −36 < 0，证实有一对共轭复数根。

9. Choosing the Best Method | 如何选择最优方法

Different situations call for different approaches. Here is a practical guide.

不同情形需要不同方法。以下是一份实用指南。

Situation	情形	Recommended method	推荐方法
Equation factors easily with integer roots	方程可轻松分解且根为整数	Factoring	因式分解法
Need vertex form for graphing	需要顶点式作图	Completing the square	配方法
Any general quadratic, guarantee of success	任意一般二次方程，需确保可行	Quadratic formula	求根公式法
Only need sum/product of roots	只需根的和或积	Vieta’s formulas	韦达定理
Checking answers visually	直观检验答案	Graphing	图像法

In exams, always look for quick factorisation first, then fall back on the quadratic formula if factoring is not obvious.

在考试中，先尝试快速因式分解，若分解不明显，再使用求根公式。

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

Even strong students make preventable mistakes on quadratic equations. Watch out for the following.

即使是优秀学生也会在二次方程上犯一些可避免的错误。请留意以下事项。

Forgetting a ≠ 0: If a = 0, it is a linear equation, not quadratic.

忽略 a ≠ 0：若 a = 0，它就是一次方程，而非二次方程。
Dividing by x without considering x = 0: If c = 0, do not cancel x from both sides, or you lose the root x = 0.

除以 x 时未考虑 x = 0：若 c = 0，切勿从两边同时约去 x，否则会丢失根 x = 0。
Mistaking b signs in the formula: For ax² + bx + c = 0, the numerator is −b ± √(b² − 4ac), not b ± …

公式中 b 的符号错误：对于 ax² + bx + c = 0，分子应为 −b ± √(b² − 4ac)，而不是 b ± …
Simplifying √(p² + q²) as p + q: √(x² + 4) cannot be simplified to x + 2.

将 √(p² + q²) 错误化简为 p + q：√(x² + 4) 不能化简为 x + 2。
Missing the ± sign: When taking square roots, always include both positive and negative solutions unless the context forbids it.

漏写 ± 号：开平方时务必同时包含正负解，除非题目背景限制。

Always test your roots by substituting them back into the original equation.

务必把根代回原方程进行检验。

11. Beyond the Basics | 拓展视野

The methods covered here extend naturally to higher-degree polynomials and related topics such as inequalities.

本文所述方法可以自然延伸到高次多项式及不等式等相关主题。

For example, solving ax² + bx + c > 0 uses the roots to split the number line into intervals, and the sign of a determines whether the solution is the interval between the roots or the two outer regions.

例如，求解 ax² + bx + c > 0 时，利用根将数轴分段，a 的符号决定解集是两根之间还是两根之外。

Additionally, the quadratic formula is the foundation for the general solution of cubic equations via transformations, and it reappears in differential equations, optimisation, and numerical analysis.

此外，求根公式是三次方程一般解变换的基础，在微分方程、优化和数值分析中也会反复出现。

If x² + px + q = 0, then x = −p/2 ± √((p/2)² − q)

This vertex-based form is often easier to remember and directly leads to the solution.

这种基于顶点的形式往往更易记忆，并能直接导出解。

12. Practice Set | 练习题精选

Consolidate your understanding with these questions. Try to solve each using at least two different methods.

通过以下问题巩固理解。请尝试用至少两种不同方法求解每题。

Solve x² − 9x + 18 = 0.

求解 x² − 9x + 18 = 0。
Solve 3x² − 12x = 0.

求解 3x² − 12x = 0。
Solve 2x² + 3x − 2 = 0.

求解 2x² + 3x − 2 = 0。
Solve x² + 4x + 8 = 0, expressing answers in the form a ± bi.

求解 x² + 4x + 8 = 0，并将答案写成 a ± bi 的形式。
If the roots of x² + kx + 6 = 0 are consecutive integers, find k and the roots.

若 x² + kx + 6 = 0 的两根为连续整数，求 k 及两根。

Answers: 1) x = 3, 6. 2) x = 0, 4. 3) x = ½, −2. 4) x = −2 ± 2i. 5) k = ±5; if roots are −2 and −3, k = 5; if roots are 2 and 3, k = −5.

答案：1) x = 3, 6。2) x = 0, 4。3) x = ½, −2。4) x = −2 ± 2i。5) k = ±5；若根为 −2 和 −3，则 k = 5；若根为 2 和 3，则 k = −5。

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Complete Guide to Solving Quadratic Equations | 一元二次方程的求解方法全归纳