📚 Methods for Solving Quadratic Equations | 二次方程求解方法
A quadratic equation is a polynomial equation of degree 2, typically written in the form ax² + bx + c = 0, where a ≠ 0. Solving such equations is a fundamental skill in algebra and appears extensively in IB Mathematics. This article introduces the main methods for finding solutions, compares their advantages, and highlights common pitfalls.
二次方程是最高次数为 2 的多项式方程，通常写作 ax² + bx + c = 0 的形式，其中 a ≠ 0。求解这类方程是代数学习的基本技能，在 IB 数学中广泛出现。本文将介绍求解二次方程的主要方法，比较它们的优缺点，并指出常见易错点。
1. The Standard Form | 标准形式
Before choosing a method, ensure the equation is written 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. For example, 2x² − 5x + 3 = 0 is in standard form, while x² = 3x − 1 must be rearranged to x² − 3x + 1 = 0.
在选择方法之前，先确保方程写成标准形式 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。例如，2x² − 5x + 3 = 0 是标准形式，而 x² = 3x − 1 需要重新整理为 x² − 3x + 1 = 0。
2. Solving by Factoring | 因式分解法
Factoring uses the zero-product property: if the product of two expressions is zero, then at least one of them must be zero. To solve x² − 5x + 6 = 0, we factor it as (x − 2)(x − 3) = 0, giving x = 2 or x = 3. This method is fast when the factors are simple integers or rational numbers.
因式分解法利用零积性质：如果两个表达式的乘积为零，则至少有一个为零。解 x² − 5x + 6 = 0 时，可分解为 (x − 2)(x − 3) = 0，得到 x = 2 或 x = 3。当因式为简单的整数或有理数时，这种方法非常快捷。
For a quadratic with a ≠ 1, such as 2x² + 5x + 3 = 0, look for two numbers whose product is a × c = 6 and whose sum is b = 5. The numbers 2 and 3 work, so rewrite the middle term: 2x² + 2x + 3x + 3 = 0, then factor by grouping: (2x + 3)(x + 1) = 0. Hence x = −3/2 or x = −1.
对于 a ≠ 1 的二次方程，例如 2x² + 5x + 3 = 0，寻找两个数，使它们的乘积等于 a × c = 6，且它们的和等于 b = 5。数字 2 和 3 满足条件，因此改写中间项：2x² + 2x + 3x + 3 = 0，然后分组因式分解：(2x + 3)(x + 1) = 0。于是 x = −3/2 或 x = −1。
3. Solving by Completing the Square | 配方法
Completing the square transforms the equation ax² + bx + c = 0 into the form a(x − h)² + k = 0. For example, solve x² + 6x + 5 = 0. Move the constant: x² + 6x = −5. Add (6/2)² = 9 to both sides: x² + 6x + 9 = 4. This gives (x + 3)² = 4, so x + 3 = ±2, yielding x = −1 or x = −5.
配方法将方程 ax² + bx + c = 0 转化为 a(x − h)² + k = 0 的形式。例如，解 x² + 6x + 5 = 0。移常数项：x² + 6x = −5。两边加上 (6/2)² = 9：x² + 6x + 9 = 4。于是 (x + 3)² = 4，因此 x + 3 = ±2，得到 x = −1 或 x = −5。
If a ≠ 1, divide by a first. For 2x² − 8x + 3 = 0, divide by 2 to get x² − 4x + 3/2 = 0. Then x² − 4x = −3/2. Add (−4/2)² = 4: x² − 4x + 4 = 5/2, so (x − 2)² = 5/2, hence x = 2 ± √(5/2).
若 a ≠ 1，先除以 a。对于 2x² − 8x + 3 = 0，除以 2 得到 x² − 4x + 3/2 = 0。然后 x² − 4x = −3/2。加上 (−4/2)² = 4：x² − 4x + 4 = 5/2，即 (x − 2)² = 5/2，所以 x = 2 ± √(5/2)。
4. The Quadratic Formula | 求根公式法
The quadratic formula solves any quadratic equation directly. For ax² + bx + c = 0, the solutions are given by:
求根公式可以直接求解任何二次方程。对于 ax² + bx + c = 0，解为：
x = (−b ± √(b² − 4ac)) / (2a)
Here the expression b² − 4ac is called the discriminant, denoted Δ. If Δ > 0, there are two distinct real roots; if Δ = 0, there is one repeated real root; if Δ < 0, there are no real roots, but two complex conjugate roots.
其中 b² − 4ac 称为判别式，记作 Δ。若 Δ > 0，有两个不同的实数根；若 Δ = 0，有一个重根；若 Δ < 0，没有实数根，但有两个共轭复根。
For example, solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2. Then x = (−3 ± √(9 + 16)) / 4 = (−3 ± 5) / 4, giving x = 1/2 or x = −2.
例如，解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。则 x = (−3 ± √(9 + 16)) / 4 = (−3 ± 5) / 4，得到 x = 1/2 或 x = −2。
5. Solving by Factoring with Special Patterns | 特殊模式因式分解
Special products can speed up factoring. The difference of squares: x² − a² = (x − a)(x + a). For example, x² − 9 = 0 gives x = ±3. A perfect square trinomial: x² ± 2ax + a² = (x ± a)². For instance, x² − 6x + 9 = 0 becomes (x − 3)² = 0, so x = 3 is a double root.
特殊乘积可以加速因式分解。平方差公式：x² − a² = (x − a)(x + a)。例如，x² − 9 = 0 得 x = ±3。完全平方三项式：x² ± 2ax + a² = (x ± a)²。例如，x² − 6x + 9 = 0 变为 (x − 3)² = 0，所以 x = 3 是二重根。
6. Solving by Taking Square Roots | 开平方法
If the equation has the form (x − h)² = k, or ax² + c = 0, you can solve directly by taking square roots. For (x − 4)² = 25, take roots: x − 4 = ±5, hence x = 9 or x = −1. For 3x² − 15 = 0, write x² = 5, then x = ±√5.
如果方程具有形式 (x − h)² = k 或 ax² + c = 0，可以直接通过开平方求解。对于 (x − 4)² = 25，开方得 x − 4 = ±5，因此 x = 9 或 x = −1。对于 3x² − 15 = 0，写成 x² = 5，则 x = ±√5。
Remember that the square root of a positive number has two values: one positive and one negative. Do not forget the ± sign, as omitting it loses a solution.
请记住，正数的平方根有两个值：一正一负。不要忘记 ± 号，遗漏它会丢失一个解。
7. Using the Discriminant to Choose a Method | 用判别式选择方法
The discriminant Δ = b² − 4ac helps decide the most efficient method. If Δ is a perfect square, the roots are rational, and factoring or the formula both work well. If Δ is not a perfect square but Δ > 0, the roots are irrational and conjugate; completing the square or the formula is appropriate. If Δ < 0, the solutions are complex; use the formula with i = √(−1).
判别式 Δ = b² − 4ac 有助于选择最有效的方法。如果 Δ 是完全平方数，根为有理数，因式分解或公式法都很好。如果 Δ 不是完全平方数但 Δ > 0，根为无理数且互为共轭；适合用配方法或公式法。如果 Δ < 0，解为复数；使用含有 i = √(−1) 的公式。
Additionally, if b = 0, the equation is ax² + c = 0 and square roots are simplest. If c = 0, the equation is ax² + bx = 0, which factors as x(ax + b) = 0, giving x = 0 or x = −b/a.
此外，若 b = 0，方程为 ax² + c = 0，开平方法最简便。若 c = 0，方程为 ax² + bx = 0，可分解为 x(ax + b) = 0，得到 x = 0 或 x = −b/a。
8. Sum and Product of Roots | 根与系数的关系
For a quadratic equation ax² + bx + c = 0 with roots r₁ and r₂, the sum and product of the roots are:
对于二次方程 ax² + bx + c = 0，设其根为 r₁ 和 r₂，则两根之和与两根之积为：
r₁ + r₂ = −b/a, r₁ · r₂ = c/a
These relations allow you to check solutions quickly. For x² − 5x + 6 = 0, the roots 2 and 3 have sum 5 and product 6, matching −b/a and c/a. They also help construct a quadratic equation when the roots are known: x² − (sum)x + (product) = 0.
这些关系可以快速检验解的正确性。对于 x² − 5x + 6 = 0，根 2 和 3 的和为 5，积为 6，与 −b/a 和 c/a 吻合。已知根时也可构造二次方程：x² − (两根之和)x + (两根之积) = 0。
9. Graphical Interpretation | 图形解释
The real solutions of ax² + bx + c = 0 correspond to the x-intercepts of the parabola y = ax² + bx + c. If the parabola crosses the x-axis at two points, the equation has two distinct real roots; if it touches the x-axis at one point, there is one double root; if it does not touch the x-axis, there are no real roots.
二次方程 ax² + bx + c = 0 的实数解对应抛物线 y = ax² + bx + c 与 x 轴的交点。如果抛物线与 x 轴相交于两点，方程有两个不同的实数根；如果与 x 轴相切于一点，则有一个重根；如果不接触 x 轴，则没有实数根。
The vertex of the parabola lies at x = −b/(2a), which is also the axis of symmetry. This value gives the average of the two roots when they are real.
抛物线的顶点位于 x = −b/(2a)，这也是对称轴。当根为实数时，这个值等于两根的平均值。
10. Common Mistakes and How to Avoid Them | 常见错误与避免方法
One common mistake is forgetting to set the equation to zero before factoring. For example, x² = 3x should not be divided by x to get x = 3, because that loses the root x = 0. Instead, write x² − 3x = 0 and factor x(x − 3) = 0.
一个常见错误是在因式分解前忘记将方程化为零。例如，x² = 3x 不应两边除以 x 得到 x = 3，因为这样丢失了根 x = 0。而应写成 x² − 3x = 0，分解为 x(x − 3) = 0。
Another error is misapplying the quadratic formula with negative coefficients. Always substitute signs carefully: for x² − 2x − 3 = 0, b = −2, so −b = 2, and the formula gives x = (2 ± √(4 + 12)) / 2 = 3 or −1.
另一个错误是在使用求根公式时对负系数的处理不当。务必小心代入符号：对于 x² − 2x − 3 = 0，b = −2，所以 −b = 2，公式给出 x = (2 ± √(4 + 12)) / 2 = 3 或 −1。
Finally, when completing the square, remember to add the same value to both sides, and if a ≠ 1, divide every term by a first.
最后，在配方时，记得两边加同一个值；若 a ≠ 1，先每一项都除以 a。
11. Worked Examples | 例题精讲
Example 1: Solve 3x² − 10x + 3 = 0 by factoring. Look for two numbers with product 9 and sum −10: they are −1 and −9. Rewrite: 3x² − x − 9x + 3 = 0, then x(3x − 1) − 3(3x − 1) = 0, giving (3x − 1)(x − 3) = 0. Thus x = 1/3 or x = 3.
例 1：用因式分解法解 3x² − 10x + 3 = 0。寻找两个数，积为 9，和为 −10：它们是 −1 和 −9。改写：3x² − x − 9x + 3 = 0，然后 x(3x − 1) − 3(3x − 1) = 0，得到 (3x − 1)(x − 3) = 0。因此 x = 1/3 或 x = 3。
Example 2: Solve 2x² + 4x + 5 = 0. Here Δ = 16 − 40 = −24 < 0, so the roots are complex. Using the formula: x = (−4 ± √(−24)) / 4 = (−4 ± 2i√6) / 4 = −1 ± (i√6)/2.
例 2：解 2x² + 4x + 5 = 0。这里 Δ = 16 − 40 = −24 < 0，因此根为复数。使用公式：x = (−4 ± √(−24)) / 4 = (−4 ± 2i√6) / 4 = −1 ± (i√6)/2。
12. Choosing the Best Method | 选择最佳方法
In IB examinations, read the question carefully. If the problem asks for exact values, factoring, completing the square, or the quadratic formula are preferred. If it asks for roots to a given degree of accuracy, the formula or a graphical calculator may be used. If the quadratic is already factored, use the zero-product property immediately.
在 IB 考试中，请仔细阅读题目。若要求精确值，优先使用因式分解、配方或求根公式。若要求保留一定精度的根，可以使用公式或图形计算器。如果二次方程已经因式分解，立即使用零积性质。
Mastering all methods gives you flexibility and helps verify answers. Always check your solutions by substituting them back into the original equation.
掌握所有方法能让你灵活应变，并帮助验证答案。始终将解代回原方程进行检验。
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