📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are a core topic in IGCSE Mathematics. Mastering the methods of solving them is essential for exam success. In this article, you will learn the standard form, three solving methods, the discriminant, and how to interpret solutions graphically.
二次方程是 IGCSE 数学的核心内容。掌握其解法对考试成功至关重要。本文将带你学习二次方程的标准形式、三种解法、判别式，以及如何从图像上理解解的含义。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation is an equation that can be written in the form ax² + bx + c = 0, where a, b, c are constants and a ≠ 0. Here, x is the variable, a is the coefficient of x², b is the coefficient of x, and c is the constant term.
二次方程是可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。这里 x 是变量，a 是 x² 的系数，b 是 x 的系数，c 是常数项。
The highest power of the variable is 2, which is why it is called "quadratic". For example, 2x² + 3x − 5 = 0 is a quadratic equation, while x³ + 2x = 0 is not.
变量的最高次数是 2，因此称为“二次”。例如，2x² + 3x − 5 = 0 是二次方程，而 x³ + 2x = 0 不是。
Some equations may need rearranging first to get them into standard form. Always collect all terms on one side of the equals sign, with 0 on the other side.
有些方程需要先整理才能得到标准形式。始终把所有项移到等号一侧，使另一侧为 0。
2. Solving by Factorisation | 因式分解法
Factorisation is often the quickest method when the quadratic has simple integer roots. You factorise the left-hand side into two brackets and then use the zero-product property.
当二次方程具有简单的整数根时，因式分解法通常是最快的。你将等号左边分解成两个括号，然后利用零乘积性质。
Zero-product property: If A × B = 0, then A = 0 or B = 0.
零乘积性质：若 A × B = 0，则 A = 0 或 B = 0。
tiply to 6 and add to −5. They are −2 and −3, so (x − 2)(x − 3) = 0. Hence x = 2 or x = 3.
例：解 x² − 5x + 6 = 0。寻找两个数相乘得 6，相加得 −5。它们是 −2 和 −3，所以 (x − 2)(x − 3) = 0。因此 x = 2 或 x = 3。
Always check if the equation can be simplified or divided by a common factor first, especially when a, b, c share a common factor.
务必先检查方程能否化简或除以公因数，尤其是当 a、b、c 有公因数时。
3. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This method is useful for solving any quadratic equation, especially when factorisation is difficult.
配方法将二次方程改写为 a(x + p)² + q 的形式。此方法可用于解任何二次方程，尤其当因式分解困难时。
For the equation x² + bx + c = 0, add and subtract (b/2)² to complete the square.
对于方程 x² + bx + c = 0，加上并减去 (b/2)² 来完成配方。
x² + bx = (x + b/2)² − (b/2)²
² − 9 − 7 = 0, so (x + 3)² = 16. Taking square roots gives x + 3 = ±4, so x = 1 or x = −7.
例：解 x² + 6x − 7 = 0。先写出 x² + 6x = (x + 3)² − 9。然后方程变为 (x + 3)² − 9 − 7 = 0，即 (x + 3)² = 16。开平方得 x + 3 = ±4，因此 x = 1 或 x = −7。
ut first: ax² + bx + c = a[x² + (b/a)x] + c, then complete the square inside the brackets.
如果 x² 的系数不是 1，先将其提出：ax² + bx + c = a[x² + (b/a)x] + c，然后在括号内配方。
4. Solving by the Quadratic Formula | 求根公式法
The quadratic formula works for every quadratic equation. It is derived from completing the square and is given by:
求根公式适用于所有二次方程。它由配方法推导得出，公式如下：
x = [−b ± √b² − 4ac] / (2a)
tute the values of a, b, and c into the formula. Simplify carefully, especially the square root and the denominator.
对于方程 ax² + bx + c = 0，将 a、b、c 的值代入公式。小心化简，特别是根号和分母。
. Here a = 2, b = −4, c = −3. Then b² − 4ac = (−4)² − 4(2)(−3) = 16 + 24 = 40. So x = [4 ± √40] / 4 = [4 ± 2√10] / 4 = 1 ± √10/2.
例：解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。则 b² − 4ac = (−4)² − 4(2)(−3) = 16 + 24 = 40。所以 x = [4 ± √40] / 4 = [4 ± 2√10] / 4 = 1 ± √10/2。
Use the formula when the equation cannot be factorised easily or when the roots are required to a certain degree of accuracy.
当方程不易因式分解或要求根保留一定精度时，使用求根公式。
5. The Discriminant and the Nature of Roots | 判别式与根的性质
t, denoted by Δ, is the part of the quadratic formula under the square root: Δ = b² − 4ac.
判别式用 Δ 表示，是求根公式中根号内的部分：Δ = b² − 4ac。
It tells us the number and type of roots without solving the equation:
它无需解方程即可告诉我们根的个数和类型：
- Δ > 0: two distinct real roots
- Δ > 0：两个不同的实数根
- Δ = 0: one repeated real root (two equal roots)
- Δ = 0：一个重根（两个相等的实数根）
- Δ < 0: no real roots (two complex roots)
- Δ < 0：没有实数根（两个复数根）
= 0, Δ = (−6)² − 4(3)(2) = 36 − 24 = 12 > 0, so the equation has two distinct real roots.
例：对于 3x² − 6x + 2 = 0，Δ = (−6)² − 4(3)(2) = 36 − 24 = 12 > 0，所以方程有两个不同的实数根。
If Δ is a perfect square, the roots are rational; if not, the roots are irrational.
如果 Δ 是完全平方数，则根为有理数；否则根为无理数。
6. Sum and Product of Roots | 根的和与积
quadratic equation ax² + bx + c = 0 with roots α and β, the following relationships hold:
对于二次方程 ax² + bx + c = 0，若其根为 α 和 β，则有以下关系：
α + β = −b/a and αβ = c/a
α + β = −b/a 且 αβ = c/a
These formulas allow you to find the sum and product of the roots without solving the equation. They are useful for checking answers and for constructing new equations.
这些公式允许你不解方程就求出根的和与积。它们可用于检查答案以及构造新的方程。
Example: For 2x² + 8x − 5 = 0, the sum of roots = −8/2 = −4, and the product = −5/2.
例：对于 2x² + 8x − 5 = 0，根的和 = −8/2 = −4，根的积 = −5/2。
You can also write a quadratic with given roots α and β as x² − (α+β)x + αβ = 0.
你也可以根据给定的根 α 和 β 写出二次方程：x² − (α+β)x + αβ = 0。
7. Quadratic Graphs and the Roots | 二次函数图像与根
re the parabola crosses the x-axis are exactly the roots of the equation ax² + bx + c = 0.
二次函数 y = ax² + bx + c 的图像是一条抛物线。抛物线与 x 轴交点的横坐标正好是方程 ax² + bx + c = 0 的根。
If Δ = 0, it touches the x-axis at one point. If Δ < 0, it does not intersect the x-axis.
若 Δ > 0，抛物线与 x 轴相交于两个不同点。若 Δ = 0，它与 x 轴相切于一点。若 Δ < 0，它不与 x 轴相交。
of symmetry is x = −b/(2a), which is also the x-coordinate of the vertex of the parabola.
对称轴是 x = −b/(2a)，这也是抛物线顶点的 x 坐标。
Using the graph, you can approximate roots by reading the x-intercepts. This is a common skill tested in IGCSE papers.
利用图像，你可以通过读取 x 截距来近似求根。这是 IGCSE 考试中常见的技能。
8. Solving Quadratic Inequalities | 解二次不等式
or ax² + bx + c ≤ 0. The process uses the roots to divide the number line into intervals.
二次不等式涉及形如 ax² + bx + c > 0 或 ax² + bx + c ≤ 0 的表达式。其解法利用根将数轴分成区间。
First solve the corresponding quadratic equation to find the critical points, then test a value in each interval to determine the sign.
首先解对应的二次方程以找到临界点，然后在每个区间内取一个值测试符号。
: x < −2 gives positive; −2 < x < 3 gives negative; x > 3 gives positive. So the solution is x < −2 or x > 3.
例：解 x² − x − 6 > 0。因式分解为 (x − 3)(x + 2) > 0。临界点为 x = −2 和 x = 3。测试区间：x < −2 为正；−2 < x < 3 为负；x > 3 为正。所以解为 x < −2 或 x > 3。
Remember to flip the inequality sign when multiplying or dividing by a negative number, though this is rare in quadratic inequalities.
注意：当乘以或除以负数时，需要改变不等号方向，不过在二次不等式中这很少发生。
9. Equations Reducible to Quadratics | 可化为二次方程的方程
Some equations are not quadratic at first glance but can be transformed into quadratics using substitution.
有些方程乍一看不是二次方程，但可以通过换元转化为二次方程。
0, which factorises as (u − 1)(u − 4) = 0, giving u = 1 or 4. Since u = x², x = ±1 or ±2.
例如，x⁴ − 5x² + 4 = 0 可通过令 u = x² 求解。原方程变为 u² − 5u + 4 = 0，因式分解为 (u − 1)(u − 4) = 0，得到 u = 1 或 4。由于 u = x²，所以 x = ±1 或 ±2。
Other examples include equations with fractional powers, such as x + √x − 6 = 0, where you let t = √x.
其他例子包括含分数次幂的方程，如 x + √x − 6 = 0，此时可令 t = √x。
Always check whether the solutions obtained from the substituted variable are valid in the original equation, especially for square roots.
务必检查从换元变量得到的解在原方程中是否有效，尤其是涉及平方根时。
10. Solving Quadratic Equations by Graphical Methods | 图解法解二次方程
ical method involves plotting the curve y = ax² + bx + c and reading off the x-intercepts.
图解法的步骤是绘制曲线 y = ax² + bx + c，然后读出 x 截距。
k by drawing the horizontal line y = k and finding the x-coordinates of the intersections.
在考试中，你可能会得到一张数值表，并需要完成图像。然后用图像求解形如 ax² + bx + c = k 的方程，方法是画水平直线 y = k，找出交点横坐标。
aph of y = x² − 2x − 3 and read the x-values where the horizontal line meets the parabola.
例如，要解 x² − 2x − 3 = 1，在 y = x² − 2x − 3 的图像上画 y = 1，读出水平直线与抛物线交点的 x 值。
This method gives approximate solutions, usually correct to one decimal place depending on the scale.
此方法给出近似解，通常根据比例尺精确到一位小数。
11. Common Exam Mistakes | 常见考试错误
Many students lose marks in quadratic equations due to avoidable errors. Here are some common pitfalls:
许多学生因可避免的错误而在二次方程上丢分。以下是一些常见陷阱：
- Forgetting to rearrange the equation to standard form before applying the formula.
- 在应用公式前忘记将方程整理成标准形式。
- Misreading the signs of b or c when substituting into the formula.
- 代入公式时看错 b 或 c 的符号。
- Making arithmetic errors when simplifying the square root.
- 化简根号时犯算术错误。
- Forgetting that x² = k has two solutions: x = +√k and x = −√k.
- 忘记 x² = k 有两个解：x = +√k 和 x = −√k。
- en solving an equation like (x − 3)(x + 4) = 0, writing x = 3 and x = 4 instead of x = −4.
- 解形如 (x − 3)(x + 4) = 0 的方程时，写出 x = 3 和 x = 4，而不是 x = −4。
Always check your answers by substituting them back into the original equation.
始终将答案代回原方程进行检验。
12. Practice Questions | 练习题
Try these questions to test your understanding:
尝试以下问题来测试你的理解：
- Solve by factorisation: x² − 7x + 12 = 0
- 用因式分解法解：x² − 7x + 12 = 0
- Solve by completing the square: x² + 4x − 1 = 0
- 用配方法解：x² + 4x − 1 = 0
- Use the quadratic formula to solve: 3x² − 5x − 2 = 0
- 用求根公式解：3x² − 5x − 2 = 0
- Find the discriminant of 2x² + 4x + 5 = 0 and state the nature of the roots.
- 求 2x² + 4x + 5 = 0 的判别式，并说明根的性质。
- If α and β are the roots of x² − 6x + 8 = 0, find α + β and αβ.
- 若 α 和 β 是 x² − 6x + 8 = 0 的根，求 α + β 和 αβ。
- Solve the inequality: x² − 4x − 5 < 0
- 解不等式：x² − 4x − 5 < 0
Solutions: 1) x = 3 or 4. 2) x = −2 ± √5. 3) x = 2 or −1/3. 4) Δ = −24 < 0, no real roots. 5) α+β = 6, αβ = 8. 6) −1 < x < 5.
答案：1) x = 3 或 4。2) x = −2 ± √5。3) x = 2 或 −1/3。4) Δ = −24 < 0，无实数根。5) α+β = 6，αβ = 8。6) −1 < x < 5。
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