📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, geometry, and even real‑life problem solving. Understanding how to solve them efficiently is essential for exam success.
二次方程是 IGCSE 数学中最重要的话题之一。它出现在代数、图像、几何，甚至现实生活问题的解决中。高效掌握它们的解法对考试成功至关重要。
1. What is a Quadratic Equation? | 什么是二次方程
A quadratic equation is a polynomial equation of degree 2. Its general form is:
二次方程是次数为 2 的多项式方程，其一般形式为：
ax² + bx + c = 0, (a ≠ 0)
The coefficient a must be non‑zero, otherwise the equation becomes linear.
系数 a 不能为零，否则方程就变成一次方程。
b and c can be any real numbers; if c = 0, the equation has no constant term, but it is still quadratic.
b 和 c 可为任意实数；若 c = 0，方程没有常数项，但依旧是二次方程。
2. Solving by Factorization | 因式分解法
Factorization is often the quickest method when the quadratic has simple integer roots. The idea is to rewrite the quadratic as a product of two linear factors, then set each factor equal to zero.
当二次方程具有简单的整数根时，因式分解通常是最快的方法。其思路是将二次式改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.
例：解 x² − 5x + 6 = 0。
(x − 2)(x − 3) = 0
Set each factor to zero: x − 2 = 0 ⇒ x = 2, and x − 3 = 0 ⇒ x = 3.
令每个因式为零：x − 2 = 0 ⇒ x = 2，及 x − 3 = 0 ⇒ x = 3。
So the solution set is {2, 3}.
因此解集为 {2, 3}。
When the coefficient a is not 1, use the “grouping” method or trial and error.
当系数 a 不等于 1 时，可使用“分组法”或尝试法。
3. Solving by Completing the Square | 配方法
Completing the square transforms the quadratic into a perfect square plus a constant. This method is useful when the equation cannot be factorized easily and gives a clear view of the vertex of the parabola.
配方法将二次式转化为一个完全平方加上一个常数。当方程不易因式分解时，这个方法很有用，并能清晰地看出抛物线的顶点。
General steps for x² + bx + c = 0:
对于 x² + bx + c = 0 的一般步骤：
Move c to the other side: x² + bx = −c.
把 c 移到另一边：x² + bx = −c。
Add (b/2)² to both sides: x² + bx + (b/2)² = (b/2)² − c.
两边加上 (b/2)²：x² + bx + (b/2)² = (b/2)² − c。
Write the left side as a square: (x + b/2)² = (b/2)² − c.
左边写成完全平方：(x + b/2)² = (b/2)² − c。
Example: Solve x² + 6x − 7 = 0 by completing the square.
例：用配方法解 x² + 6x − 7 = 0。
(x + 3)² = 16 ⇒ x + 3 = ±4 ⇒ x = 1 or x = −7
4. Solving by the Quadratic Formula | 求根公式法
The quadratic formula works for all quadratic equations, including those that cannot be factorized. It is an essential tool to memorise.
求根公式适用于所有二次方程，包括那些不能因式分解的方程。它是必须记忆的重要工具。
x = (−b ± √b² − 4ac) / (2a)
The expression b² − 4ac is called the discriminant.
表达式 b² − 4ac 称为判别式。
Always check if the discriminant is positive, zero, or negative before calculating the roots.
计算根之前，先判断判别式是正、零还是负。
Example: Solve 2x² + 3x − 2 = 0 using the formula.
例：用公式法解 2x² + 3x − 2 = 0。
a = 2, b = 3, c = −2 ⇒ x = (−3 ± √9 + 16) / 4 = (−3 ± 5) / 4
We get x = 0.5 or x = −2.
得到 x = 0.5 或 x = −2。
5. Solving by Graphing | 图像法
Graphing a quadratic function y = ax² + bx + c gives a parabola. The x‑intercepts (where y = 0) are the solutions of the equation ax² + bx + c = 0.
绘制二次函数 y = ax² + bx + c 的图像得到一条抛物线。与 x 轴的交点（y = 0 处）就是方程 ax² + bx + c = 0 的解。
If the parabola crosses the x‑axis at two points, there are two real roots.
若抛物线与 x 轴有两个交点，则有两个实根。
If it touches at one point, there is one repeated root.
若只切于一点，则有一个重根。
If it does not intersect the x‑axis, there are no real roots.
若不与 x 轴相交，则没有实根。
Graphical solutions are approximate, but they help check the accuracy of algebraic methods.
图像解是近似的，但有助于检验代数方法的准确性。
6. The Discriminant | 判别式
inant, Δ = b² − 4ac, determines the nature of the roots without solving the full equation.
判别式 Δ = b² − 4ac 无需解完整方程即可判断根的性质。
| Value of Δ | Nature of Roots |
| Δ > 0 | Two distinct real roots (rational if Δ is a perfect square) |
| Δ = 0 | One repeated real root |
| Δ < 0 | No real roots (complex roots) |
This classification is frequently tested in the IGCSE non‑calculator paper.
这个分类在 IGCSE 非计算器试卷中经常出现。
7. Sum and Product of Roots | 根的和与积
For a quadratic equation ax² + bx + c = 0, with roots α and β:
对于二次方程 ax² + bx + c = 0，设根为 α 和 β：
α + β = −b/a, αβ = c/a
These relationships allow you to find new equations whose roots are functions of α and β.
这些关系让你能够构造以 α 和 β 的某种函数为根的新方程。
For example, if roots are 2 and 5, then the equation is x² − 7x + 10 = 0.
例如，若根为 2 和 5，则方程为 x² − 7x + 10 = 0。
8. Solving Quadratic Inequalities | 解二次不等式
Quadratic inequalities are solved by first finding the roots of the corresponding equation, then testing intervals on a number line.
解二次不等式时，先求出对应方程的根，然后在数轴上测试区间。
Example: Solve x² − 4x < 0.
例：解 x² − 4x < 0。
First, solve x² − 4x = 0 ⇒ x(x − 4) = 0 ⇒ x = 0 or x = 4.
先解 x² − 4x = 0 ⇒ x(x − 4) = 0 ⇒ x = 0 或 x = 4。
Test intervals: for 0 < x < 4, the expression is negative; therefore the solution is 0 < x < 4.
测试区间：当 0 < x < 4 时，表达式为负；因此解为 0 < x < 4。
9. Word Problems | 应用题
Many real‑world situations produce quadratic equations. Typical problems involve areas, projectile motion, and consecutive numbers.
许多现实情境会产生二次方程。典型问题涉及面积、抛体运动和连续整数。
Example: The length of a rectangle is 3 cm more than its width, and its area is 40 cm². Find the dimensions.
例：长方形的长比宽多 3 cm，面积为 40 cm²。求尺寸。
Let width = x, length = x + 3 ⇒ x(x + 3) = 40 ⇒ x² + 3x − 40 = 0
Factorizing: (x + 8)(x − 5) = 0 ⇒ x = −8 (discard) or x = 5. So width = 5 cm and length = 8 cm.
因式分解：(x + 8)(x − 5) = 0 ⇒ x = −8（舍去）或 x = 5。因此宽为 5 cm，长为 8 cm。
10. Common Mistakes | 常见错误
Forgetting to set the equation equal to zero before factorizing or using the formula.
在因式分解或使用公式前忘记将方程化为等于零的形式。
Losing the negative sign when extracting the square root: √expression² always gives ± expression.
取平方根时丢掉负号：√表达式² 总是给出 ± 表达式。
Confusing the quadratic formula with linear equations; ensure you use the correct coefficients for a, b, c.
将求根公式与一次方程混淆；务必正确使用 a、b、c 的系数。
In inequalities, reversing the inequality sign when multiplying or dividing by a negative number.
在不等式中，当乘以或除以负数时忘记改变不等号方向。
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