Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

A quadratic equation is a polynomial equation of degree 2, expressed in the general form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. Mastering the techniques for solving quadratic equations is one of the most important skills in IGCSE Mathematics, as these questions appear regularly in both Paper 1 and Paper 4.

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，其中 a、b、c 是常数且 a ≠ 0。掌握二次方程的解法是 IGCSE 数学中最关键的技能之一，试卷 1 和试卷 4 中都会经常出现这类题目。

1. The Standard Form | 二次方程的标准形式

Before solving any quadratic equation, you must ensure it is written in standard form: ax² + bx + c = 0. In this form, a is the coefficient of x², b is the coefficient of x, and c is the constant term. For example, in the equation 3x² – 2x + 5 = 0, we have a = 3, b = -2 and c = 5.

在解任何二次方程之前，必须确保它已写成标准形式：ax² + bx + c = 0。在此形式中，a 是 x² 的系数，b 是 x 的系数，c 是常数项。例如，在方程 3x² – 2x + 5 = 0 中，a = 3，b = -2，c = 5。

Sometimes the equation is not given in standard form. For instance, x(2x – 3) = 5 must first be expanded to 2x² – 3x = 5, then rearranged to 2x² – 3x – 5 = 0 before solving.

有时方程并非以标准形式给出。例如，x(2x – 3) = 5 需要先展开为 2x² – 3x = 5，再移项整理为 2x² – 3x – 5 = 0 才能求解。

General form: ax² + bx + c = 0 (a ≠ 0)

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic expression can be written as the product of two linear factors. The principle is: if p × q = 0, then either p = 0 or q = 0. This is known as the zero product property.

当二次表达式可以写成两个一次因式的乘积时，因式分解法通常是最快的求解方法。其原理是：如果 p × q = 0，那么 p = 0 或 q = 0。这称为零积性质。

Consider the equation x² + 5x + 6 = 0. We look for two numbers that multiply to give 6 (the constant term) and add to give 5 (the coefficient of x). The numbers 2 and 3 work, so we write (x + 2)(x + 3) = 0. Therefore, x + 2 = 0 or x + 3 = 0, giving x = -2 or x = -3.

考虑方程 x² + 5x + 6 = 0。我们需要找两个数，它们相乘等于 6（常数项），相加等于 5（x 的系数）。数字 2 和 3 满足条件，因此写成 (x + 2)(x + 3) = 0。所以 x + 2 = 0 或 x + 3 = 0，解得 x = -2 或 x = -3。

When the coefficient a is greater than 1, factorisation requires more care. For example, with 2x² – 5x + 2 = 0, the factors are (2x – 1)(x – 2) = 0, giving x = ½ or x = 2.

当系数 a 大于 1 时，因式分解需要更加细心。例如，对于 2x² – 5x + 2 = 0，其因式为 (2x – 1)(x – 2) = 0，解得 x = ½ 或 x = 2。

Special case — difference of two squares: a² – b² = (a – b)(a + b). For example, x² – 9 = 0 becomes (x – 3)(x + 3) = 0, so x = 3 or x = -3.

特殊情形——平方差公式：a² – b² = (a – b)(a + b)。例如，x² – 9 = 0 可化为 (x – 3)(x + 3) = 0，因此 x = 3 或 x = -3。

If (x – m)(x – n) = 0, then x = m or x = n

3. Solving by the Quadratic Formula | 公式法

When factorisation is difficult or impossible, the quadratic formula provides a universal solution. For any quadratic equation ax² + bx + c = 0, the solutions are given by the formula below. This formula can be applied to every quadratic equation, whether the roots are rational, irrational or even non-real.

当因式分解困难甚至无法进行时，求根公式提供了一种通用的解法。对于任意二次方程 ax² + bx + c = 0，其解由下面的公式给出。该公式适用于所有二次方程，无论根是有理数、无理数还是非实数。

x = (-b ± √(b² – 4ac)) / 2a

Let us solve 2x² + 3x – 5 = 0 using the formula. Here a = 2, b = 3 and c = -5. Substituting into the formula gives x = (-3 ± √(3² – 4 × 2 × (-5))) / (2 × 2) = (-3 ± √(9 + 40)) / 4 = (-3 ± √49)/4 = (-3 ± 7)/4. This yields x = 1 or x = -5/2, which is -2.5.

我们用公式法求解 2x² + 3x – 5 = 0。这里 a = 2，b = 3，c = -5。代入公式得 x = (-3 ± √(3² – 4 × 2 × (-5))) / (2 × 2) = (-3 ± √(9 + 40)) / 4 = (-3 ± √49)/4 = (-3 ± 7)/4。由此得到 x = 1 或 x = -5/2，即 -2.5。

Always write down the values of a, b and c before substituting, and be extremely careful with negative signs. The ± symbol means there are two possible solutions: one using the plus sign and one using the minus sign. Using a calculator for the arithmetic is allowed on Paper 4, but you must show the substitution clearly.

代入前务必先写出 a、b、c 的值，并且要极其小心负号。± 符号表示有两个可能的解：一个取加号，一个取减号。在试卷 4 中可以使用计算器进行运算，但必须清楚写出代入过程。

4. Solving by Completing the Square | 配方法

Completing the square rewrites the quadratic expression as a perfect square plus a constant. The process starts from x² + bx and adds (b/2)² to create (x + b/2)². This method is particularly useful for finding the turning point of a parabola and solving equations without factorisation.

配方法将二次表达式改写成一个完全平方式加上一个常数的形式。该过程从 x² + bx 开始，加上 (b/2)² 构成 (x + b/2)²。这种方法在求抛物线顶点坐标以及不解因式直接求解时特别有用。

Consider x² + 6x – 7 = 0. The coefficient of x is 6, so half of it is 3, and 3² = 9. We write x² + 6x – 7 as (x + 3)² – 9 – 7 = 0, which simplifies to (x + 3)² – 16 = 0. Then (x + 3)² = 16, so x + 3 = ±4, giving x = 1 or x = -7.

考虑 x² + 6x – 7 = 0。x 的系数是 6，取其一半为 3，且 3² = 9。我们将 x² + 6x – 7 改写为 (x + 3)² – 9 – 7 = 0，化简得 (x + 3)² – 16 = 0。于是 (x + 3)² = 16，所以 x + 3 = ±4，解得 x = 1 或 x = -7。

When the coefficient of x² is not 1, first factor it out from the first two terms. For example, 2x² – 4x + 1 = 0 becomes 2[x² – 2x] + 1 = 0, then 2[(x – 1)² – 1] + 1 = 0, giving 2(x – 1)² – 2 + 1 = 0, so 2(x – 1)² – 1 = 0.

当 x² 的系数不为 1 时，先从前面两项中提出该系数。例如，2x² – 4x + 1 = 0 可写成 2[x² – 2x] + 1 = 0，进而化为 2[(x – 1)² – 1] + 1 = 0，得到 2(x – 1)² – 2 + 1 = 0，即 2(x – 1)² – 1 = 0。

x² + bx = (x + b/2)² – (b/2)²

5. The Discriminant | 判别式

The discriminant is the expression b² – 4ac inside the square root of the quadratic formula. It tells us the nature of the roots without actually solving the equation. The value of the discriminant determines how many real solutions a quadratic equation has.

判别式是求根公式中根号内的表达式 b² – 4ac。它可以在不解方程的情况下判断根的性质。判别式的值决定了二次方程有多少个实数解。

If b² – 4ac > 0, the equation has two distinct real roots. This means the graph of the quadratic crosses the x-axis at two different points. If b² – 4ac = 0, there is exactly one repeated root, and the graph touches the x-axis at one point, known as the vertex. If b² – 4ac < 0, there are no real roots, meaning the graph does not cross the x-axis at all.

如果 b² – 4ac > 0，方程有两个不相等的实数根。这意味着二次函数图象与 x 轴相交于两个不同的点。如果 b² – 4ac = 0，方程恰好有一个重根，图象与 x 轴相切于一点，即顶点处。如果 b² – 4ac < 0，方程没有实数根，图象完全不与 x 轴相交。

Value of b² – 4ac	Nature of roots
Positive (greater than 0)	Two distinct real roots
Zero	One repeated real root
Negative (less than 0)	No real roots

For example, the equation x² – 4x + 3 = 0 has discriminant (-4)² – 4 × 1 × 3 = 16 – 12 = 4 > 0, so it has two distinct real roots. The equation x² + 2x + 5 = 0 has discriminant 2² – 4 × 1 × 5 = 4 – 20 = -16 < 0, so it has no real roots.

例如，方程 x² – 4x + 3 = 0 的判别式为 (-4)² – 4 × 1 × 3 = 16 – 12 = 4 > 0，因此它有两个不相等的实数根。方程 x² + 2x + 5 = 0 的判别式为 2² – 4 × 1 × 5 = 4 – 20 = -16 < 0，因此它没有实数根。

6. Quadratic Graphs and Their Roots | 二次函数图象与根

The graph of a quadratic function y = ax² + bx + c is a smooth curve called a parabola. When a > 0, the parabola opens upward, and when a < 0, it opens downward. The x-coordinates of the points where the curve crosses the x-axis are exactly the roots of the equation ax² + bx + c = 0.

二次函数 y = ax² + bx + c 的图象是一条平滑曲线，称为抛物线。当 a > 0 时，抛物线开口向上；当 a < 0 时，抛物线开口向下。曲线与 x 轴交点的横坐标正是方程 ax² + bx + c = 0 的根。

The axis of symmetry of the parabola is the vertical line x = -b/(2a), and the turning point (vertex) lies on this axis. The y-coordinate of the vertex can be found by substituting x = -b/(2a) into the function. This connects nicely to completing the square, since the completed-square form y = a(x – h)² + k directly reveals the vertex at (h, k).

抛物线的对称轴是竖直线 x = -b/(2a)，顶点位于这条对称轴上。将 x = -b/(2a) 代入函数即可求出顶点的 y 坐标。这与配方法紧密相连，因为配方形式 y = a(x – h)² + k 直接给出顶点坐标 (h, k)。

For instance, the function y = x² – 2x – 3 factorises to y = (x – 3)(x + 1), so the roots are x = 3 and x = -1. The axis of symmetry is x = (3 + (-1))/2 = 1, and the vertex is at (1, -4).

例如，函数 y = x² – 2x – 3 可因式分解为 y = (x – 3)(x + 1)，因此根为 x = 3 和 x = -1。对称轴为 x = (3 + (-1))/2 = 1，顶点坐标为 (1, -4)。

7. Word Problems with Quadratic Equations | 二次方程应用题

Quadratic equations frequently arise in real-world contexts, such as calculating areas, projectile motion, and number problems. The key steps are to define a variable, translate the problem into a quadratic equation, solve it, and then interpret the answers in the context of the original question.

二次方程经常出现在实际应用场景中，如面积计算、抛体运动和数字问题。关键步骤是：设定变量、将问题转化为二次方程、求解方程，然后结合原问题的实际意义解释答案。

Example: The area of a rectangle is 40 cm², and its length is 3 cm longer than its width. Let the width be x cm. Then the length is (x + 3) cm, so x(x + 3) = 40, which expands to x² + 3x – 40 = 0. Factorising gives (x + 8)(x – 5) = 0, so x = -8 or x = 5. Since a width cannot be negative, the width is 5 cm and the length is 8 cm.

例：一个矩形的面积为 40 cm²，其长比宽长 3 cm。设宽为 x cm，则长为 (x + 3) cm，所以 x(x + 3) = 40，展开得 x² + 3x – 40 = 0。因式分解得 (x + 8)(x – 5) = 0，所以 x = -8 或 x = 5。由于宽度不能为负数，因此宽为 5 cm，长为 8 cm。

Always check whether your solutions make sense in the given context. A negative length, time or distance is usually invalid, so such solutions should be rejected even though they satisfy the algebraic equation.

务必检查解在具体情境下是否合理。负数长度、时间或距离通常都是无效的，即使它们在代数上满足方程，也应予以舍弃。

8. Common Mistakes to Avoid | 常见易错点

One frequent error is attempting to solve a quadratic equation before rearranging it into standard form. For example, x² = 3x must be rewritten as x² – 3x = 0 before factorising; dividing both sides by x would lose the root x = 0. Never divide by a variable that could be zero.

一个常见错误是在将方程化为标准形式之前就急于求解。例如，x² = 3x 必须先改写为 x² – 3x = 0 再进行因式分解；如果两边同时除以 x，就会丢失 x = 0 这个根。绝不能除以一个可能为零的变量。

Another common mistake is sign errors when substituting negative values into the quadratic formula. Use brackets around negative numbers, as in b = -3, so write (-3)² rather than -3², which would incorrectly give -9. Also, do not forget that √(b² – 4ac) produces both a positive and a negative result — omitting the ± sign leaves you with only half the solution.

另一个常见错误是在将负数代入求根公式时出现符号错误。用括号把负数括起来，例如 b = -3 时，应写成 (-3)² 而不是 -3²，后者会错误地得到 -9。同时，不要忘记 √(b² – 4ac) 会产生正负两个结果——漏掉 ± 符号只会得到一半的解。

A third common error is confusing the roots with the turning point. The roots are where the graph crosses the x-axis, while the vertex is the maximum or minimum point. They are related but distinct concepts, and questions often ask for both separately.

第三个常见错误是混淆根与顶点。根是图象与 x 轴的交点，而顶点是图象的最大值或最小值点。它们虽然相关但概念不同，考试中常分别设问。

9. Exam Practice and Strategy | 备考建议与应试策略

When facing a quadratic equation in the exam, first check if it is already in standard form. Then choose the most efficient method: factorise if possible; otherwise use the quadratic formula. Completing the square is especially useful when the question asks for the vertices or when a is small.

考试中遇到二次方程时，先检查它是否已经是标准形式。然后选择最高效的方法：能因式分解就分解；否则使用求根公式。当题目要求顶点坐标或 a 值较小时，配方法尤为适用。

Always verify your answer by substituting each root back into the original equation. For example, if you solve x² – 5x + 6 = 0 and obtain x = 2 and x = 3, check that 2² – 5(2) + 6 = 4 – 10 + 6 = 0 and 3² – 5(3) + 6 = 9 – 15 + 6 = 0. This simple check can catch many careless errors.

务必通过代入检验来验证答案。例如，解 x² – 5x + 6 = 0 得到 x = 2 和 x = 3 后，检查 2² – 5(2) + 6 = 4 – 10 + 6 = 0 以及 3² – 5(3) + 6 = 9 – 15 + 6 = 0。这个简单检验能发现许多粗心错误。

Practice mixed problem sets that require you to choose the appropriate method. On Paper 2 (non-calculator), factorisation is often the intended method, while on Paper 4, the quadratic formula may be more convenient. Show all steps clearly to earn full method marks even if your final arithmetic is wrong.

练习混合题型，训练自己选择合适方法的能力。在试卷 2（不允许使用计算器）中，因式分解常常是预期方法；而在试卷 4 中，求根公式可能更方便。清楚地写出每一步骤，即使最终运算出错，也能获得部分方法分。

Quadratic formula: x = (-b ± √(b² – 4ac)) / 2a

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Solving Quadratic Equations | 解二次方程