Solving Quadratic Equations | 解二次方程指南

📚 Solving Quadratic Equations | 解二次方程指南

A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in exams, in real-life problem solving, and in higher-level study. This guide will take you through every method you need, from recognition to factorisation, completing the square, and the quadratic formula.

二次方程是IGCSE数学中最重要的话题之一。它出现在考试、实际生活问题以及更高级的学习中。本指南将带你掌握从识别到因式分解、配方法以及二次公式的每一种必备方法。

1. What is a Quadratic Equation | 什么是二次方程

A quadratic equation is an equation where the highest power of the variable is 2. The general form is usually written as ax² + bx + c = 0, where a ≠ 0. For example, x² – 5x + 6 = 0 is a quadratic equation.

二次方程是变量最高次数为2的方程。一般形式通常写成 ax² + bx + c = 0，其中 a ≠ 0。例如，x² – 5x + 6 = 0 就是一个二次方程。

If a = 0, then the equation becomes linear, not quadratic, so the condition a ≠ 0 is essential.

如果 a = 0，那么方程就变成一次方程，而不是二次方程，所以 a ≠ 0 这个条件至关重要。

Quadratic equations can have two real roots, one repeated real root, or no real roots, depending on the discriminant.

根据判别式的不同，二次方程可以有两个实数根、一个重实数根，或者没有实数根。

2. Standard Form | 标准形式

Before solving, a quadratic equation must be written in standard form: ax² + bx + c = 0. All terms are on one side, and the other side is zero.

在求解之前，二次方程必须写成标准形式：ax² + bx + c = 0。所有项都在等式一边，另一边为零。

For example, x² = 3x – 2 must be rearranged to x² – 3x + 2 = 0. Only then can we apply factorisation or the quadratic formula.

例如，x² = 3x – 2 必须重新整理为 x² – 3x + 2 = 0。只有这样才能应用因式分解或二次公式。

Remember to collect like terms and simplify carefully. A common error is forgetting to move all terms to one side.

记住要合并同类项并仔细化简。一个常见错误是忘记把所有项移到一边。

3. Factorising Quadratics (a = 1) | 二次因式分解（a = 1）

When the coefficient of x² is 1, we look for two numbers that multiply to give c and add to give b. For x² + 7x + 12 = 0, the numbers are 3 and 4 because 3 × 4 = 12 and 3 + 4 = 7.

当x²的系数为1时，我们寻找两个数，它们相乘得到 c，相加得到 b。对于 x² + 7x + 12 = 0，这两个数是3和4，因为3 × 4 = 12且3 + 4 = 7。

So the factorised form is (x + 3)(x + 4) = 0. Always check by expanding: (x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12.

因此因式分解的形式是 (x + 3)(x + 4) = 0。总是通过展开来验证：(x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12。

If c is negative, one number is positive and the other is negative. For x² – x – 20 = 0, the numbers are -5 and 4 because -5 × 4 = -20 and -5 + 4 = -1.

如果c是负数，那么一个数是正数，另一个是负数。对于 x² – x – 20 = 0，这两个数是-5和4，因为-5 × 4 = -20且-5 + 4 = -1。

4. Solving by Factorisation | 用因式分解求解

Once the quadratic is factorised into two brackets, we use the zero product property: if (px + q)(rx + s) = 0, then either px + q = 0 or rx + s = 0.

一旦二次方程被分解为两个括号相乘的形式，我们使用零乘积性质：如果 (px + q)(rx + s) = 0，那么要么 px + q = 0，要么 rx + s = 0。

For example, solve x² – 5x + 6 = 0. Factorise to (x – 2)(x – 3) = 0. Then x – 2 = 0 gives x = 2, and x – 3 = 0 gives x = 3.

例如，解 x² – 5x + 6 = 0。因式分解得 (x – 2)(x – 3) = 0。于是 x – 2 = 0 得 x = 2，x – 3 = 0 得 x = 3。

Always state both solutions clearly. In exams, you may lose marks if you only write x = 2 and omit x = 3.

始终清楚地写出两个解。在考试中，如果你只写x = 2而漏掉x = 3，可能会失分。

Check your answers by substituting back into the original equation. For x = 2: 4 – 10 + 6 = 0. For x = 3: 9 – 15 + 6 = 0. Both work.

将答案代回原方程进行检验。对于x = 2：4 – 10 + 6 = 0。对于x = 3：9 – 15 + 6 = 0。都成立。

5. Difference of Two Squares | 平方差

A special form of quadratic is the difference of two squares: x² – a² = (x – a)(x + a). There is no x term.

二次方程的一种特殊形式是平方差：x² – a² = (x – a)(x + a)。它没有一次项。

For example, x² – 25 = 0 can be written as (x – 5)(x + 5) = 0. Therefore x = 5 or x = -5.

例如，x² – 25 = 0 可以写成 (x – 5)(x + 5) = 0。因此 x = 5 或 x = -5。

This also works for expressions like 4x² – 9 = (2x – 3)(2x + 3). Notice that 4x² = (2x)² and 9 = 3².

这种方法也适用于类似4x² – 9 = (2x – 3)(2x + 3)的表达式。注意4x² = (2x)²且9 = 3²。

6. Completing the Square | 配方法

Completing the square writes a quadratic in the form (x + p)² + q. This is useful for solving equations and finding the vertex of a parabola.

配方法将二次方程写成 (x + p)² + q 的形式。这有助于解方程和求抛物线的顶点。

For x² + bx, take half of b and square it. For example, x² + 6x becomes (x + 3)² – 9. Because (x + 3)² = x² + 6x + 9, we subtract 9.

对于 x² + bx，取b的一半并平方。例如，x² + 6x 变为 (x + 3)² – 9。因为 (x + 3)² = x² + 6x + 9，所以减去9。

To solve x² + 6x – 7 = 0, rewrite as (x + 3)² – 9 – 7 = 0, so (x + 3)² = 16. Then x + 3 = ±4, so x = 1 or x = -7.

要解 x² + 6x – 7 = 0，将其改写为 (x + 3)² – 9 – 7 = 0，因此 (x + 3)² = 16。于是 x + 3 = ±4，所以 x = 1 或 x = -7。

This method works for all quadratics, even when factorisation is difficult.

这种方法适用于所有二次方程，即使因式分解很困难。

7. The Quadratic Formula | 二次公式

The quadratic formula solves any quadratic equation ax² + bx + c = 0. The formula is:

二次公式可以求解任何二次方程 ax² + bx + c = 0。公式为：

x = (−b ± √(b² − 4ac)) ÷ (2a)

You do not need to factorise. Just identify a, b, c and substitute.

你不需要因式分解。只需识别a、b、c并代入。

For example, solve 2x² – 4x – 3 = 0. Here a = 2, b = -4, c = -3.

例如，解 2x² – 4x – 3 = 0。这里 a = 2，b = -4，c = -3。

Substitute: x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4. This simplifies to x = 1 ± (√10)/2.

代入：x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4。化简得 x = 1 ± (√10)/2。

Always write the two solutions separately if required.

如果要求分开写，请分别写出两个解。

8. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: Δ = b² – 4ac. It tells us how many real roots the equation has.

判别式是二次公式中根号下的部分：Δ = b² – 4ac。它告诉我们方程有多少个实数根。

If Δ > 0, there are two distinct real roots.
如果 Δ > 0，方程有两个不同的实数根。
If Δ = 0, there is one repeated real root.
如果 Δ = 0，方程有一个重实数根。
If Δ < 0, there are no real roots.
如果 Δ < 0，方程没有实数根。

For example, for x² – 4x + 4 = 0, Δ = 16 – 16 = 0, so it has one repeated root x = 2.

例如，对于 x² – 4x + 4 = 0，Δ = 16 – 16 = 0，所以它有一个重根 x = 2。

The discriminant also helps in exam questions where you are asked to find the range of k for which the equation has real roots.

判别式也有助于解决考试中要求找出使方程有实数根的k的取值范围的问题。

9. Word Problems | 应用题

Many exam problems require translating words into a quadratic equation. Look for keywords like ‘product’, ‘area’, ‘square of a number’, or ‘consecutive integers’.

许多考试问题需要将文字转化为二次方程。注意关键词，如“乘积”、“面积”、“一个数的平方”或“连续整数”。

Example: The product of two consecutive positive integers is 42. Let the smaller integer be x. Then x(x + 1) = 42, so x² + x – 42 = 0.

例如：两个连续正整数的乘积是42。设较小整数为x。则 x(x + 1) = 42，即 x² + x – 42 = 0。

Factorise to (x + 7)(x – 6) = 0. Since x is positive, x = 6. The two integers are 6 and 7.

因式分解得 (x + 7)(x – 6) = 0。因为x为正数，所以 x = 6。这两个整数是6和7。

Always check that your answer fits the original problem. Negative roots may be rejected if the problem asks for a positive value.

始终检查答案是否符合题意。如果问题要求正数，负根可能需要舍去。

10. Common Mistakes | 常见错误

Mistake 1: Not writing the equation in standard form before solving. Always rearrange to ax² + bx + c = 0 first.

错误1：在求解前没有将方程写成标准形式。务必先整理为 ax² + bx + c = 0。

Mistake 2: Incorrectly factoring a quadratic with a ≠ 1. For example, 2x² + 5x + 2 ≠ (2x + 1)(x + 2) is actually correct, but many students misapply the signs. Check by expanding.

错误2：当a ≠ 1时错误地进行因式分解。例如，2x² + 5x + 2 = (2x + 1)(x + 2) 实际上是正确的，但许多学生会弄错符号。用展开来检验。

Mistake 3: Forgetting the ± sign when using the quadratic formula or square roots.

错误3：在使用二次公式或平方根时忘记±号。

Mistake 4: Dividing both sides by x when x could be zero. For example, x² = 3x cannot be solved by dividing by x, because you lose the root x = 0. Instead, rearrange to x(x – 3) = 0.

错误4：两边同时除以x，而x可能为零。例如，x² = 3x 不能通过除以x来求解，因为会失去根x = 0。应该整理为 x(x – 3) = 0。

Mistake 5: Misreading the discriminant. Remember b² – 4ac uses the coefficient of x, including its sign.

错误5：误读判别式。记住 b² – 4ac 使用的是x的系数，包括其符号。

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Solving Quadratic Equations | 解二次方程指南