T-2-1071: Quadratic Equations | 二次方程

📚 T-2-1071: Quadratic Equations | 二次方程

Quadratic equations are one of the most fundamental topics in IGCSE Mathematics. They appear in a wide range of exam questions, from simple factorisation to real-world problem solving. Mastering this topic builds a strong foundation for advanced algebra and calculus.

二次方程是 IGCSE 数学中最基础也最重要的主题之一。它们出现在大量考题中，从简单的因式分解到实际应用问题。掌握这个主题能够为后续的代数与微积分学习打下坚实基础。

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

A quadratic equation is an equation in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\) and \(c\) are constants and \(a \neq 0\). The highest power of the variable is 2.

二次方程是形如 \(ax^2 + bx + c = 0\) 的方程，其中 \(a\)、\(b\) 和 \(c\) 是常数，且 \(a \neq 0\)。变量的最高次数是 2。

ay² + by + c = 0

A typical example is \(2x^2 – 3x + 1 = 0\). This is a quadratic because the highest exponent of \(x\) is 2.

一个典型的例子是 \(2x^2 – 3x + 1 = 0\)。这是一个二次方程，因为 \(x\) 的最高指数是 2。

2. Standard Form and Key Terms | 标准形式与关键术语

The standard form of a quadratic equation is written as \(ax^2 + bx + c = 0\). Here, \(a\) is called the leading coefficient, \(b\) is the linear coefficient, and \(c\) is the constant term.

二次方程的标准形式写为 \(ax^2 + bx + c = 0\)。其中，\(a\) 称为二次项系数，\(b\) 称为一次项系数，\(c\) 称为常数项。

If an equation is not in this form, you should rearrange it before solving. For example, \(2x^2 = 5x – 3\) becomes \(2x^2 – 5x + 3 = 0\).

如果方程不在标准形式，求解前必须先整理。例如，\(2x^2 = 5x – 3\) 可变为 \(2x^2 – 5x + 3 = 0\)。

Term	Meaning
Coefficient	A number multiplied by a variable.
Constant term	A value that does not contain a variable.

In the equation \(x^2 + 5x + 6 = 0\), \(a = 1\), \(b = 5\), and \(c = 6\).

在方程 \(x^2 + 5x + 6 = 0\) 中，\(a = 1\)，\(b = 5\)，\(c = 6\)。

3. The Zero Product Property | 零乘积性质

The zero product property states that if two numbers multiply to give zero, then at least one of them must be zero. Formally, if \(p \times q = 0\), then \(p = 0\) or \(q = 0\).

零乘积性质指出：如果两个数相乘等于零，那么这两个数中至少有一个为零。形式化地说，如果 \(p \times q = 0\)，则 \(p = 0\) 或 \(q = 0\)。

This property is the key to solving quadratic equations by factorisation. Once a quadratic expression is written as a product of two brackets, you can set each bracket equal to zero separately.

该性质是使用因式分解法解二次方程的关键。一旦把二次表达式写成两个括号的乘积，就可以分别令每个括号等于零来求解。

If (x – 2)(x + 3) = 0, then x = 2 or x = -3

Notice that substituting either value back into the factorised equation gives zero, confirming that both are valid solutions.

注意到将任一解代回因式方程都会得到零，这验证了它们都是有效解。

4. Solving by Factorisation | 因式分解法求解

Factorisation is often the quickest method for solving quadratic equations when the roots are integers. The goal is to rewrite \(ax^2 + bx + c = 0\) as \((px + q)(rx + s) = 0\).

当根为整数时，因式分解通常是解二次方程最快的方法。目标是把 \(ax^2 + bx + c = 0\) 改写为 \((px + q)(rx + s) = 0\) 的形式。

Example: Solve \(x^2 + 7x + 10 = 0\).

例题：解方程 \(x^2 + 7x + 10 = 0\)。

We need two numbers that multiply to 10 and add to 7. These numbers are 2 and 5. Therefore the factorised form is \((x + 2)(x + 5) = 0\).

我们需要找到两个数，它们乘积为 10，和为 7。这两个数是 2 和 5。因此因式分解形式为 \((x + 2)(x + 5) = 0\)。

Using the zero product property, \(x + 2 = 0\) gives \(x = -2\), and \(x + 5 = 0\) gives \(x = -5\).

利用零乘积性质，由 \(x + 2 = 0\) 得到 \(x = -2\)，由 \(x + 5 = 0\) 得到 \(x = -5\)。

Always check your factors by expanding them back to the original expression.

始终通过展开来检查因式是否正确，确保得到原表达式。

5. Solving by Completing the Square | 配方法求解

Completing the square rewrites a quadratic equation in the form \((x + m)^2 = n\). This is useful when the equation does not factorise easily.

配方法将二次方程改写为 \((x + m)^2 = n\) 的形式。当方程不易因式分解时，这种方法非常有用。

Starting with \(x^2 + 6x – 7 = 0\), we move the constant term to the other side: \(x^2 + 6x = 7\).

以 \(x^2 + 6x – 7 = 0\) 为例，先将常数项移到另一边：\(x^2 + 6x = 7\)。

Take half of the coefficient of \(x\), which is 3, and square it to get 9. Add 9 to both sides: \(x^2 + 6x + 9 = 16\).

取 \(x\) 系数的一半，即 3，平方得到 9。将 9 加到两边：\(x^2 + 6x + 9 = 16\)。

The left side becomes a perfect square: \((x + 3)^2 = 16\). Taking square roots gives \(x + 3 = \pm 4\), so \(x = 1\) or \(x = -7\).

左侧变成完全平方 \((x + 3)^2 = 16\)。开平方得到 \(x + 3 = \pm 4\)，因此 \(x = 1\) 或 \(x = -7\)。

6. The Quadratic Formula | 二次方程求根公式

For any quadratic equation \(ax^2 + bx + c = 0\), the solutions can be found using the quadratic formula.

对于任意二次方程 \(ax^2 + bx + c = 0\)，其解可以通过求根公式得到。

x = (-b ± √(b² – 4ac)) / (2a)

This formula works for all quadratics, including those with irrational or complex roots. You need to memorise it carefully.

这个公式适用于所有二次方程，包括具有无理根或复数根的情况。你需要小心地记住它。

Example: Solve \(2x^2 – 3x – 2 = 0\) using the quadratic formula.

例题：用求根公式解 \(2x^2 – 3x – 2 = 0\)。

Here \(a = 2\), \(b = -3\), \(c = -2\). Substituting into the formula gives \(x = (3 ± √(9 + 16)) / 4 = (3 ± 5)/4\).

这里 \(a = 2\)，\(b = -3\)，\(c = -2\)。代入公式得 \(x = (3 ± √(9 + 16)) / 4 = (3 ± 5)/4\)。

Thus \(x = 8/4 = 2\) or \(x = -2/4 = -1/2\).

因此 \(x = 8/4 = 2\) 或 \(x = -2/4 = -1/2\)。

7. The Discriminant | 判别式

The expression \(b² – 4ac\) inside the quadratic formula is called the discriminant. It determines the nature of the roots.

求根公式中的 \(b² – 4ac\) 称为判别式。它决定根的性质。

If the discriminant is positive, the equation has two distinct real roots. If it is zero, there is exactly one repeated real root. If it is negative, there are no real roots (two complex roots).

如果判别式为正，方程有两个不相等实数根。如果为零，方程有一个重复实数根。如果为负，方程没有实数根（有两个复数根）。

Discriminant \(b² – 4ac\)	Nature of roots
> 0	Two distinct real roots
= 0	One repeated real root
< 0	No real roots

For the equation \(x^2 + 4x + 5 = 0\), the discriminant is \(16 – 20 = -4\), so there are no real roots.

对于方程 \(x^2 + 4x + 5 = 0\)，判别式为 \(16 – 20 = -4\)，因此没有实数根。

8. Solving Word Problems with Quadratics | 二次方程应用题

Many real-world problems can be modelled using quadratic equations. Typical examples include area problems, projectile motion, and number reasoning.

许多现实问题可以用二次方程建模。典型例子包括面积问题、抛体运动和数字推理。

Example: The area of a rectangle is 52 cm², and its length is 5 cm longer than its width. Find the width.

例题：一个矩形的面积是 52 平方厘米，它的长比宽长 5 厘米。求宽。

Let the width be \(x\). Then the length is \(x + 5\). The area is \(x(x + 5) = 52\).

设宽为 \(x\)，则长为 \(x + 5\)。面积为 \(x(x + 5) = 52\)。

Expanding gives \(x^2 + 5x – 52 = 0\). Factorising, we look for two numbers multiplying to -52 and adding to 5. The numbers are 13 and -4, so \((x + 13)(x – 4) = 0\).

展开得 \(x^2 + 5x – 52 = 0\)。因式分解时，寻找两个数乘积为 -52，和为 5。这两个数是 13 和 -4，所以 \((x + 13)(x – 4) = 0\)。

The solutions are \(x = -13\) and \(x = 4\). Since a width cannot be negative, the width is 4 cm.

解为 \(x = -13\) 和 \(x = 4\)。由于宽度不能为负数，所以宽为 4 厘米。

9. Graphs of Quadratic Functions | 二次函数图像

A quadratic function \(y = ax^2 + bx + c\) always produces a parabola. The sign of \(a\) determines whether it opens upward (\(a > 0\)) or downward (\(a < 0\)).

二次函数 \(y = ax^2 + bx + c\) 的图像始终是抛物线。\(a\) 的符号决定开口方向：\(a > 0\) 时开口向上，\(a < 0\) 时开口向下。

The solutions of the equation \(ax^2 + bx + c = 0\) correspond to the \(x\)-intercepts of the parabola. If the discriminant is negative, the graph does not cross the \(x\)-axis.

方程 \(ax^2 + bx + c = 0\) 的解对应抛物线 \(x\) 轴交点的横坐标。如果判别式为负，图像不与 \(x\) 轴相交。

The axis of symmetry is given by x = -b / (2a)

For \(y = x^2 – 4x + 3\), the roots are \(x = 1\) and \(x = 3\), and the axis of symmetry is \(x = 2\).

对于 \(y = x^2 – 4x + 3\)，根为 \(x = 1\) 和 \(x = 3\)，对称轴为 \(x = 2\)。

10. Common Mistakes to Avoid | 常见错误避免

One common mistake is forgetting to rearrange the equation to standard form before solving. For example, solving \(x^2 = 3x\) by dividing both sides by \(x\) loses the solution \(x = 0\).

一个常见错误是在求解前忘记将方程整理成标准形式。例如，解 \(x^2 = 3x\) 时两边同时除以 \(x\) 会丢失 \(x = 0\) 这个解。

Another mistake is misapplying the quadratic formula by not using brackets correctly. Always write \(x = (-b ± √(b² – 4ac)) / (2a)\) clearly.

另一个错误是使用求根公式时括号运用不当。务必清楚写出 \(x = (-b ± √(b² – 4ac)) / (2a)\)。

When completing the square, pupils often forget to add the squared value to both sides of the equation.

在使用配方法时，学生常常忘记将平方后的数加到等号两边。

Finally, always check whether your answers satisfy the original equation, especially for word problems where negative values may be meaningless.

最后，务必检查答案是否满足原方程，尤其在应用题中负值可能没有实际意义。

11. Practice Questions | 练习题目

Here are a few quick questions to test your understanding.

这里有几道快速练习来检验你的理解。

Solve \(x^2 – 9 = 0\).

解方程 \(x^2 – 9 = 0\)。
Solve \(3x^2 + 5x – 2 = 0\) by factorisation.

使用因式分解法解 \(3x^2 + 5x – 2 = 0\)。
Find the discriminant of \(2x^2 + 4x + 9 = 0\) and state the nature of the roots.

求 \(2x^2 + 4x + 9 = 0\) 的判别式，并说明根的性质。

For the first question, factorisation gives \((x – 3)(x + 3) = 0\), so \(x = 3\) or \(x = -3\).

第一题因式分解得 \((x – 3)(x + 3) = 0\)，所以 \(x = 3\) 或 \(x = -3\)。

For the second question, the factors are \((3x – 1)(x + 2) = 0\), giving \(x = 1/3\) or \(x = -2\).

第二题因式为 \((3x – 1)(x + 2) = 0\)，可得 \(x = 1/3\) 或 \(x = -2\)。

For the third question, the discriminant is \(16 – 72 = -56\), so there are no real roots.

第三题判别式为 \(16 – 72 = -56\)，所以没有实数根。

12. Summary | 总结

Quadratic equations are a core part of the IGCSE Mathematics syllabus. You should be able to identify the standard form, solve using factorisation, completing the square or the quadratic formula, and interpret the discriminant.

二次方程是 IGCSE 数学教学大纲的核心部分。你应当能够识别标准形式，使用因式分解、配方或求根公式求解，并理解判别式的意义。

Regular practice is essential. Start with simple factorisation, move on to completing the square, and finally master the quadratic formula. Remember to check your answers and avoid common mistakes.

定期练习必不可少。从简单的因式分解开始，然后学习配方法，最后熟练掌握求根公式。记住要检查答案并避免常见错误。

With a solid understanding of the ideas above, you will be well prepared for any quadratic equation question on the IGCSE exam.

如果你能透彻理解上述概念，你就能从容应对 IGCSE 考试中的任何二次方程题目。

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T-2-1071: Quadratic Equations | 二次方程