Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

A quadratic equation is one of the most heavily tested topics in IGCSE Mathematics. It appears in almost every paper, whether as a direct solve question, a word problem, or part of a graph-sketching exercise. Mastering the different methods of solution is essential for building confidence and securing top grades.

二次方程是IGCSE数学中考查频率最高的知识点之一。几乎每张试卷都会出现，无论是直接求解题、文字应用题，还是函数图像题的一部分。掌握不同的解法对于建立信心和取得高分至关重要。

1. The Standard Form | 标准形式

A quadratic equation is any equation that can be written in the form shown below. The value of \(a\) must never be zero, because if \(a = 0\) the equation becomes linear, not quadratic.

二次方程是指可以写成下面这种形式的方程。\(a\) 的值绝不能为零，因为如果 \(a = 0\)，方程就变成一次方程而不是二次方程了。

ax² + bx + c = 0, where a ≠ 0

The letters \(a\), \(b\), and \(c\) are real numbers, and \(x\) is the unknown variable. For example, \(3x² – 5x + 2 = 0\) is a quadratic equation with \(a = 3\), \(b = -5\), and \(c = 2\). Notice that the terms are usually arranged in descending powers of \(x\).

字母 \(a\)、\(b\)、\(c\) 是实数，\(x\) 是未知数。例如，\(3x² – 5x + 2 = 0\) 就是一个二次方程，其中 \(a = 3\)、\(b = -5\)、\(c = 2\)。请注意，各项通常按 \(x\) 的降幂排列。

Before attempting any solution method, always rewrite the equation so that one side equals zero. This is the most important first step, and many errors in exams come from students trying to factor or apply the formula without rearranging the equation correctly.

在尝试任何解法之前，务必先将方程整理为一侧等于零的形式。这是最重要的第一步，许多考试错误都源于学生在未正确整理方程的情况下就尝试因式分解或套用公式。

2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic has simple integer factors. The underlying idea is that if the product of two expressions equals zero, then at least one of them must be zero. This is known as the zero product property.

当二次方程具有简单的整数因式时，因式分解是最快捷的方法。其核心思想是：如果两个表达式的乘积等于零，那么至少其中一个必须为零。这称为零乘积性质。

If (x – p)(x – q) = 0, then x = p or x = q

For example, to solve \(x² + 5x + 6 = 0\), look for two numbers that multiply to 6 and add to 5. The numbers 2 and 3 satisfy both conditions, so the equation becomes \((x + 2)(x + 3) = 0\). Hence \(x = -2\) or \(x = -3\).

例如，解 \(x² + 5x + 6 = 0\) 时，寻找两个相乘等于 6 且相加等于 5 的数。数字 2 和 3 满足这两个条件，因此方程变为 \((x + 2)(x + 3) = 0\)。所以 \(x = -2\) 或 \(x = -3\)。

When the coefficient of \(x²\) is not 1, the process is slightly longer. For \(2x² + 7x + 3 = 0\), you need two numbers that multiply to \(2 × 3 = 6\) and add to 7. Those numbers are 1 and 6. Rewrite the middle term as \(2x² + x + 6x + 3 = 0\), then factor by grouping: \(x(2x + 1) + 3(2x + 1) = 0\), giving \((2x + 1)(x + 3) = 0\). Therefore \(x = -½\) or \(x = -3\).

当 \(x²\) 的系数不为 1 时，过程稍长。对于 \(2x² + 7x + 3 = 0\)，需要找两个数相乘等于 \(2 × 3 = 6\) 且相加等于 7。这两个数是 1 和 6。将中间项改写为 \(2x² + x + 6x + 3 = 0\)，然后分组因式分解：\(x(2x + 1) + 3(2x + 1) = 0\)，得到 \((2x + 1)(x + 3) = 0\)。因此 \(x = -½\) 或 \(x = -3\)。

Always check your factors by expanding them mentally. A quick expansion will confirm that you have not swapped the signs or missed a constant factor.

务必通过心算展开来检查因式是否正确。快速展开可以确认你没有弄错符号或漏掉常数因子。

3. Completing the Square | 配方法

Completing the square is a powerful technique that transforms a quadratic into a perfect square plus a constant. It is especially useful for solving equations that do not factorise neatly and for finding turning points of quadratic graphs.

配方法是一种强大的技巧，它将二次式转化为完全平方加上一个常数。它特别适用于解无法整齐因式分解的方程，以及求二次函数图像的顶点。

To complete the square for \(x² + bx\), take half of \(b\), square it, add it, and then subtract it. For example:

对 \(x² + bx\) 配方时，取 \(b\) 的一半，然后平方，加上这个平方项并再减去它。例如：

x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7

To solve \(x² + 6x + 2 = 0\), rewrite it as \((x + 3)² = 7\). Then take the square root of both sides: \(x + 3 = ±√7\), so \(x = -3 ± √7\). This gives two exact answers without using a calculator.

要解 \(x² + 6x + 2 = 0\)，将其改写为 \((x + 3)² = 7\)。然后对两边开平方：\(x + 3 = ±√7\)，因此 \(x = -3 ± √7\)。这样不用计算器就能得到两个精确答案。

When the coefficient of \(x²\) is not 1, factor it out first. For \(2x² – 8x + 5 = 0\), write \(2(x² – 4x) + 5 = 0\), then complete the square inside the brackets: \(2[(x – 2)² – 4] + 5 = 2(x – 2)² – 8 + 5 = 2(x – 2)² – 3\). Setting this equal to zero gives \(2(x – 2)² = 3\), so \((x – 2)² = 1.5\) and \(x = 2 ± √1.5\).

当 \(x²\) 的系数不为 1 时，先将其提出来。对于 \(2x² – 8x + 5 = 0\)，写成 \(2(x² – 4x) + 5 = 0\)，然后在括号内配方：\(2[(x – 2)² – 4] + 5 = 2(x – 2)² – 8 + 5 = 2(x – 2)² – 3\)。令其等于零，得到 \(2(x – 2)² = 3\)，所以 \((x – 2)² = 1.5\)，\(x = 2 ± √1.5\)。

Completing the square is also the foundation for deriving the quadratic formula, so understanding this method deeply will help you remember how the formula works.

配方法也是推导二次求根公式的基础，因此深入理解这种方法有助于你记住公式的来龙去脉。

4. The Quadratic Formula | 二次求根公式

The quadratic formula is the most reliable method because it works for every quadratic equation, whether factorisable or not. You should memorise it exactly as shown below.

二次求根公式是最可靠的方法，因为它适用于所有二次方程，无论是否能因式分解。你应该准确记住下面的公式。

x = (−b ± √(b² − 4ac)) ÷ 2a

To use the formula, first rewrite the equation in standard form \(ax² + bx + c = 0\). Then substitute the values of \(a\), \(b\), and \(c\). For example, solve \(2x² + 3x – 5 = 0\): here \(a = 2\), \(b = 3\), \(c = -5\). Substitute into the formula:

使用公式时，先将方程改写为标准形式 \(ax² + bx + c = 0\)。然后将 \(a\)、\(b\)、\(c\) 的值代入。例如，解 \(2x² + 3x – 5 = 0\)：这里 \(a = 2\)、\(b = 3\)、\(c = -5\)。代入公式：

x = (−3 ± √(3² − 4 × 2 × (−5))) ÷ (2 × 2) = (−3 ± √(9 + 40)) ÷ 4 = (−3 ± 7) ÷ 4

So \(x = (−3 + 7) ÷ 4 = 1\) or \(x = (−3 – 7) ÷ 4 = -2.5\). Notice that the same result can be obtained by factorising \((2x + 5)(x – 1) = 0\).

所以 \(x = (−3 + 7) ÷ 4 = 1\) 或 \(x = (−3 – 7) ÷ 4 = -2.5\)。请注意，通过因式分解 \((2x + 5)(x – 1) = 0\) 也能得到相同结果。

When you use the formula, be careful with negative signs. The expression \(-b\) reverses the sign of \(b\), and \(4ac\) includes the sign of \(c\). Writing out every step reduces the chance of arithmetic slips.

使用公式时，要特别注意负号。\(-b\) 会反转 \(b\) 的符号，而 \(4ac\) 中的 \(c\) 要带上自身符号。逐步写出计算过程能减少算术失误的概率。

5. The Discriminant | 判别式

The expression inside the square root of the quadratic formula, \(b² − 4ac\), is called the discriminant. It tells us how many real roots a quadratic equation has without solving it fully.

二次求根公式根号内的表达式 \(b² − 4ac\) 称为判别式。它可以告诉我们二次方程有多少个实数根，而无需完整求解。

If \(b² − 4ac > 0\), the equation has two distinct real roots.
如果 \(b² − 4ac > 0\)，方程有两个不相等的实数根。
If \(b² − 4ac = 0\), the equation has exactly one real root (a repeated root).
如果 \(b² − 4ac = 0\)，方程只有一个实数根（重根）。
If \(b² − 4ac < 0\), the equation has no real roots.
如果 \(b² − 4ac < 0\)，方程没有实数根。

For example, in \(x² – 4x + 9 = 0\), the discriminant is \(16 – 36 = -20\). Since this is negative, the graph never crosses the \(x\)-axis, and the equation has no real solutions. In \(x² – 6x + 9 = 0\), the discriminant is \(36 – 36 = 0\), so the graph touches the \(x\)-axis at exactly one point, namely \(x = 3\).

例如，在 \(x² – 4x + 9 = 0\) 中，判别式为 \(16 – 36 = -20\)。因为该值为负，图像与 \(x\) 轴无交点，方程无实数解。在 \(x² – 6x + 9 = 0\) 中，判别式为 \(36 – 36 = 0\)，因此图像与 \(x\) 轴相切于一点，即 \(x = 3\)。

Exam questions often ask you to find the range of values of \(k\) for which an equation has two roots. Simply set the discriminant greater than zero and solve the resulting inequality. For instance, if \(x² + kx + 4 = 0\) has two real roots, then \(k² – 16 > 0\), so \(k > 4\) or \(k < -4\).

考试中常出现求参数 \(k\) 的取值范围使得方程有两个根的题型。只需令判别式大于零并解所得不等式。例如，若 \(x² + kx + 4 = 0\) 有两个实数根，则 \(k² – 16 > 0\)，所以 \(k > 4\) 或 \(k < -4\)。

6. The Sum and Product of Roots | 根的和与积

For a quadratic equation \(ax² + bx + c = 0\), there are elegant relationships between the coefficients and the roots. If the roots are \(α\) and \(β\), then:

对于二次方程 \(ax² + bx + c = 0\)，系数与根之间存在巧妙的关系。若两个根为 \(α\) 和 \(β\)，则：

α + β = −b ÷ a, and α × β = c ÷ a

These relationships are useful when you are given the roots and asked to construct an equation, or when you need to find expressions such as \(α² + β²\). Since \(α² + β² = (α + β)² – 2αβ\), you can substitute the sum and product directly.

这些关系在已知根求方程，或需要求 \(α² + β²\) 等表达式时非常有用。因为 \(α² + β² = (α + β)² – 2αβ\)，你可以直接代入和与积进行计算。

For example, if the roots of \(2x² – 5x + 1 = 0\) are \(α\) and \(β\), then \(α + β = 5/2\) and \(αβ = 1/2\). Therefore \(α² + β² = (5/2)² – 2 × (1/2) = 25/4 – 1 = 21/4\).

例如，若 \(2x² – 5x + 1 = 0\) 的根为 \(α\) 和 \(β\)，则 \(α + β = 5/2\)，\(αβ = 1/2\)。因此 \(α² + β² = (5/2)² – 2 × (1/2) = 25/4 – 1 = 21/4\)。

7. Solving by Graphs | 图像解法

A quadratic equation \(ax² + bx + c = 0\) can be solved graphically by sketching the curve \(y = ax² + bx + c\) and reading off the \(x\)-coordinates of the points where the curve crosses the \(x\)-axis. These intersections are exactly the roots of the equation.

二次方程 \(ax² + bx + c = 0\) 可以通过图像法求解：画出曲线 \(y = ax² + bx + c\)，读出曲线与 \(x\) 轴交点的 \(x\) 坐标。这些交点就是方程的根。

If the quadratic has a positive \(a\), the graph is a U-shaped parabola with a minimum point. If \(a\) is negative, the graph is an upside-down U with a maximum point. The turning point has coordinates given by completing the square:

当二次项系数 \(a\) 为正时，图像是开口向上的抛物线，有最小值点；当 \(a\) 为负时，图像是开口向下的抛物线，有最大值点。顶点坐标可通过配方法求得：

Turning point = (−b ÷ 2a, c − b² ÷ 4a)

For example, the graph of \(y = x² – 4x + 3\) crosses the \(x\)-axis at \(x = 1\) and \(x = 3\), so the equation \(x² – 4x + 3 = 0\) has roots 1 and 3. The vertex is at \(x = 2\), and substituting gives \(y = -1\), so the turning point is \((2, -1)\).

例如，\(y = x² – 4x + 3\) 的图像与 \(x\) 轴交于 \(x = 1\) 和 \(x = 3\)，因此方程 \(x² – 4x + 3 = 0\) 的根为 1 和 3。顶点在 \(x = 2\) 处，代入得 \(y = -1\)，所以顶点为 \((2, -1)\)。

Graphical methods are especially helpful for checking your algebraic answers. Compare the sign of the \(y\)-values around each suspected root: the curve should change sign as it passes through a root.

图像法特别适合用来检验代数答案。观察每个疑似根附近的 \(y\) 值符号：曲线经过根时应该会变号。

8. Quadratic Word Problems | 二次方程应用题

Many real-world situations produce quadratic equations. A classic IGCSE problem involves a rectangle whose length is 3 metres longer than its width, with an area of 28 square metres. If the width is \(x\) metres, then \(x(x + 3) = 28\), which expands to \(x² + 3x – 28 = 0\). Factorising gives \((x + 7)(x – 4) = 0\), so \(x = 4\) (the negative root \(-7\) is rejected because length cannot be negative).

许多现实情境会导出二次方程。一个经典的IGCSE题目是：一个矩形的长比宽多 3 米，面积为 28 平方米。若宽为 \(x\) 米，则 \(x(x + 3) = 28\)，展开得 \(x² + 3x – 28 = 0\)。因式分解得 \((x + 7)(x – 4) = 0\)，所以 \(x = 4\)（负根 \(-7\) 因长度不能为负而舍去）。

Projectile motion is another common context. A ball thrown upward follows a height equation \(h = -5t² + 20t + 15\), where \(h\) is the height in metres and \(t\) is time in seconds. To find when the ball hits the ground, set \(h = 0\) and solve \(-5t² + 20t + 15 = 0\). Divide throughout by \(-5\) to get \(t² – 4t – 3 = 0\), then use the quadratic formula to find the positive value of \(t\).

抛体运动是另一个常见情境。一个向上抛出的球遵循高度方程 \(h = -5t² + 20t + 15\)，其中 \(h\) 以米为单位，\(t\) 以秒为单位。要求球何时落地，令 \(h = 0\) 并解 \(-5t² + 20t + 15 = 0\)。整体除以 \(-5\) 得 \(t² – 4t – 3 = 0\)，然后使用二次求根公式求 \(t\) 的正值。

When solving word problems, always define your variable, translate the conditions into an equation, solve it, and then check whether each answer makes sense in the original context. Reject any answer that is impossible.

解应用题时，务必先定义变量，将条件转化为方程，求解后再检查每个答案是否在原始情境中合理。舍去任何不可能的答案。

9. Common Mistakes and How to Avoid Them | 常见错误与规避方法

One of the most frequent errors is failing to rearrange the equation to standard form before solving. For example, if a question gives \(x² = 4x + 12\), many students incorrectly write \(x = ±√(4x + 12)\). Instead, you must subtract \(4x + 12\) from both sides to obtain \(x² – 4x – 12 = 0\), then factorise to \((x – 6)(x + 2) = 0\).

最常犯的错误之一是未将方程整理为标准形式就急于求解。例如，若题目给出 \(x² = 4x + 12\)，许多学生会错误地写成 \(x = ±√(4x + 12)\)。正确做法是两边同时减去 \(4x + 12\)，得到 \(x² – 4x – 12 = 0\)，然后因式分解为 \((x – 6)(x + 2) = 0\)。

Another common mistake is losing a root when taking square roots. Remember that \(x² = 25\) has two solutions: \(x = 5\) and \(x = -5\). Always include the \(\pm\) sign when solving equations of this type.

另一个常见错误是在开平方时丢失一个根。切记 \(x² = 25\) 有两个解：\(x = 5\) 和 \(x = -5\)。解这类方程时一定要加上 \(\pm\) 号。

Students also frequently sign errors when substituting into the quadratic formula. Write the formula clearly, bracket every negative value, and perform each arithmetic operation step by step. Checking your answers by substituting back into the original equation is a quick way to catch mistakes.

学生在代入二次求根公式时也常出现符号错误。请将公式写清楚，对每个负数加上括号，并逐步进行算术运算。将答案代回原方程进行检验是快速发现错误的有效方法。

10. Exam Tips and Quick Checklist | 考试技巧与快速清单

In the exam, read each question carefully to determine which method is expected. The phrase “solve by factorisation” means you must show factorising steps; the phrase “give your answers correct to 2 decimal places” suggests using the quadratic formula or completing the square with a calculator.

考试时，仔细阅读题目以判断期望使用哪种方法。短语 “solve by factorisation” 意味着你必须展示因式分解的步骤；短语 “give your answers correct to 2 decimal places” 则暗示应使用二次求根公式或配方法配合计算器。

Always rearrange to \(ax² + bx + c = 0\) before solving.
求解前务必整理成 \(ax² + bx + c = 0\) 的形式。
Check whether the equation factorises: look for factors of \(ac\) that sum to \(b\).
检查方程是否能因式分解：寻找 \(ac\) 的因式中相加等于 \(b\) 的一对。
If factorisation is not possible, use the quadratic formula.
如果无法因式分解，则使用二次求根公式。
After finding roots, substitute them back into the original equation to verify.
求出根后，代回原方程进行验证。
In word problems, reject solutions that do not make sense in the context.
在应用题中，舍去在情境中不合理的解。
Use the discriminant to predict the number of roots before solving.
在求解前使用判别式预判根的个数。

With regular practice, quadratic equations will become one of the most rewarding topics in IGCSE Mathematics. The key is to be systematic: standard form first, choose the right method, show clear working, and always check your results.

通过定期练习，二次方程将成为IGCSE数学中最容易得分的主题之一。关键是按部就班：先化为标准形式，选择合适的方法，展示清晰的过程，并始终检验你的结果。

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Solving Quadratic Equations | 解二次方程