Quadratic Equations: Factorising, Completing the Square and the Formula | IGCSE 数学：二次方程的三种解法

📚 Quadratic Equations: Factorising, Completing the Square and the Formula | IGCSE 数学：二次方程的三种解法

Quadratic equations are at the heart of the IGCSE Mathematics syllabus. Whether you are solving simple equations by factorising, rearranging perfect squares, or applying the quadratic formula, a clear method will help you avoid careless errors and earn full marks on algebra questions.

二次方程是 IGCSE 数学课程的核心内容。无论是用因式分解法解简单方程、通过配方法整理完全平方，还是套用求根公式，掌握清晰的方法都能帮助你避免粗心错误，并在代数题中拿到满分。

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

c are constants and a ≠ 0. If a = 0, the equation becomes linear, so we exclude that case.

二次方程是含有一个变量 x 的次数为 2 的多项式方程，最高次项是 x²。它通常写成一般式 ax² + bx + c = 0，其中 a、b、c 是常数且 a ≠ 0。如果 a = 0，方程就退化为一次方程，因此我们排除这种情况。

The word 'quadratic' comes from the Latin 'quadratus' meaning square, because the variable is squared. Quadratic equations arise in geometry, physics, finance and many real-life situations, such as calculating areas, projectile paths and profit functions.

“quadratic（二次）”一词源自拉丁语“quadratus”，意为“平方”，因为变量被平方。二次方程出现在几何、物理、金融以及许多现实生活情境中，例如计算面积、抛体运动轨迹和利润函数。

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2. The Standard Form ax² + bx + c = 0 | 标准形式 ax² + bx + c = 0

e side so that the other side is zero. For example, 3x² + 7 = 5x becomes 3x² − 5x + 7 = 0.

在尝试任何解法之前，你都应先把方程整理成标准形式。将所有项移到等式同一边，使另一边为 0。例如，3x² + 7 = 5x 应整理为 3x² − 5x + 7 = 0。

ax² + bx + c = 0

Keeping this structure clear is essential because every method below depends on identifying the coefficients a, b and c correctly. If terms are on the wrong side, the signs will cause mistakes.

保持这种结构清晰非常重要，因为下面的每一种方法都依赖于正确识别系数 a、b 和 c。如果项放错了边，符号就会导致错误。

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3. Solving by Factorising | 因式分解法

Factorising is usually the fastest method when the quadratic has integer roots. You look for two numbers that multiply to give ac and add to give b. Then split the middle term, factor by grouping, and set each bracket equal to zero.

当二次方程有整数根时，因式分解法通常是最快的方法。你需要找到两个数，使它们的乘积等于 ac，和等于 b。然后拆分中间项，分组分解，并令每个括号等于 0。

ne root. This works because if a product equals zero, at least one factor must equal zero.

如果方程可以写成 (px + q)(rx + s) = 0，那么要么 px + q = 0，要么 rx + s = 0。每个一次因式给出一个根。这样做的依据是：如果乘积为 0，那么至少有一个因式必须为 0。

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4. Worked Example: Factorising Method | 因式分解法例题

re, x² + 7x + 12 = (x + 3)(x + 4) = 0. Setting each factor to zero gives x = −3 or x = −4.

解方程 x² + 7x + 12 = 0。我们需要两个数，使它们的乘积为 12、和为 7：它们是 3 和 4。因此 x² + 7x + 12 = (x + 3)(x + 4) = 0。令每个因式为 0，得 x = −3 或 x = −4。

uping: 2x(x + 3) − 1(x + 3) = 0, so (2x − 1)(x + 3) = 0. The roots are x = 1/2 and x = −3.

对于 2x² + 5x − 3 = 0，ac = 2 × (−3) = −6。我们需要两个数，使它们的乘积为 −6、和为 5：它们是 6 和 −1。拆分中间项：2x² + 6x − x − 3 = 0。分组分解：2x(x + 3) − 1(x + 3) = 0，所以 (2x − 1)(x + 3) = 0。根为 x = 1/2 和 x = −3。

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5. Solving by Completing the Square | 配方法

Completing the square is useful when the quadratic does not factorise easily or when you need to find the vertex of the parabola. The idea is to rewrite ax² + bx + c in the form a(x + p)² + q.

当二次方程不容易因式分解，或需要求抛物线的顶点时，配方法非常有用。其思路是将 ax² + bx + c 改写成 a(x + p)² + q 的形式。

same idea works when the leading coefficient is not 1, but you must factor out the leading coefficient from the x terms first.

对于首项系数为 1 的二次式 x² + bx，加上并减去 (b/2)²。这样得到 x² + bx + (b/2)² − (b/2)² = (x + b/2)² − (b/2)²。当首项系数不为 1 时，思路相同，但你必须先从含 x 的项中提出首项系数。

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6. Worked Example: Completing the Square | 配方法例题

² = 9 to get x² + 6x + 9 − 9 + 2 = 0, so (x + 3)² − 7 = 0. Then (x + 3)² = 7, so x + 3 = ±√7. Therefore x = −3 ± √7.

用配方法解 x² + 6x + 2 = 0。加上并减去 (6/2)² = 9，得到 x² + 6x + 9 − 9 + 2 = 0，所以 (x + 3)² − 7 = 0。然后 (x + 3)² = 7，所以 x + 3 = ±√7。因此 x = −3 ± √7。

ify to 2(x + 2)² − 8 + 5 = 0, so 2(x + 2)² = 3. Divide by 2: (x + 2)² = 3/2. Thus x = −2 ± √3/2 = −2 ± √6/2.

对于 2x² + 8x + 5 = 0，先将含 x 的项提出 2：2(x² + 4x) + 5 = 0。在括号内加上并减去 (4/2)² = 4：2[(x + 2)² − 4] + 5 = 0。化简得 2(x + 2)² − 8 + 5 = 0，所以 2(x + 2)² = 3。除以 2：(x + 2)² = 3/2。因此 x = −2 ± √3/2 = −2 ± √6/2。

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7. The Quadratic Formula | 求根公式

+ bx + c = 0. It is derived by completing the square on the general form. The formula is:

求根公式可以给出任意二次方程 ax² + bx + c = 0 的根。它是由一般式的配方法推导出来的。公式为：

x = (−b ± √b² − 4ac) / 2aPublished by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

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