IGCSE Maths: Quadratic Equations from Factorising to the Discriminant | IGCSE 数学：从因式分解到判别式的二次方程

📚 IGCSE Maths: Quadratic Equations from Factorising to the Discriminant | IGCSE 数学：从因式分解到判别式的二次方程

A quadratic equation is a polynomial equation in which the highest power of the unknown is 2. It can always be arranged into the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.

二次方程是未知数最高次数为 2 的多项式方程。它总可以整理成标准形式 ax² + bx + c = 0，其中 a、b、c 为常数且 a ≠ 0。

In IGCSE Mathematics, quadratic equations appear across algebra, graphs, measurement and word problems. You must be able to solve them using different methods and explain the meaning of the roots.

在 IGCSE 数学中，二次方程广泛出现在代数、图像、测量和文字题中。你必须会用不同方法求解，并解释根的含义。

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

The coefficient a is the leading coefficient, b is the linear coefficient, and c is the constant term. If a = 0, the equation becomes linear, not quadratic.

系数 a 是二次项系数，b 是一次项系数，c 是常数项。如果 a = 0，方程就变成一次方程，而不是二次方程。

Before solving, always rearrange the equation so that one side is 0. For example, 3x – x² = 4 should be rearranged to -x² + 3x – 4 = 0, or equivalently x² – 3x + 4 = 0 after multiplying by -1.

求解前，务必先将方程整理为一边为 0。例如 3x – x² = 4 应整理为 -x² + 3x – 4 = 0，或乘以 -1 后得到 x² – 3x + 4 = 0。

It is useful to recognise a quadratic immediately from its highest power. This helps you choose the fastest solution method and avoids accidental linear-equation mistakes.

快速识别二次方程的最高次数是很有用的。这能帮助你选择最快的解法，并避免误用一次方程的方法。

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2. Solving by Factorising | 因式分解法求解

If a quadratic expression factorises easily, this is often the fastest method. For x² + bx + c, look for two numbers whose sum is b and product is c.

如果二次式容易因式分解，这通常是最快的方法。对于 x² + bx + c，要找到两个数，它们的和为 b，积为 c。

Once you write the equation as (px + q)(rx + s) = 0, apply the zero-product property: if a product is zero, at least one factor must be zero. Set each bracket equal to 0 and solve.

把方程写成 (px + q)(rx + s) = 0 后，使用零乘积性质：如果乘积为 0，则至少一个因式为 0。令每个括号等于 0 再求解。

Worked example: Solve x² – 5x + 6 = 0. Factorise as (x – 2)(x – 3) = 0, so x – 2 = 0 or x – 3 = 0. The solutions are x = 2 and x = 3.

例题：解 x² – 5x + 6 = 0。因式分解为 (x – 2)(x – 3) = 0，所以 x – 2 = 0 或 x – 3 = 0。解为 x = 2 和 x = 3。

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3. Completing the Square | 配方法求解

Completing the square is useful when the quadratic does not factorise with integers. The key identity is x² + bx = (x + b/2)² – (b/2)².

当二次式不能用整数因式分解时，配方法非常有用。核心恒等式是 x² + bx = (x + b/2)² – (b/2)²。

For ax² + bx + c = 0 with a ≠ 1, first divide every term by a. Then move the constant to the right and add (b/2a)² to both sides.

对于 a ≠ 1 的 ax² + bx + c = 0，先每一项除以 a。然后把常数移到右边，两边加上 (b/2a)²。

Worked example: Solve x² + 6x + 2 = 0. Write x² + 6x = -2. Add (6/2)² = 9 to both sides: (x + 3)² = 7. Therefore x + 3 = ±√7, so x = -3 ± √7.

例题：解 x² + 6x + 2 = 0。写成 x² + 6x = -2。两边加上 (6/2)² = 9：(x + 3)² = 7。因此 x + 3 = ±√7，所以 x = -3 ± √7。

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4. The Quadratic Formula | 求根公式

The quadratic formula is x = (-b ± √(b² – 4ac)) / (2a). It works for every quadratic equation and is obtained by completing the square on the general form.

求根公式是 x = (-b ± √(b² – 4ac)) / (2a)。它对所有二次方程都适用，由一般形式配方得到。

When using the formula, first identify a, b and c carefully, including negative signs. Use brackets around each value and calculate the discriminant b² – 4ac before the square root.

使用公式时，先仔细确定 a、b、c，并注意负号。计算时用括号把每个值括起来，并在开方前先计算判别式 b² – 4ac。

Worked example: Solve 2x² – 4x – 3 = 0. Here a = 2, b = -4, c = -3. The formula gives x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = 1 ± √10/2.

例题：解 2x² – 4x – 3 = 0。这里 a = 2，b = -4，c = -3。公式给出 x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = 1 ± √10/2。

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5. The Discriminant and Nature of Roots | 判别式与根的性质

The discriminant D = b² – 4ac tells you the nature of the roots without solving the equation. It is the expression under the square root in the quadratic

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