Solving Quadratic Equations | 求解二次方程

📚 Solving Quadratic Equations | 求解二次方程

Quadratic equations are one of the most important topics in Edexcel A-Level Mathematics. They appear directly in pure mathematics papers, but they also underpin curve sketching, optimisation, mechanics and many applied problems. You need to be confident using factorising, completing the square, the quadratic formula and the discriminant, as well as recognising hidden quadratics and common exam traps.

二次方程是 Edexcel A-Level 数学中最重要的主题之一。它们不仅直接出现在纯数学试卷中，也是曲线作图、最优化、力学以及许多应用题的基础。你需要熟练运用因式分解法、配方法、二次公式和判别式，同时能够识别隐含二次方程和常见考试陷阱。

1. Recognising a Quadratic Equation | 识别二次方程

A quadratic equation in one variable is an equation of the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. If a = 0, the x² term disappears and the equation becomes linear, so the coefficient of x² must be non-zero.

一元二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。若 a = 0，x² 项消失，方程退化为一次方程，因此 x² 的系数必须不为零。

ax² + bx + c = 0, a ≠ 0

The values of x that make the equation true are called its roots or solutions. A quadratic equation can have two distinct real roots, one repeated real root, or no real roots. In A-Level Mathematics, complex roots are not covered until Further Mathematics, so ‘no real roots’ is a complete description at this stage.

使方程成立的 x 值称为方程的根或解。二次方程可能有两个相异实根、一个重根，或没有实根。在 A-Level 数学中，复数根要到 Further Mathematics 才会涉及，因此现阶段“无实根”就是完整描述。

2. Solving by Factorising | 因式分解法

Factorising works when the quadratic expression can be written as a product of two linear factors. For example, x² + 5x + 6 = (x + 2)(x + 3), so the equation x² + 5x + 6 = 0 gives x + 2 = 0 or x + 3 = 0.

当二次式可以写成两个一次因式的乘积时，可以使用因式分解法。例如 x² + 5x + 6 = (x + 2)(x + 3)，因此方程 x² + 5x + 6 = 0 推出 x + 2 = 0 或 x + 3 = 0。

(x + 2)(x + 3) = 0 ⇒ x = −2 or x = −3

Always rearrange the equation so one side equals zero before factorising. If you divide both sides by a common factor involving x, a root may be lost. For example, from x² = 3x, you can write x² − 3x = 0 and factorise as x(x − 3) = 0, giving x = 0 and x = 3.

因式分解前务必先将方程整理为一边等于零。如果两边同时除以含 x 的公因式，可能会丢失根。例如由 x² = 3x，应写成 x² − 3x = 0，再因式分解为 x(x − 3) = 0，得到 x = 0 和 x = 3。

For non-monic quadratics, such as 2x² − 3x − 2 = 0, look for two numbers that multiply to ac = −4 and add to b = −3. The numbers are −4 and 1, so 2x² − 4x + x − 2 = 2x(x − 2) + 1(x − 2) = (2x + 1)(x − 2).

对于非首一二次式，例如 2x² − 3x − 2 = 0，寻找两个数，它们相乘等于 ac = −4，相加等于 b = −3。这两个数是 −4 和 1，因此 2x² − 4x + x − 2 = 2x(x − 2) + 1(x − 2) = (2x + 1)(x − 2)。

3. Completing the Square | 配方法

Completing the square transforms the quadratic into the form a(x + p)² + q = 0, which makes the vertex and roots easier to identify. For x² + bx + c, add and subtract (b/2)² to form a perfect square.

配方法将二次式转化为 a(x + p)² + q = 0 的形式，便于确定顶点和根。对于 x² + bx + c，加减 (b/2)² 可构成完全平方。

x² + bx + c = (x + b/2)² − (b/2)² + c

To solve x² − 6x + 4 = 0, complete the square: (x − 3)² − 9 + 4 = 0, so (x − 3)² = 5. Taking square roots gives x − 3 = ±√5, hence x = 3 ± √5.

解 x² − 6x + 4 = 0 时，配方得 (x − 3)² − 9 + 4 = 0，因此 (x − 3)² = 5。两边开平方得 x − 3 = ±√5，所以 x = 3 ± √5。

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 bracket: 2[(x + 2)² − 4] + 5 = 0, which simplifies to 2(x + 2)² − 8 + 5 = 0, so 2(x + 2)² = 3.

当 x² 的系数不是 1 时，先将其提出。对于 2x² + 8x + 5 = 0，写成 2(x² + 4x) + 5 = 0，然后对括号内配方：2[(x + 2)² − 4] + 5 = 0，化简为 2(x + 2)² − 8 + 5 = 0，因此 2(x + 2)² = 3。

4. The Quadratic Formula | 二次公式

The quadratic formula solves ax² + bx + c = 0 for any real coefficients. It is derived by completing the square and is given in the Edexcel formula booklet, so you should know how to use it efficiently.

二次公式可求解任意实系数方程 ax² + bx + c = 0。它由配方法导出，并收录在 Edexcel 公式手册中，因此你应当熟练掌握其用法。

x = (−b ± √(b² − 4ac)) / (2a)

Use the formula when factorising is difficult or when exact answers are required. Substitute a, b and c carefully, paying special attention to negative signs. For example, in 2x² + 3x − 2 = 0, we have a = 2, b = 3 and c = −2.

当因式分解困难或需要精确答案时，使用公式。代入 a、b、c 时要小心负号。例如在 2x² + 3x − 2 = 0 中，a = 2，b = 3，c = −2。

x = (−3 ± √(3² − 4×2×(−2))) / (2×2) = (−3 ± √25) / 4 = (−3 ± 5) / 4

This gives x = (−3 + 5)/4 = 1/2 and x = (−3 − 5)/4 = −2. These are the two exact roots. In an exam, show the substitution line before simplifying, because method marks are available even if arithmetic goes wrong later.

由此得到 x = (−3 + 5)/4 = 1/2 和 x = (−3 − 5)/4 = −2。这是两个精确根。在考试中，先展示代入过程再化简，因为即使后续计算出错，方法分仍然可得。

5. The Discriminant | 判别式

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

判别式是二次公式中平方根下的表达式：D = b² − 4ac。它无需解方程即可判断根的个数和性质。

D = b² − 4ac

If D > 0, there are two distinct real roots. If D = 0, there is one repeated real root, also called a double root. If D < 0, there are no real roots.

若 D > 0，有两个相异实根；若 D = 0，有一个重根，也称二重根；若 D < 0，无实根。

D > 0: two distinct real roots | 两个相异实根
D = 0: one repeated real root | 一个重根
D < 0: no real roots | 无实根

This is especially useful in questions that ask for the number of roots or for conditions on a parameter such as k. For example, kx² + 2x + 1 = 0 has a repeated root when D = 2² − 4×k×1 = 4 − 4k = 0, so k = 1.

这在考试题中尤其有用，例如要求判断根的个数或求参数 k 的取值范围。例如 kx² + 2x + 1 = 0 有重根时，D = 2² − 4×k×1 = 4 − 4k = 0，因此 k = 1。

6. Choosing an Efficient Method | 选择高效方法

If the quadratic factorises quickly with integer coefficients, factorising is fastest. If not, completing the square or the quadratic formula is more reliable.

如果二次式能快速分解为整系数因式，因式分解最快。否则配方法或二次公式更可靠。

Completing the square is preferred when you also need the vertex or the maximum/minimum value of a quadratic function. The formula is best for a direct algebraic solution when exact factors are not obvious.

当还需要抛物线顶点或二次函数的最大/最小值时，优先考虑配方法。当无法直接看出因式时，二次公式最适合进行直接的代数求解。

In Edexcel exams, working must show method. Simply writing answers from a calculator may not earn full marks, especially when surds are involved. Present clear substitution and intermediate steps.

在 Edexcel 考试中，必须展示解题过程。仅用计算器给出答案可能无法得满分，尤其是涉及根式时。要清晰写出代入过程与中间步骤。

7. Graphical Interpretation of Roots | 根的图像解释

The solutions of ax² + bx + c = 0 are the x-coordinates where the graph of y = ax² + bx + c crosses the x-axis. These points are called x-intercepts.

方程 ax² + bx + c = 0 的解是抛物线 y = ax² + bx + c 与 x 轴交点的横坐标，这些点称为 x 截距。

If the graph cuts the x-axis at two points, the quadratic has two distinct real roots. If the graph touches the x-axis at one point, the quadratic has a repeated root. If the graph never intersects the x-axis, the roots are not real.

若图像与 x 轴相交于两点，方程有两个相异实根；若图像与 x 轴交于一点，方程有一个重根；若图像不与 x 轴相交，则没有实根。

This geometric view helps check your answers. For example, if the discriminant is negative but you calculate two real roots, you have made a sign error somewhere.

这种几何观点有助于检查答案。例如，如果判别式为负，但你计算出两个实根，那么一定在符号上出了错。

8. Hidden Quadratics | 隐含二次方程

Some equations are not initially quadratic, but can be rearranged into a quadratic in a new variable. A common type is ax⁴ + bx² + c = 0, where letting y = x² gives ay² + by + c = 0.

有些方程一开始不是二次方程，但通过换元可化为新变量的二次方程。常见类型是 ax⁴ + bx² + c = 0，令 y = x² 得到 ay² + by + c = 0。

After solving for y, substitute back to find x and remember to reject any negative values of y if y = x² and x is real. This is a standard source of extra roots that must be removed.

解出 y 后，代回求 x；若 y = x² 且 x 为实数，则需舍去 y 的负值。这是产生增根的常见来源，必须剔除。

For example, solve x⁴ − 5x² + 4 = 0. Let y = x², so y² − 5y + 4 = 0, giving (y − 1)(y − 4) = 0. Thus y = 1 or y = 4. Since y = x², we get x² = 1 or x² = 4, so x = ±1 and x = ±2.

例如解 x⁴ − 5x² + 4 = 0。令 y = x²，则 y² − 5y + 4 = 0，因式分解得 (y − 1)(y − 4) = 0。因此 y = 1 或 y = 4。由于 y = x²，得到 x² = 1 或 x² = 4，所以 x = ±1 和 x = ±2。

9. Roots and Coefficient Relationships | 根与系数关系

For a quadratic equation ax² + bx + c = 0 with roots α and β, the sum of roots is α + β = −b/a and the product is αβ = c/a.

对于根为 α 和 β 的二次方程 ax² + bx + c = 0，根之和为 α + β = −b/a，根之积为 αβ = c/a。

α + β = −b/a, αβ = c/a

These are useful for forming quadratic equations from given roots or for finding unknown coefficients. For example, if the roots are 3 and −2, the sum is 1 and the product is −6, so the equation is x² − x − 6 = 0.

这些关系可用于由给定根构造二次方程，或求未知系数。例如，若根为 3 和 −2，则和为 1，积为 −6，因此方程为 x² − x − 6 = 0。

10. Common Errors and Exam Tips | 常见错误与应试技巧

A common mistake is forgetting to rearrange the

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Solving Quadratic Equations | 求解二次方程