📚 Solving Quadratic Equations: Factorising, Completing the Square, and the Quadratic Formula | 二次方程求解：因式分解、配方法与求根公式
Solving quadratic equations is one of the most important algebra skills in IGCSE and A-Level mathematics. A quadratic equation is an equation of degree 2 in one variable, and it appears in coordinate geometry, sequences, modelling questions and many examination papers. This revision guide explains three standard methods: factorising, completing the square, and the quadratic formula. It also shows how to choose the fastest method, how the discriminant tells you the number of real roots, and how to avoid the errors that cost the most marks.
解二次方程是 IGCSE 与 A-Level 数学中最重要的代数技能之一。二次方程是含一个未知数、次数为 2 的方程，它广泛出现在坐标几何、数列、建模题以及大量试卷中。本复习指南讲解三种标准方法：因式分解法、配方法与求根公式法，并说明如何选择最快的方法、判别式如何判断实根个数，以及如何避免最容易失分的错误。
1. Recognising a Quadratic Equation | 识别二次方程
≠ 0 is essential, because if a = 0 then the x² term disappears and the equation becomes linear. The constants a, b and c are called the coefficients of the quadratic, and b or c may be zero, but a can never be zero.
含一个未知数的二次方程，是指任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是常数且 a ≠ 0。条件 a ≠ 0 非常关键，因为如果 a = 0，x² 项就消失了，方程变成一次方程。常数 a、b、c 称为二次方程的系数，其中 b 或 c 可以为零，但 a 永远不能为零。
o not look quadratic at first and must be rearranged before you can see the standard form. You should also recognise disguised quadratics, such as x⁴ - 5x² + 4 = 0, which is really a quadratic in x².
例如，x² + 3x + 2 = 0 中 a = 1、b = 3、c = 2；2x² - 5 = 0 中 a = 2、b = 0、c = -5；-x² + 4x = 0 中 a = -1、b = 4、c = 0。有些方程一开始看起来不像二次方程，必须先整理才能看出标准形式。你还需要识别伪装的二次方程，例如 x⁴ - 5x² + 4 = 0，它实际上是关于 x² 的二次方程。
2. Writing the Standard Form ax² + bx + c = 0 | 写出标准形式 ax² + bx + c = 0
ample, 3x + x² = 10 becomes x² + 3x - 10 = 0. This standard form makes it easy to identify a, b and c, which are needed in the quadratic formula and in many factorising steps.
解任何二次方程之前，都要把各项移到一边，使另一边为零，并按 x 的降幂排列。例如，3x + x² = 10 应改写为 x² + 3x - 10 = 0。这种标准形式使识别 a、b、c 更加容易，而这些值在求根公式和许多因式分解步骤中都会用到。
Be careful with signs when a term crosses the equals sign. If the leading coefficient is negative, you may multiply the whole equation by -1 to make a positive, which often makes factorising easier. Multiplying every term by -1 does not change the solutions, because the equation is still equal to zero. However, never multiply only one side or only some terms, as this changes the equation completely.
当项移过等号时，要特别小心符号变化。如果二次项系数为负，可以把整个方程乘以 -1，使 a 变为正数，这通常会让因式分解更容易。方程每一项都乘以 -1 不会改变解，因为方程仍然等于零。但是绝不能只乘一边，或只乘部分项，那样会彻底改变方程。
ax² + bx + c = 0, a ≠ 0
3. Solving by Factorising | 因式分解法
that multiply to 6 and add to 5: these are 2 and 3. The equation becomes (x + 2)(x + 3) = 0. Since the product of the two brackets is zero, at least one bracket must be zero, so x + 2 = 0 or x + 3 = 0.
当二次式可以写成两个一次括号的乘积时，因式分解法最快。对于 x² + 5x + 6 = 0，找出两个相乘得 6、相加得 5 的数：2 和 3。方程变为 (x + 2)(x + 3) = 0。因为两个括号的乘积为零，所以至少有一个括号为零，因此 x + 2 = 0 或 x + 3 = 0。
en factorise in pairs to get 2x(x + 3) + 1(x + 3) = 0, so (2x + 1)(x + 3) = 0. The solutions are x = -1/2 and x = -3.
由此得到解 x = -2 和 x = -3。当 a 不为 1 时，因式分解稍难。对于 2x² + 7x + 3 = 0，可拆分中间项：2x² + 6x + x + 3 = 0，再分组分解得 2x(x + 3) + 1(x + 3) = 0，即 (2x + 1)(x + 3) = 0，解为 x = -1/2 和 x = -3。
x² - 4x = 0 factorises as x(x - 4) = 0, giving x = 0 and x = 4. Never divide both sides by x in the second example, because x = 0 is a valid solution and would be lost.
两种特殊模式经常出现。平方差形式 x² - 9 = 0 可分解为 (x - 3)(x + 3) = 0，得 x = 3 和 x = -3。有公因式的情况如 x² - 4x = 0，可分解为 x(x - 4) = 0，得 x = 0 和 x = 4。第二个例子中绝不能两边同时除以 x，因为 x = 0 是有效解，会被丢掉。
4. Completing the Square: The Technique | 配方法：基本技巧
Completing the square rewrites a quadratic expression in the form (x + p)² + q. The key idea is that x² + bx becomes (x + b/2)² - (b/2)². You add and subtract the square of half the coefficient of x, so the value of the expression does not change. This form is very useful because it shows the minimum or maximum value of the quadratic and helps you solve equations that do not factorise.
配方法把二次式改写为 (x + p)² + q 的形式。关键在于 x² + bx 可以写成 (x + b/2)² - (b/2)²。你加上再减去一次项系数一半的平方，因此表达式的值不变。这种形式非常有用，因为它能直接看出二次式的最小值或最大值，也有助于解无法因式分解的方程。
+ 5, write 2(x² + 4x) + 5, then complete the square inside the bracket: 2[(x + 2)² - 4] + 5 = 2(x + 2)² - 3.
例如，x² + 6x + 1 可写成 (x + 3)² - 9 + 1，化简为 (x + 3)² - 8。当 a 不为 1 时，先从前两项中提取 a。对于 2x² + 8x + 5，写成 2(x² + 4x) + 5，再在括号内配方：2[(x + 2)² - 4] + 5 = 2(x + 2)² - 3。
x² + bx + c = (x + b/2)² - (b/2)² + c
5. Solving Equations by Completing the Square | 用配方法解方程
as (x + 3)² - 8 = 0, so (x + 3)² = 8. Taking the square root of both sides gives x + 3 = ±√8, because both the positive and negative square roots are possible. Since √8 = 2√2, the exact solutions are x = -3 + 2√2 and x = -3 - 2√2.
方程化为配方式后，可以通过开平方求解。对于 x² + 6x + 1 = 0，改写为 (x + 3)² - 8 = 0，所以 (x + 3)² = 8。两边开平方得 x + 3 = ±√8，因为正负平方根都可能。由于 √8 = 2√2，精确解为 x = -3 + 2√2 和 x = -3 - 2√2。
5 = 0, use the completed-square form 2(x + 2)² - 3 = 0, so (x + 2)² = 3/2. Then x + 2 = ±√3/2, which can be written as x = -2 ± √6/2 in exact form. Completing the square always works, even when the roots are irrational, so it is a reliable method when factorising fails but you need an exact answer.
如果 a 不为 1，先除以 a 或提取 a。对于 2x² + 8x + 5 = 0，利用配方形式 2(x + 2)² - 3 = 0，得 (x + 2)² = 3/2，于是 x + 2 = ±√3/2，精确形式可写成 x = -2 ± √6/2。配方法总是可行，即使根是无理数也能用，因此当因式分解失败又需要精确答案时，它是可靠的方法。
6. The Quadratic Formula | 求根公式
tandard form ax² + bx + c = 0, identify a, b and c with their signs, then substitute into the formula. The formula comes from completing the square on the general equation ax² + bx + c = 0, so it is not a separate trick but a general result.
求根公式可以解任何二次方程，尤其适用于因式分解困难或无法分解的情况。先把方程写成标准形式 ax² + bx + c = 0，带着符号确定 a、b、c，再代入公式。该公式正是对一般方程 ax² + bx + c = 0 配方得到的，因此它不是孤立的技巧，而是一个一般结论。
x = (-b ± √b² - 4ac) / (2a)
b = 3 and c = -2. Substitute carefully: x = (-3 ± √(3² - 4 × 2 × (-2))) / (2 × 2) = (-3 ± √9 + 16) / 4 = (-3 ± √25 1/2 or x = -8/4 = -2. Always simplify the square root before finding the two final values.
例如解 2x² + 3x - 2 = 0。这里 a = 2、b = 3、c = -2。仔细代入：x = (-3 ± √(3² - 4 × 2 × (-2))) / (2 × 2) = (-3 ± √9 + 16) / 4 = (-3 ± √25) / 4 = (-3 ± 5) / 4。因此 x = 2/4 = 1/2 或 x = -8/4 = -2。一定要先化简根号，再求两个最终值。
7. The Discriminant and the Number of Real Roots | 判别式与实根个数
quation has before you finish solving. If Δ > 0, there are two distinct real roots. If Δ = 0, there is exactly one real root, called a repeated root. If Δ < 0, there are no real roots, because the square root of a negative number is not real.
求根公式中根号下的表达式称为判别式，记作 Δ = b² - 4ac。它能在解完之前告诉你方程有多少个实根。若 Δ > 0，有两个不同的实根；若 Δ = 0，恰有一个实根，称为重根；若 Δ < 0，没有实根，因为负数在实数范围内不能开平方。
+ 5 = 0, Δ = 4 - 20 = -16 < 0, so there are no real roots. On a graph, the number of real roots equals the number of times the curve y = ax² + bx + c crosses the x-axis.
对于 x² + 5x + 6 = 0，Δ = 25 - 24 = 1 > 0，所以有两个实根 -2 和 -3。对于 x² - 6x + 9 = 0，Δ = 36 - 36 = 0，所以有一个重根 x = 3。对于 x² + 2x + 5 = 0，Δ =
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