📚 Algebra 7: Quadratic Equations | 代数7：二次方程
In IGCSE Algebra 7, we move into one of the most powerful topics in secondary mathematics: quadratic equations. A quadratic equation is any polynomial equation of the second degree. Mastering the ways to solve, analyse and apply quadratics gives you essential tools for graphs, functions, and real‑world modelling. This article covers factorising, completing the square, the quadratic formula, the discriminant, graphical interpretation, inequalities and word problems – all carefully linked to the Edexcel IGCSE specification.
在 IGCSE 代数 7 中，我们进入中学数学最重要的主题之一：二次方程。二次方程即二次多项式方程。掌握二次方程的求解、分析和应用方法，将为你学习图像、函数和现实建模提供关键工具。本文涵盖因式分解法、配方法、二次公式、判别式、图像解法、不等式和应用题——所有内容均紧扣 Edexcel IGCSE 大纲。
1. What is a Quadratic Equation? | 什么是二次方程？
nts and a ≠ 0. The expression ax² is the quadratic term, bx is the linear term, and c is the constant term.
一元二次方程可以写成标准形式 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。ax² 是二次项，bx 是一次项，c 是常数项。
Quadratic equations arise naturally from problems involving area, projectile motion, and many geometric situations. Because the highest power of x is 2, a quadratic equation will generally have two solutions (which may be real and distinct, repeated, or not real at all).
二次方程自然地出现在面积、抛体运动和许多几何问题中。由于 x 的最高次幂是 2，二次方程通常有两个解（可能为不等实根、等根或无实根）。
The process of solving means finding the values of x that satisfy the equation. IGCSE requires fluency with three algebraic methods, plus graphical interpretation.
求解过程即是找出满足方程的 x 值。IGCSE 要求熟练掌握三种代数方法，并理解图像解法。
2. Solving by Factorising | 因式分解法求解
setting each bracket equal to zero. For example, solve x² + 5x + 6 = 0. Factorise the left‑ (x + 2)(x + 3) = 0. So either x + 2 = 0 or x + 3 = 0. The solutions are x = −2 and x = −3.
当二次式可以写成两个一次因式的乘积时，我们可以让每个因式等于零来解方程。例如，解 x² + 5x + 6 = 0。将左边因式分解：(x + 2)(x + 3) = 0。于是有 x + 2 = 0 或 x + 3 = 0。解为 x = −2 和 x = −3。
our factors by expanding: (x+2)(x+3)=x²+3x+2x+6=x²+5x+6, which confirms the factorisation.
进行因式分解前，务必将方程整理为 ax² + bx + c = 0 的形式。通过展开可以检验因式分解是否正确：(x+2)(x+3)=x²+3x+2x+6=x²+5x+6。
m: 2x² + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3)=0. Solutions are x = −1/2 and x = −3.
若 a ≠ 1，则需要更仔细地分解。对于 2x² + 7x + 3 = 0，找两个数，积为 (2×3)=6，和为 7。这两个数是 6 和 1。将一次项拆分：2x² + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3)=0。解为 x = −1/2 和 x = −3。
3. Completing the Square | 配方法
Completing the square transforms a quadratic into a perfect square plus a constant. The method is essential for understanding the derivation of the quadratic formula and for rewriting quadratic functions in vertex form. For a monic quadratic x² + bx, the completed square is (x + b/2)² − (b/2)².
配方法将二次式转化为一个完全平方加上一个常数。这一方法对于理解二次公式的推导以及将二次函数化为顶点式至关重要。对于首项系数为 1 的二次式 x² + bx，配方得 (x + b/2)² − (b/2)²。
ing the square. Write x² + 6x = (x + 3)² − 9. So the equation becomes (x + 3)² − 9 + 5 = 0 ⇒ (x + 3)² − 4 = 0 ⇒ (x + 3)² = 4. Taking square roots: x + 3 = ±2, giving x = −3 ± 2, so x = −1 or x = −5.
示例：用配方法解 x² + 6x + 5 = 0。配方 x² + 6x = (x + 3)² − 9。于是方程变为 (x + 3)² − 9 + 5 = 0 ⇒ (x + 3)² − 4 = 0 ⇒ (x + 3)² = 4。开平方：x + 3 = ±2，得 x = −3 ± 2，即 x = −1 或 x = −5。
= (x+2)² − 4, so 2[(x+2)² − 4] + 5 = 2(x+2)² − 8 + 5 = 2(x+2)² − 3 = 0. Then solve for x.
当 x² 的系数不为 1 时，首先将该系数从 x² 与 x 项中提出来。对于 2x² + 8x + 5 = 0，写成 2[x² + 4x] + 5 = 0，在括号内配方：x² + 4x = (x+2)² − 4，于是 2[(x+2)² − 4] + 5 = 2(x+2)² − 8 + 5 = 2(x+2)² − 3 = 0。然后解出 x。
4. The Quadratic Formula | 二次公式法
The quadratic formula gives the solutions of ax² + bx + c = 0 directly:
x = (−b ± √b² − 4ac) / (2a)
二次公式可直接给出 ax² + bx + c = 0 的解：
x = (−b ± √b² − 4ac) / (2a)
For 2x² + 5x − 3 = 0, a = 2, b = 5, c = −3. Then x = (−5 ± √(25 − 4(2)(−3))) / (4) = (−5 ± √25 + 24) / 4 = (−5 ± √49) / 4 = (−5 ± 7) / 4. So x = (2)/4 = 1/2 or x = (−12)/4 = −3.
该公式适用于任何二次方程，即使那些无法整整齐齐因式分解的方程也能用。代入 a、b、c 的值时务必仔细，特别注意正负号。对于 2x² + 5x − 3 = 0，a = 2，b = 5，c = −3。则 x = (−5 ± √(25 − 4(2)(−3))) / (4) = (−5 ± √25 + 24) / 4 = (−5 ± √49) / 4 = (−5 ± 7) / 4。因此 x = (2)/4 = 1/2 或 x = (−12)/4 = −3。
You must memorise the quadratic formula and be able to quote it accurately. IGCSE exams often ask you to give your answer correct to 3 significant figures or in surd form using the formula.
你必须熟记二次公式并能准确写出。IGCSE 试题常要求使用公式将答案保留三位有效数字或以根式形式给出。
5. The Discriminant (Δ) | 判别式 (Δ)
Δ = b² − 4ac. It tells us the nature of the roots without solving the equation completely.
判别式即平方根号内的部分：Δ = b² − 4ac。它无需完全解出方程就能告诉我们根的性质。
one real root (a repeated root). If Δ < 0, there are no real roots (the solutions are complex, which is not covered in IGCSE).
若 Δ > 0，二次方程有两个不等实根。若 Δ = 0，恰好有一个实根（重根）。若 Δ < 0，无实根（方程有复数解，IGCSE 不作要求）。
= 2. For 2x² + x + 1 = 0, Δ = 1² − 4·2·1 = 1 − 8 = −7 < 0, so there are no real solutions.
示例：对于 x² − 4x + 4 = 0，Δ = (−4)² − 4·1·4 = 16 − 16 = 0，故方程在 x = 2 处有重根。对于 2x² + x + 1 = 0，Δ = 1² − 4·2·1 = 1 − 8 = −7 < 0，因此无实数解。
Questions may ask you to find the range of values of k for which a quadratic has two distinct real roots. Set Δ > 0 and solve the inequality.
考题可能要求你找出使二次方程有两个不等实根的 k 的取值范围。此时需令 Δ > 0 并解不等式。
6. Graphical Solutions | 图像解法
h of y = ax² + bx + c is a parabola. The solutions of ax² + bx + c = 0 correspond to the x‑intercepts of the graph (where y = 0). If the parabola crosses the x‑axis twice, there are two real distinct roots; if it touches the x‑axis once, there is one repeated root; if it does not meet the x‑axis, there are no real roots.
y = ax² + bx + c 的图像是一条抛物线。方程 ax² + bx + c = 0 的解对应于图像与 x 轴的交点（即 y = 0 处）。如果抛物线与 x 轴相交于两点，则有两个不等实根；若与 x 轴相切于一点，则有一个重根；若与 x 轴无交点，则无实根。
nd where y = 0, or to add a straight line to solve a related equation like x² + x − 2 = 0.
IGCSE 题目常要求绘制合适图像或利用给定图像估算解。你可能需要描点并求出 y = 0 的位置，或添加一条直线来解相关方程，比如 x² + x − 2 = 0。
Using graphs helps develop deeper understanding of the connection between algebraic and geometric representations of quadratics.
利用图像有助于深入理解二次方程代数表示与几何表示之间的联系。
7. Word Problems Leading to Quadratic Equations | 应用题转化为二次方程
Many practical problems yield quadratic equations. Common contexts include area problems, Pythagoras’ theorem or consecutive integer problems. The key steps are: assign a variable to the unknown, translate the words into an equation, solve the quadratic, and interpret the solutions in context (discarding any that are not physically possible, such as negative lengths).
许多实际问题会归结为二次方程。常见的背景有面积问题、勾股定理或连续整数问题。关键步骤是：为未知数设定变量，将文字描述转化为方程，解二次方程，并在实际问题背景下解释解的意义（舍去没有物理意义的解，如负的长度）。
its width. Its area is 40 m². Let width be x m, then length = x + 3. Area = x(x + 3) = 40 ⇒ o x = −8 or x = 5. Since width cannot be negative, the width is 5 m and the length is 8 m.
示例：一个矩形花园的长比宽多 3 m，面积为 40 m²。设宽为 x m，则长为 x + 3。面积 = x(x + 3) = 40 ⇒ x² + 3x − 40 = 0。因式分解：(x + 8)(x − 5) = 0，因此 x = −8 或 x = 5。由于宽不能为负，故宽为 5 m，长为 8 m。
In projectile motion, the height of an object can often be modelled by a quadratic equation in time t. Setting height equal to a certain value gives a quadratic to solve for t.
在抛体运动中，物体的高度常可用关于时间 t 的二次方程建模。令高度等于某一特定值即得到一个关于 t 的二次方程。
8. Quadratic Inequalities | 二次不等式
= 1 and x = 3. Use a sign diagram or sketch the parabola y = x² − 4x + 3. The graph is a U‑shaped parabola crossing the x‑is at 1 and 3. The expression is negative between the roots, so the solution is 1 < x < 3.
二次不等式如 x² − 4x + 3 < 0 需要求出使二次式为负的 x 范围。先解相应方程 x² − 4x + 3 = 0：因式分解为 (x − 1)(x − 3) = 0，根为 x = 1 和 x = 3。用符号图或画出抛物线 y = x² − 4x + 3。图像是开口向上的抛物线，与 x 轴交于 1 和 3。二次式在两根之间为负，因此解为 1 < x < 3。
r to include the boundary points (the roots) in your solution set if the inequality allows equality.
若不等式为 ≥ 0 或 ≤ 0，且不等号允许取等，则需在解集中包含边界点（即根）。
Sketching a quick graph is often the most reliable method for solving quadratic inequalities, as it clearly shows where the parabola lies above or below the x‑axis.
快速画出简图通常是解二次不等式最可靠的方法，因为它清晰地显示了抛物线位于 x 轴上方或下方的区域。
9. Hidden Quadratics (Equations Reducible to Quadratics) | 隐藏二次方程（可化为二次方程的方程）
then the equation becomes y² − 5y + 4 = 0, a quadratic in y. Factorise: (y − 1)(y − 4) = 0 ⇒ y = 1 or y = 4. Replace y with x²: x² = 1 ⇒ x = ±1; x² = 4 ⇒ x = ±2. So there are four solutions: ±1, ±2.
一些并非明显二次的方程可以通过代换转化为二次方程。例如，x⁴ − 5x² + 4 = 0 是四次方程，但令 y = x²，则方程变为 y² − 5y + 4 = 0，这是一个关于 y 的二次方程。因式分解：(y − 1)(y − 4) = 0 ⇒ y = 1 或 y = 4。将 y 代回 x²：x² = 1 ⇒ x = ±1；x² = 4 ⇒ x = ±2。因此该方程有四个解：±1、±2。
mmon denominator x(x+1)(x−1) to clear fractions, then simplify to a quadratic equation and solve, checking for any extraneous solutions that make a denominator zero.
类似地，含有代数分式的方程有时可通过乘以公分母化为二次方程。例如：1/(x+1) + 1/(x−1) = 2/x。两边乘以公分母 x(x+1)(x−1) 去分母，化简为一个二次方程并求解，需检验是否有使分母为零的增根。
Always verify solutions in the original equation, especially after squaring or cross‑multiplying, as these steps can introduce false roots.
务必在原方程中检验所得解，尤其是在平方或交叉相乘之后，因为这些步骤可能引入多余的根。
10. Common Mistakes and Exam Tips | 常见错误与考试提示
One frequent error is forgetting that a quadratic must be equal to zero before factorising. Do not divide by x or a variable factor; instead bring all terms to one side. Another common slip is mishandling signs when substituting into the quadratic formula or completing the square – always use brackets around negative numbers.
一个常见错误是忘记在因式分解前将方程化为等于零的形式。不要两边除以 x 或含有变量的因式；而应把所有项移到一边。另一个常见疏漏是在代入二次公式或配方时弄错符号——务必在负数外加括号。
When solving word problems, always check whether your solution makes sense in the real‑world context. A negative time or length is typically discarded. Also, simplify square roots completely: √50 = 5√2, and present your final answer clearly.
在解答应用题时，一定要检查所得解在实际背景下是否有意义。负的时间或长度通常应舍去。此外，要彻底化简平方根：√50 = 5√2，并清晰呈现最终答案。
In the exam, show all steps for factorising and using the formula. Even if factorisation fails, the quadratic formula will always give the answer – be confident using it.
在考试中，展示因式分解和使用公式的所有步骤。即便因式分解不成功，二次公式总能给出答案——要大胆使用它。
e the link between discriminant and the graph: if you are given a graph that touches the x‑axis, you instantly know Δ = 0. These connections save time.
复习判别式与图像的联系：如果题目给出的图像与 x 轴相切，你立刻可以知道 Δ = 0。这些联系能节省时间。
11. Summary | 小结
Quadratic equations are central to IGCSE Algebra 7. You must be able to solve them by factorising, completing the square and the quadratic formula, interpret the discriminant, solve inequalities and recognise hidden quadratics. Regular practice of mixed problems will build the speed and accuracy needed for the exam.
二次方程是 IGCSE 代数 7 的核心内容。你必须能够用因式分解法、配方法和二次公式去解二次方程，理解判别式的意义，求解二次不等式并识别隐藏二次方程。定期进行混合练习将提高解题速度和准确性，应对考试需求。
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