📚 IGCSE Mathematics: Quadratic Equations and Graphs | IGCSE 数学：二次方程与图像
In IGCSE Mathematics, quadratic equations connect algebra, coordinate geometry and problem solving. This revision guide covers factorising, the quadratic formula, completing the square, the discriminant and graph shapes.
在 IGCSE 数学中，二次方程把代数、坐标几何和问题求解连接在一起。本复习指南涵盖因式分解、二次公式、配方法、判别式以及图像形状。
1. Quadratic Expressions and Standard Form | 二次表达式与标准式
A quadratic expression is an algebraic expression of the form ax² + bx + c, where a, b and c are constants and a ≠ 0. The term ax² is the quadratic term, bx is the linear term, and c is the constant term.
二次表达式是形如 ax² + bx + c 的代数式，其中 a、b、c 是常数且 a ≠ 0。ax² 是二次项，bx 是一次项，c 是常数项。
If a = 0, the expression becomes linear rather than quadratic. In standard form, terms are written in descending powers of x, with the x² term first.
如果 a = 0，表达式就变成一次式而不是二次式。在标准式中，各项按 x 的降幂排列，x² 项在最前。
2. Expanding Double Brackets | 展开双括号
To expand (x + p)(x + q), multiply each term in the first bracket by each term in the second. The FOIL method helps: multiply First, Outer, Inner and Last terms, then collect like terms.
展开 (x + p)(x + q) 时，用第一个括号中的每一项乘以第二个括号中的每一项。FOIL 方法有助于操作：先乘首项、外项、内项和末项，然后合并同类项。
Example: (x + 3)(x – 5) = x² – 5x + 3x – 15 = x² – 2x – 15. The middle term comes from the sum of the outer and inner products.
示例：(x + 3)(x – 5) = x² – 5x + 3x – 15 = x² – 2x – 15。中间项来自外项乘积与内项乘积之和。
(x + p)(x + q) = x² + (p + q)x + pq
3. Factorising Simple Quadratics | 简单二次式的因式分解
Factorising is the reverse of expanding. For x² + bx + c, find two numbers that multiply to give c and add to give b. If c is positive, both numbers have the same sign; if c is negative, they have opposite signs.
因式分解是展开的逆运算。对于 x² + bx + c，找到两个数，它们相乘得 c，相加得 b。如果 c 为正，两个数同号；如果 c 为负，两个数异号。
Example: x² + 7x + 10 = (x + 2)(x + 5), because 2 × 5 = 10 and 2 + 5 = 7. Always expand the brackets to check your answer.
示例：x² + 7x + 10 = (x + 2)(x + 5)，因为 2 × 5 = 10 且 2 + 5 = 7。始终展开括号来检验答案。
x² + bx + c = (x + m)(x + n), where m + n = b and mn = c
4. Difference of Two Squares | 平方差
The pattern a² – b² factorises as (a + b)(a – b). It has no middle term because the cross terms -ab and +ab cancel out.
平方差公式 a² – b² 分解为 (a + b)(a – b)。它没有中间项，因为交叉项 -ab 和 +ab 相互抵消。
Example: x² – 16 = (x + 4)(x – 4). This works for any terms that are perfect squares, such as 9x² – 25 = (3x + 5)(3x – 5).
示例：x² – 16 = (x + 4)(x – 4)。这适用于任何完全平方项，例如 9x² – 25 = (3x + 5)(3x – 5)。
a² – b² = (a + b)(a – b)
5. Solving Quadratics by Factorising | 通过因式分解解二次方程
Set the quadratic equal to zero, factorise the left side, then use the zero-product rule: if pq = 0, then p = 0 or q = 0. This gives the roots or solutions of the equation.
将二次方程设为零，对左边进行因式分解，然后使用零乘积规则：如果 pq = 0，则 p = 0 或 q = 0。这样就得到方程的根或解。
Example: solve x² – 5x + 6 = 0. Factorise as (x – 2)(x – 3) = 0, so x – 2 = 0 or x – 3 = 0, giving x = 2 or x = 3.
示例：解 x² – 5x + 6 = 0。分解为 (x – 2)(x – 3) = 0，因此 x – 2 = 0 或 x – 3 = 0，得 x = 2 或 x = 3。
Do not divide both sides by x, because this can remove a valid root at x = 0. Instead, rearrange so all terms are on one side.
不要两边同时除以 x，因为这样可能会去掉 x = 0 的有效根。相反，应把所有项移到一边。
6. The Quadratic Formula | 二次公式
If factorising is difficult, solve ax² + bx + c = 0 using the quadratic formula. The symbol ± means there are two possible answers: one using plus and one using minus.
如果因式分解困难，可用二次公式解 ax² + bx + c = 0。符号 ± 表示有两个可能的答案：一个取加号，一个取减号。
x = [-b ± √(b² – 4ac)] ÷ 2a
Example: for 2x² – 3x – 2 = 0, substitute a = 2, b = -3 and c = -2. Use brackets around negative signs to avoid errors.
示例：对于 2x² – 3x – 2 = 0，代入 a = 2、b = -3 和 c = -2。在负号周围使用括号以避免错误。
The formula gives x = [-(-3) ± √((-3)² – 4(2)(-2))] ÷ 2(2), which simplifies to x = 2 or x = -½.
公式给出 x = [-(-3) ± √((-3)² – 4(2)(-2))] ÷ 2(2)，化简后得 x = 2 或 x = -½。
7. Completing the Square | 配方法
Completing the square rewrites ax² + bx + c as a(x + h)² + k. This form makes it easier to find the turning point and solve the equation.
配方法将 ax² + bx + c 改写为 a(x + h)² + k 的形式。这种形式更容易求顶点和求解方程。
Example: x² – 6x + 4 = (x – 3)² – 9 + 4 = (x – 3)² – 5. The completed-square form shows the minimum value is -5 at x = 3.
示例：x² – 6x + 4 = (x – 3)² – 9 + 4 = (x – 3)² – 5。配方法形式表明最小值在 x = 3 处为 -5。
x² + bx = (x + b/2)² – (b/2)²
8. The Discriminant | 判别式
The discriminant Δ = b² – 4ac determines the number of real roots without solving the full equation. It is the expression under the square root in the quadratic formula.
判别式 Δ = b² – 4ac 无需完整求解方程即可判断实数根的个数。它是二次公式中平方根下的表达式。
If Δ > 0, there are two distinct real roots. If Δ = 0, there is one repeated real root. If Δ < 0, there are no real roots because the square root of a negative number is not real.
如果 Δ > 0，有两个不同的实数根。如果 Δ = 0，有一个重根。如果 Δ < 0，没有实数根，因为负数的平方根不是实数。
| Δ > 0 | Two distinct real roots | 两个不同实数根 |
| Δ = 0 | One repeated real root | 一个重根 |
| Δ < 0 | No real roots | 没有实数根 |
9. Graphs of Quadratic Functions | 二次函数图像
The graph of y = ax² + bx + c is a parabola. If a > 0, the parabola opens upward like a U-shape. If a < 0, it opens downward like an inverted U-shape.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线向上开口，呈 U 形。如果 a < 0，它向下开口，呈倒 U 形。
The y-intercept is c, because at x = 0 the value of y is c. The x-intercepts, or roots, occur where y = 0 and can be found by solving ax² + bx + c = 0.
y 轴截距为 c，因为当 x = 0 时 y = c。x 轴截距即根，出现在 y = 0 处，可通过解 ax² + bx + c = 0 得到。
A sketch should show the shape, the y-intercept, the roots and the turning point. No plotting of many points is usually needed in IGCSE exam sketches.
草图中应标出形状、y 轴截距、根和顶点。在 IGCSE 考试草图中通常不需要描很多点。
10. Turning Point and Symmetry | 顶点与对称性
The turning point is the minimum point when a > 0 and the maximum point when a < 0. Its x-coordinate is given by x = -b ÷ 2a.
顶点是 a > 0 时的最小值点，也是 a < 0 时的最大值点。它的 x 坐标为 x = -b ÷ 2a。
x = -b ÷ 2a
A parabola is symmetric about the vertical line x = -b ÷ 2a. If two roots are known, their midpoint equals the x-coordinate of the turning point.
抛物线关于竖直直线 x = -b ÷ 2a 对称。如果已知两个根，它们的中点等于顶点 x 坐标。
Substitute x = -b ÷ 2a back into y = ax² + bx + c to find the y-coordinate of the turning point.
将 x = -b ÷ 2a 代回 y = ax² + bx + c 可求得顶点的 y 坐标。
11. Quadratic Inequalities | 二次不等式
To solve ax² + bx + c > 0 or ax² + bx + c < 0, first find the roots by solving the equality. Then sketch the parabola and identify where the graph lies above or below the x-axis.
要解 ax² + bx + c > 0 或 ax² + bx + c < 0，首先解等式求出根。然后画出抛物线草图，判断图像位于 x 轴上方或下方的区间。
Example: x² – 4 > 0 factorises as (x – 2)(x + 2) > 0. Since the parabola opens upward, the solution is x < -2 or x > 2.
示例：x² – 4 > 0 分解为 (x – 2)(x + 2) > 0。由于抛物线开口向上，解为 x < -2 或 x > 2。
If a > 0 and the inequality is < 0, the solution is between the two roots. If a > 0 and the inequality is > 0, the solution lies outside the roots.
如果 a > 0 且不等式为 < 0，解在两个根之间。如果 a > 0 且不等式为 > 0，解在两个根之外。
12. Applications and Common Mistakes | 应用与常见错误
Quadratic equations model area problems, projectile motion, profit functions and optimisation. Define variables clearly before forming an equation, and write the equation in standard form.
二次方程可用于对面积问题、抛体运动、利润函数和最优化进行建模。在建立方程前要明确定义变量，并将方程写成标准式。
Common mistakes include losing a negative sign, forgetting to set the equation to zero before factorising, and selecting the wrong values for a, b and c in the quadratic formula.
常见错误包括丢失负号、因式分解前忘记将方程设为零，以及在二次公式中代入错误的 a、b、c 值。
Always check final solutions by substituting them back into the original equation. In graph questions
Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导