📚 Mastering Quadratic Equations | 掌握二次方程
Quadratic equations appear frequently in the IGCSE mathematics syllabus. They test your ability to manipulate algebraic expressions, solve equations using different strategies, and interpret graphs. A solid understanding of quadratics is essential for both Foundation and Higher Tier exams.
二次方程在IGCSE数学考纲中频繁出现。它们考查你化简代数表达式、运用不同策略解方程以及解读图像的能力。无论是基础级别还是高级别考试，扎实掌握二次方程都是必不可少的。
1. Standard Form and Key Terms | 标准形式与关键术语
A quadratic expression is written in the general form ax² + bx + c, where a, b and c are constants and a ≠ 0. If a = 0, the expression is linear, not quadratic.
二次表达式的一般形式为 ax² + bx + c，其中 a、b、c 是常数，且 a ≠ 0。若 a = 0，则该表达式是一次式，而非二次式。
The term ax² is called the quadratic term, bx is the linear term, and c is the constant term. In a quadratic equation, the expression is set equal to zero:
ax² 称为二次项，bx 称为一次项，c 称为常数项。在二次方程中，表达式等于零：
ax² + bx + c = 0, a ≠ 0
When solving quadratic equations, you are finding the values of x that make the equation true. These values are called the roots or solutions of the equation.
解二次方程时，你需要找到使方程成立的 x 值，这些值称为方程的根或解。
2. Expanding Double Brackets | 展开双括号
Before factorising quadratics, you must be able to expand brackets. For example, (x + 2)(x + 3) can be expanded using the distributive law:
在因式分解二次式之前，你必须能够展开括号。例如，使用分配律展开 (x + 2)(x + 3)：
(x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6
Notice that the constant term is the product 2 × 3 = 6, and the coefficient of x is the sum 2 + 3 = 5.
注意，常数项是 2 × 3 = 6，x 的系数是 2 + 3 = 5。
When the coefficient of x² is greater than 1, such as (2x + 1)(x + 4), expand each term carefully:
当 x² 的系数大于1时，例如 (2x + 1)(x + 4)，需要仔细逐项展开：
(2x + 1)(x + 4) = 2x² + 8x + x + 4 = 2x² + 9x + 4
- Multiply each term in the first bracket by each term in the second bracket. | 用第一个括号中的每一项分别乘以第二个括号中的每一项。
- Combine like terms before simplifying further. | 合并同类项后再进一步化简。
3. Factorising Quadratic Trinomials | 因式分解二次三项式
Factorising is the reverse of expanding. To factorise x² + 5x + 6, look for two numbers that multiply to 6 and add to 5. These are 2 and 3, so:
因式分解是展开的逆运算。要分解 x² + 5x + 6，找出两个数相乘得6且相加得5。这两个数是2和3，因此：
x² + 5x + 6 = (x + 2)(x + 3)
For expressions like x² - 4x - 12, the two numbers must multiply to -12 and add to -4. The numbers are -6 and 2, so:
对于像 x² - 4x - 12 这样的表达式，两个数相乘得 -12 且相加得 -4。这两个数是 -6 和 2，因此：
x² - 4x - 12 = (x - 6)(x + 2)
If a ≠ 1, for example 2x² + 7x + 3, you can use the "splitting the middle term" method. Multiply a and c 3 = 6. Find two numbers whose product is 6 and sum is 7: 1 and 6. Rewrite the middle term:
如果 a≠1，例如 2x² + 7x + 3，可以使用“拆中项”法。将 a 与 c 相乘：2 × 3 = 6。找出两个数，其积为6且和为7：即1和6。重写中间项：
2x² + 7x + 3 = 2x² + x + 6x + 3
Then group and factorise:
然后分组并因式分解：
2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)
4. Solving by Factorising | 因式分解求解法
can be factorised, you can use the zero-product property: if AB = 0, then A = 0 or B = 0.
如果二次方程可以因式分解，你可以运用零积性质：如果 AB = 0，那么 A = 0 或 B = 0。
For example, solve x² + 5x + 6 = 0. Factorise first:
例如，解方程 x² + 5x + 6 = 0。首先因式分解：
x² + 5x + 6 = (x + 2)(x + 3) = 0
Then set each bracket equal to zero:
然后令每个括号等于零：
x + 2 = 0 ⇒ x = -2
x + 3 = 0 ⇒ x = -3
and x = -3. Always check your roots by substituting them back into the original equation.
因此根为 x = -2 和 x = -3。始终将根代回原方程检验。
5. The Quadratic Formula | 二次公式法
uadratic equation, even when factorising is difficult or impossible. For ax² + bx + c = 0,
二次公式可以求解任何二次方程，即使难以因式分解或无法因式分解时也适用。对于 ax² + bx + c = 0，
x = (-b ± √b² - 4ac) / (2a)
rmula gives two solutions because of the ± sign. For example, solve 2x² - 3x - 2 = 0. Here a = 2, b = -3, and c = -2.
由于 ± 符号，该公式给出两个解。例如，解 2x² - 3x - 2 = 0。这里 a = 2，b = -3，c = -2。
x = (3 ± √((-3)² - 4 × 2 × (-2))) / (2 × 2) = (3 ± √9 + 16) / 4 = (3 ± 5) / 4
Therefore x = (3 + 5)/4 = 2, or x = (3 - 5)/4 = -0.5. The roots are 2 and -0.5.
因此 x = (3 + 5)/4 = 2，或 x = (3 - 5)/4 = -0.5。根为2和-0.5。
6. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. This is useful for finding turning points and solving equations.
配方法将二次式改写为 (x + p)² + q 的形式。这在求顶点和求解方程时非常有用。
Start with x² + bx. Take half of b, square it, and add and subtract it. For x² + 6x + 2:
从 x² + bx 开始。取 b 的一半，计算平方，然后加上并减去该值。对于 x² + 6x + 2：
x² + 6x + 2 = (x + 3)² - 9 + 2 = (x + 3)² - 7
Notice that (x + 3)² expands to x² + 6x + 9, so we subtract 9 to keep the expression equivalent.
注意 (x + 3)² 展开为 x² + 6x + 9，因此我们减去9以保持表达式等价。
To solve x² + 6x + 2 = 0, rewrite as (x + 3)² - 7 = 0, then:
要解 x² + 6x + 2 = 0，将其改写为 (x + 3)² - 7 = 0，然后：
(x + 3)² = 7 ⇒ x + 3 = ±√7 ⇒ x = -3 ± √7
So the exact solutions are x = -3 + √7 and x = -3 - √7.
因此精确解为 x = -3 + √7 和 x = -3 - √7。
7. The Discriminant and Nature of Roots | 判别式与根的性质
The expression b² - 4ac inside the quadratic formula is called the discriminant. It determines how many real roots a quadratic equation has.
二次公式中的表达式 b² - 4ac 称为判别式，它决定二次方程有多少个实根。
| Discriminant (Δ) | 判断式（Δ） | Nature of roots | 根的性质 |
| Δ = b² - 4ac > 0 | Two distinct real roots | 两个不相等的实数根 |
| Δ = 0 | One repeated real root (equal roots) | 一个重根（相等的实数根） |
| Δ < 0 | No real roots | 没有实数根 |
0 has Δ = 16 - 20 = -4 < 0, so it has no real roots. Its graph does not cross the x-axis.
例如，方程 x² + 4x + 5 = 0 的判别式 Δ = 16 - 20 = -4 < 0，因此没有实数根。其图像不与 x 轴相交。
8. Graphs of Quadratic Functions | 二次函数的图像
The graph of y = ax² + bx + c is a parabola. If a > 0, it opens upwards and has a minimum point; if a < 0, it opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，开口向上，有最小值点；如果 a < 0，开口向下，有最大值点。
The x-intercepts are the roots of the equation. If the discriminant is positive, the graph crosses the x-axis at two points; if zero, it touches at one point; if negative, it does not cross.
x 轴截距是方程的根。若判别式为正，图像与 x 轴交于两点；若为零，则与 x 轴相切于一点；若为负，则不相交。
- The y-intercept is the point (0, c). | y 轴截距是点 (0, c)。
- The line of symmetry is x = -b/(2a). | 对称轴是 x = -b/(2a)。
Sketch the graph by marking the roots, the y-intercept, and the turning point.
画图时标出根、y 轴截距和顶点。
9. Finding the Turning Point | 求顶点坐标
can be found by completing the square. For y = (x + p)² + q, the turning point is (-p, q).
二次函数图像的顶点可以通过配方法求得。对于 y = (x + p)² + q，顶点为 (-p, q)。
rite y = x² - 2x + 3 as y = (x - 1)² + 2. Therefore the turning point is (1, 2), and since a = 1 > 0, it is a minimum point.
例如，将 y = x² - 2x + 3 改写为 y = (x - 1)² + 2。因此顶点为 (1, 2)，因为 a = 1 > 0，所以该点是最低点。
Alternatively, the x-coordinate of the turning point is -b/(2a). Substitute this value into the equation to find the y-coordinate.
另一种方法是，顶点的 x 坐标为 -b/(2a)，将此值代入方程可求得 y 坐标。
For y = 2x² + 4x - 1, x = -4/(4) = -1, y = 2(-1)² + 4(-1) - 1 = -3, so vertex = (-1, -3)
对于 y = 2x² + 4x - 1，x = -4/4 = -1，y = 2(-1)² + 4(-1) - 1 = -3，所以顶点为 (-1, -3)。
10. Problem Solving with Quadratics | 二次方程应用题
Many real-world problems can be modelled by quadratic equations. For example, the area of a rectangle, projectile motion, or number puzzles involving consecutive integers.
许多现实问题可以用二次方程建模，例如矩形面积、抛体运动或连续整数数字谜题。
Example: The product of two consecutive positive integers is 56. Let the integers be n and n + 1. Then:
例子：两个连续正整数的乘积是56。设整数为 n 和 n + 1，则：
n(n + 1) = 56 ⇒ n² + n - 56 = 0
Factorise or use the formula:
用因式分解或公式法求解：
n² + n - 56 = (n + 8)(n - 7) = 0
ositive, n = 7 and n + 1 = 8. Always interpret the solution in the context of the problem.
因此 n = -8 或 n = 7。由于整数为正，所以 n = 7，n + 1 = 8。务必结合问题情境解释答案。
When solving word problems, define your variable, write an equation, solve it, and then check that the answer makes sense in the original context.
解决应用题时，先设定变量，写出方程，求解，然后检查答案在原始情境中是否有意义。
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