Mastering Quadratic Equations | 掌握二次方程

📚 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。

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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。

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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。

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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: 2 × 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)

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4. Solving by Factorising | 因式分解求解法

If a quadratic equation 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

So the roots are x = -2 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。

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5. The Quadratic Formula | 二次公式法

The quadratic formula can solve any quadratic 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)

This formula 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。

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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。

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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

For example, the equation x² + 4x + 5 = 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。

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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。

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9. Finding the Turning Point | 求顶点坐标

The turning point of a quadratic graph 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)。

For example, rewrite 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)。

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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

So n = -8 or n = 7. Since the integers are positive, 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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