Solving Quadratic Equations and Graphs | 解二次方程与图像

📚 Solving Quadratic Equations and Graphs | 解二次方程与图像

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, word problems, and higher-level calculus. This article gives you a complete, exam-focused guide to solving quadratic equations and understanding their graphs for the Edexcel IGCSE syllabus.

二次方程是 IGCSE 数学中最重要的话题之一。它们出现在代数、图像、应用题以及更高阶的微积分中。本文针对 Edexcel IGCSE 考纲，为你提供一份完整且紧扣考点的“解二次方程与函数图像”指南。

1. Expanding and Factorising Quadratics | 展开与因式分解二次式

Before solving, you must be fluent in expanding and factorising. A quadratic expression has the general form ax² + bx + c, where a, b, c are constants and a ≠ 0.

在解方程之前，你必须熟练掌握展开和因式分解。二次表达式的一般形式为 ax² + bx + c，其中 a、b、c 为常数，且 a ≠ 0。

Expanding means removing brackets. For example: (x + 3)(x – 2) = x² + x – 6. This is the distributive law.

展开意味着去掉括号。例如：(x + 3)(x – 2) = x² + x – 6，也就是使用分配律。

Factorising is the reverse process. To factorise x² + bx + c, find two numbers that multiply to c and add to b.

因式分解是相反的过程。对于 x² + bx + c，需要找到两个数，它们相乘等于 c，相加等于 b。

Example: Factorise x² – 5x + 6. We need numbers that multiply to 6 and add to -5. These are -2 and -3, so x² – 5x + 6 = (x – 2)(x – 3).

例如：分解 x² – 5x + 6。需要找两个数相乘为 6、相加为 -5。它们是 -2 和 -3，因此 x² – 5x + 6 = (x – 2)(x – 3)。

When a ≠ 1, use the method of grouping or the “ac” method. For instance, factorise 2x² + 7x + 3. Multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: 1 and 6. Rewrite 7x as 1x + 6x, then factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

当 a ≠ 1 时，可以使用分组法或 “ac” 法。例如，分解 2x² + 7x + 3。先算 a × c：2 × 3 = 6。找到两个数相乘为 6、相加为 7：1 和 6。把 7x 改写为 1x + 6x，再分组分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。

2. Solving by Factorisation | 用因式分解法解方程

A quadratic equation has the form ax² + bx + c = 0. The key principle is: if A × B = 0, then A = 0 or B = 0.

二次方程的形式为 ax² + bx + c = 0。关键原理是：如果 A × B = 0，那么 A = 0 或 B = 0。

Example: Solve x² – 5x + 6 = 0. Factorise first: (x – 2)(x – 3) = 0. Therefore x – 2 = 0 or x – 3 = 0, giving x = 2 or x = 3.

例如：解方程 x² – 5x + 6 = 0。先因式分解：(x – 2)(x – 3) = 0。因此 x – 2 = 0 或 x – 3 = 0，得到 x = 2 或 x = 3。

Check your solutions by substituting them back into the original equation. For x = 2: 4 – 10 + 6 = 0. For x = 3: 9 – 15 + 6 = 0. Both work.

将解代回原方程进行检验。当 x = 2 时：4 – 10 + 6 = 0。当 x = 3 时：9 – 15 + 6 = 0。两个都成立。

Sometimes you must first rearrange the equation so that one side equals zero. For example, x² = 4x gives x² – 4x = 0, then x(x – 4) = 0, so x = 0 or x = 4. Never divide by x, because you lose the solution x = 0.

有时需要先移项使一边等于零。例如，x² = 4x 化为 x² – 4x = 0，再得 x(x – 4) = 0，所以 x = 0 或 x = 4。切勿两边同时除以 x，因为你会丢失 x = 0 这个解。

3. Solving by the Quadratic Formula | 用求根公式解方程

When factorisation is difficult or impossible, use the quadratic formula. For ax² + bx + c = 0, the solutions are given by:

当因式分解困难或无法进行时，使用求根公式。对于 ax² + bx + c = 0，解为：

x = (−b ± √(b² − 4ac)) / 2a

Here “±” means you calculate two values, one using “+” and one using “−”.

此处的 “±” 表示你需要计算两个值，一个用 “+”，另一个用 “−”。

Example: Solve 2x² + 3x – 5 = 0 using the formula. Here a = 2, b = 3, c = −5.

例如：用公式解 2x² + 3x – 5 = 0。这里 a = 2，b = 3，c = −5。

Substitute into the formula: x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2) = (−3 ± √(9 + 40)) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4.

代入公式：x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2) = (−3 ± √(9 + 40)) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4。

So x = (−3 + 7) / 4 = 1, or x = (−3 − 7) / 4 = −2.5.

因此 x = (−3 + 7) / 4 = 1，或 x = (−3 − 7) / 4 = −2.5。

Always write down the values of a, b, c before substituting. This avoids careless mistakes.

代入前务必先写出 a、b、c 的值，这能避免粗心错误。

4. Solving by Completing the Square | 用配方法解方程

Completing the square rewrites a quadratic in the form a(x + p)² + q. This is especially useful for finding turning points and solving equations.

配方法将二次式改写为 a(x + p)² + q 的形式。这对于找到顶点和解方程特别有用。

For x² + bx, add and subtract (b/2)². Example: x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7.

对于 x² + bx，需要加上并减去 (b/2)²。例如：x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7。

To solve (x + 3)² − 7 = 0, rearrange to (x + 3)² = 7, then square root both sides: x + 3 = ±√7, so x = −3 ± √7.

解方程 (x + 3)² − 7 = 0 时，移项得 (x + 3)² = 7，然后两边开平方：x + 3 = ±√7，因此 x = −3 ± √7。

If the coefficient of x² is not 1, factor it out first. For example, 2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3.

如果 x² 的系数不为 1，先把它提取出来。例如，2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3。

You can solve from here: 2(x − 2)² − 3 = 0 → (x − 2)² = 3/2 → x = 2 ± √(3/2).

由此继续求解：2(x − 2)² − 3 = 0 → (x − 2)² = 3/2 → x = 2 ± √(3/2)。

5. The Discriminant | 判别式

The discriminant is the part under the square root in the quadratic formula: Δ = b² − 4ac. It tells us how many roots a quadratic equation has.

判别式是求根公式中根号下的部分：Δ = b² − 4ac。它告诉我们二次方程有多少个根。

If Δ > 0, the equation has two distinct real roots. In the graph, the curve crosses the x-axis at two points.

如果 Δ > 0，方程有两个不同的实数根。在图像上，曲线与 x 轴有两个交点。

If Δ = 0, the equation has exactly one repeated real root. The graph touches the x-axis at one point (the turning point lies on the axis).

如果 Δ = 0，方程有一个重复的实数根。图像与 x 轴相切于一点（顶点在轴上）。

If Δ < 0, the equation has no real roots. The graph does not intersect the x-axis.

如果 Δ < 0，方程没有实数根。图像不与 x 轴相交。

Discriminant 判别式	Number of real roots 实数根个数	Graph shape 图像形状
Δ > 0	2 distinct roots 两个不同根	Crosses x-axis twice 穿过 x 轴两次
Δ = 0	1 repeated root 一个重根	Touches x-axis once 与 x 轴相切一次
Δ < 0	0 real roots 无实数根	Never meets x-axis 与 x 轴无交点

Examination questions often ask you to find the range of k for which a quadratic has real roots. Set Δ ≥ 0 and solve the resulting inequality.

考试常要求你求 k 的取值范围，使得某个二次方程有实数根。只需令 Δ ≥ 0 并解这个不等式。

6. Quadratic Graphs and Roots | 二次函数的图像与根

A quadratic function y = ax² + bx + c produces a parabola. If a > 0, the parabola opens upwards (a “U” shape). If a < 0, it opens downwards (an “n” shape).

二次函数 y = ax² + bx + c 的图像是一条抛物线。当 a > 0 时，开口向上（呈 “U” 形）；当 a < 0 时，开口向下（呈 “n” 形）。

The x-intercepts (where y = 0) are the roots of the equation ax² + bx + c = 0. Solving the equation finds these points exactly.

与 x 轴的交点（即 y = 0 处）就是方程 ax² + bx + c = 0 的根。解这个方程就能精确求出这些交点。

The y-intercept is simply the constant term c, because when x = 0, y = c.

与 y 轴的交点就是常数项 c，因为当 x = 0 时，y = c。

Example: Sketch y = x² − 2x − 3. Factorise: (x − 3)(x + 1) = 0, so roots are x = 3 and x = −1. The y-intercept is −3 (at x = 0, y = −3). Since a = 1 > 0, the parabola opens upward.

例如：画出 y = x² − 2x − 3 的图像。因式分解得 (x − 3)(x + 1) = 0，所以根为 x = 3 和 x = −1。与 y 轴的交点为 −3（因为 x = 0 时，y = −3）。由于 a = 1 > 0，抛物线开口向上。

7. Turning Point and Axis of Symmetry | 顶点与对称轴

The graph of a quadratic is symmetric about a vertical line called the axis of symmetry. Its equation is x = −b/(2a). This is also the x-coordinate of the turning point.

二次函数图像关于一条竖直线对称，称为对称轴。其方程为 x = −b/(2a)。这也是顶点的 x 坐标。

The turning point (vertex) has coordinates:

顶点坐标为：

(−b/(2a), c − b²/(4a))

Alternatively, if the quadratic is written in completed square form y = a(x − h)² + k, then the vertex is (h, k) and the axis of symmetry is x = h.

另一种方法：如果二次式写成配方法形式 y = a(x − h)² + k，则顶点为 (h, k)，对称轴为 x = h。

For y = x² − 2x − 3, a = 1, b = −2, c = −3. Axis: x = −(−2)/(2×1) = 1. Vertex y-value: (1)² − 2(1) − 3 = −4. So the vertex is (1, −4).

对于 y = x² − 2x − 3，a = 1，b = −2，c = −3。对称轴：x = −(−2)/(2×1) = 1。顶点的 y 值：1² − 2(1) − 3 = −4。因此顶点为 (1, −4)。

Remember: the turning point is a minimum if a > 0 and a maximum if a < 0.

请记住：当 a > 0 时顶点是最小值；当 a < 0 时顶点是最大值。

8. Solving Quadratic Inequalities | 解二次不等式

You may also be asked to solve inequalities such as x² − x − 6 > 0. First factorise: (x − 3)(x + 2) > 0.

有时题目要求解二次不等式，例如 x² − x − 6 > 0。首先因式分解：(x − 3)(x + 2) > 0。

Sketch the parabola. The roots are x = 3 and x = −2. Since a = 1 > 0, the parabola opens upward, so it is above the x-axis outside the interval between the roots.

画出抛物线草图。根为 x = 3 和 x = −2。因为 a = 1 > 0，抛物线开口向上，因此在两根之间的区间之外，图像位于 x 轴上方。

Therefore the solution is x < −2 or x > 3. If the inequality were “< 0”, the solution would be −2 < x < 3.

因此解为 x < −2 或 x > 3。如果不等式是 “< 0”，解则为 −2 < x < 3。

Always be careful with the direction of the inequality when the coefficient of x² is negative. Consider 6 − x − x² > 0. Multiply by −1 to get x² + x − 6 < 0, then proceed as above.

当 x² 系数为负时，要小心不等号方向。例如 6 − x − x² > 0，两边乘以 −1 得 x² + x − 6 < 0，再按上述方法求解。

9. Applications and Worded Problems | 应用与实际应用题

Quadratic equations arise in areas, projectile motion, and number problems. The key is to translate the words into an equation.

二次方程常见于面积问题、抛体运动和数字问题。关键是把文字翻译成方程。

Example: A rectangle has length (x + 3) cm and width (x − 1) cm. Its area is 45 cm². Find x.

例如：一个矩形的长为 (x + 3) cm，宽为 (x − 1) cm，面积为 45 cm²。求 x。

Write the area equation: (x + 3)(x − 1) = 45. Expand: x² + 2x − 3 = 45, so x² + 2x − 48 = 0. Factorise: (x + 8)(x − 6) = 0. Hence x = −8 or x = 6. Since a length cannot be negative, x = 6.

列出面积方程：(x + 3)(x − 1) = 45。展开：x² + 2x − 3 = 45，即 x² + 2x − 48 = 0。因式分解：(x + 8)(x − 6) = 0。所以 x = −8 或 x = 6。由于长度不能为负，x = 6。

Check: length = 9 cm, width = 5 cm, area = 45 cm². Always check if the answers make sense in the context.

检验：长 = 9 cm，宽 = 5 cm，面积 = 45 cm²。务必检查答案在情境中是否合理。

For projectile motion problems such as h = −5t² + 20t + 25, the height h is given in metres and time t in seconds. To find when it hits the ground, set h = 0 and solve.

对于抛体运动问题，例如 h = −5t² + 20t + 25，高度 h 以米为单位，时间 t 以秒为单位。求落地时间时，令 h = 0 并解方程。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Common mistakes include: forgetting to rearrange the equation to zero before factorising; dividing by a variable and losing roots; incorrect signs when using the quadratic formula; and ignoring the value of √(b² − 4ac) when it is not a perfect square.

常见错误包括：在因式分解前忘记移项使方程等于零；除以变量导致丢根；使用求根公式时符号出错；以及忽略 √(b² − 4ac) 不是完全平方的情况。

Exam tips: write your solutions clearly; show all working steps; use your calculator only for arithmetic, not for finding factors; and always check your answers by substitution.

考试技巧：清晰书写答案；展示所有计算步骤；计算器只用于算术，不用来找因式；最后务必代回原方程检验。

If a = 1, look for two numbers with product c and sum b.

若 a = 1，寻找乘积为 c、和为 b 的两个数。
If you are solving a word problem, state the units and reject negative lengths, times, or distances.

解应用题时，写出单位，并舍去负的长度、时间或距离。
Learn the quadratic formula by heart — it will be needed when factorisation fails.

牢记求根公式——因式分解失败时会用到。

With regular practice on factorising, using the quadratic formula, completing the square, and sketching graphs, you will master quadratics for the IGCSE exam. Remember that every method should give the same roots, so use them to check one another.

通过经常练习因式分解、使用求根公式、配方以及画图，你将掌握 IGCSE 考试中的二次方程内容。请记住，所有方法给出的根应该相同，因此可以用不同方法互相验算。

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Solving Quadratic Equations and Graphs | 解二次方程与图像