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Solving Quadratic Equations | 解二次方程

2026-08-30 00:21 · 本文部分内容由AI辅助创作

📚 Solving Quadratic Equations | 解二次方程

p you unlock many other areas, including graphs, functions, and problem-solving. This revision guide covers every method you need for the Edexcel specification, with clear examples and common pitfalls.

二次方程是 IGCSE 数学中最重要的主题之一。掌握形如 ax² + bx + c = 0 的方程解法,可以帮助你更好地理解图像、函数和实际问题解决。本复习指南覆盖 Edexcel 考纲要求的每一种方法,并配有清晰的例子和常见错误提醒。


1. What is a Quadratic Equation? | 什么是二次方程

A quadratic equation is a polynomial equation of degree 2. The highest power of the unknown variable is 2. The general form is:

二次方程是最高次数为 2 的多项式方程,未知数的最高幂为 2。其一般形式为:

ax² + bx + c = 0, where a ≠ 0

, the equation becomes linear, not quadratic. The values of x that satisfy the equation are called roots or solutions. A quadratic equation can have two distinct real roots, one repeated real root, or no real roots.

其中 a、b、c 为常数,且 a ≠ 0。若 a = 0,方程退化为一次方程。满足方程的 x 值称为根或解。二次方程可以有两个不同的实根、一个重根,或者没有实根。


2. The Standard Form and Key Features | 标准形式与关键特征

one side, equal to zero. Expanding brackets, collecting like terms, and moving terms across the equals sign are common steps.

在求解之前,必须把方程整理成标准形式 ax² + bx + c = 0,即所有项都在等号一侧,另一侧为 0。去括号、合并同类项、移项是常见步骤。

e graph is a parabola. The sign of a determines the shape: if a > 0, the parabola opens upwards; if a < 0, it opens downwards. The roots of the equation are the x-coordinates where the curve crosses the x-axis.

对于二次曲线 y = ax² + bx + c,其图像是抛物线。a 的符号决定开口方向:a > 0 时开口向上,a < 0 时开口向下。方程的根就是曲线与 x 轴交点的横坐标。


3. Solving by Factorisation | 因式分解法

Factorisation is usually the fastest method when the quadratic has simple integer roots. To solve by factorisation, you rewrite the quadratic as a product of two linear expressions. Then use the fact that if the product is zero, at least one of the factors must be zero.

因式分解法通常是最快的方法,适用于根为简单整数的情况。求解时,将二次式改写为两个一次式的乘积,然后利用“若乘积为零,则至少有一个因式为零”的性质。

We look for two numbers that multiply to 6 and add to -5. Those numbers are -2 and -3. So:

例如,解 x² - 5x + 6 = 0。我们要找两个数,乘积为 6,和为 -5。这两个数是 -2 和 -3。因此:

(x - 2)(x - 3) = 0

Then x - 2 = 0 or x - 3 = 0, giving x = 2 or x = 3.

于是 x - 2 = 0 或 x - 3 = 0,得到 x = 2 或 x = 3。

When the coefficient a is not 1, you may need the “ac method”: find two numbers whose product is ac and whose sum is b, then split the middle term and factorise by grouping.

当 a 不为 1 时,可以使用 “ac 法”:找两个数,使它们的乘积为 ac,和为 b,然后拆中项并分组分解。


4. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This is useful for solving equations, finding maximum or minimum points, and sketching graphs. For Edexcel IGCSE, you must be able to write a quadratic expression in completed-square form and use it to solve equations.

配方法将二次式改写为 a(x + p)² + q 的形式。这可用于解方程、求最大值或最小值点,以及画图。Edexcel IGCSE 要求你能将二次式化为完全平方形式,并利用它解方程。

The method works as follows, for x² + bx + c = 0:

方法步骤如下(以 x² + bx + c = 0 为例):

  1. Take half of b, square it, then add and subtract this value.

    取 b 的一半,平方,然后加上并减去这个值。

  2. Rewrite the first three terms as a perfect square.

    将前三项改写为完全平方。

  3. Simplify the constant terms.

    简化常数项。

Example: Solve x² + 6x + 4 = 0 by completing the square. Half of 6 is 3, and 3² = 9. So:

例:用配方法解 x² + 6x + 4 = 0。6 的一半是 3,3² = 9。因此:

(x + 3)² - 9 + 4 = 0 → (x + 3)² - 5 = 0

Then (x + 3)² = 5, so x + 3 = ±√5, hence x = -3 ± √5.

于是 (x + 3)² = 5,所以 x + 3 = ±√5,因此 x = -3 ± √5。

When a ≠ 1, first factor out a from the x² and x terms, then complete the square inside the brackets.

当 a ≠ 1 时,先从 x² 和 x 项中提出 a,再在括号内配方。


5. The Quadratic Formula | 二次公式

those that cannot be factorised easily. For ax² + bx + c = 0, the solutions are given by:

二次公式适用于所有二次方程,包括不易因式分解的情况。对于 ax² + bx + c = 0,解为:

x = (-b ± √b² - 4ac) / (2a)

You should memorise this formula. To use it, substitute the values of a, b, and c after ensuring the equation is in standard form. Remember that a is the coefficient of x², b is the coefficient of x, and c is the constant term.

这个公式需要背熟。使用前确保方程已化为标准形式,然后代入 a、b、c 的值。注意 a 是 x² 的系数,b 是 x 的系数,c 是常数项。

Example: Solve 2x² - 4x - 3 = 0. Here a = 2, b = -4, c = -3. Substituting:

例:解 2x² - 4x - 3 = 0。此时 a = 2,b = -4,c = -3。代入:

x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4

Simplifying: x = (4 ± 2√10) / 4 = 1 ± (√10)/2. Give your answer to a suitable degree of accuracy if required.

化简:x = (4 ± 2√10) / 4 = 1 ± (√10)/2。若要求保留一定精度,请按要求给出答案。


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6. The Discriminant | 判别式

f the roots without fully solving the equation. For a quadratic equation ax² + bx + c = 0:

判别式是二次公式中根号内的表达式 b² - 4ac。它可以在不解方程的情况下告诉我们根的性质。对于二次方程 ax² + bx + c = 0:

Discriminant b² - 4acNature of roots根的性质
b² - 4ac > 0Two distinct real roots两个不同的实根
b² - 4ac = 0One repeated real root一个重根(两个相等实根)
b² - 4ac < 0No real roots无实根

6 < 0, so there are no real roots. The curve y = 3x² + 2x + 5 never crosses the x-axis.

例:对于 3x² + 2x + 5 = 0,b² - 4ac = 4 - 60 = -56 < 0,因此没有实根。曲线 y = 3x² + 2x + 5 从不与 x 轴相交。

If b² - 4ac is a perfect square and the coefficients are rational, then the quadratic factorises over the integers or rationals.

若 b² - 4ac 是完全平方数,且系数为有理数,则二次式可以在整数或有理数范围内因式分解。


7. Roots and Turning Points | 根与转向点

The roots of a quadratic equation are also the x-intercepts of the corresponding graph. The turning point (vertex) of the parabola lies midway between the two roots. This is useful when sketching or when finding the minimum or maximum value.

二次方程的根也是对应图像的 x 轴截距。抛物线的转向点(顶点)位于两个根的正中间。这在画图或求最小值/最大值时非常有用。

g point is x = (p + q) / 2. Substitute this x into the quadratic to find the y-coordinate.

若根为 x = p 和 x = q,则转向点的 x 坐标为 x = (p + q) / 2。将 x 代入二次式可得到 y 坐标。

Alternatively, if you have completed the square as a(x + p)² + q, the turning point is (-p, q). If a > 0, this is a minimum; if a < 0, it is a maximum.

或者,若已将二次式配方为 a(x + p)² + q,则转向点为 (-p, q)。若 a > 0,则为最小值点;若 a < 0,则为最大值点。

2 has turning point (3, 2), and since the coefficient is positive, it is a minimum point.

例:y = (x - 3)² + 2 的转向点为 (3, 2),由于系数为正,所以是最小值点。


8. Word Problems | 应用题

Many examination questions present a real-life situation or a geometrical problem that leads to a quadratic equation. You must translate the words into symbols, form an equation, solve it, and then interpret the answer in context. Discard any negative roots if they do not make sense in the situation.

许多考试题目会给出实际生活场景或几何问题,从而导出二次方程。你需要将文字转化为符号,建立方程,求解,然后结合情境解释答案。如果负根在情境中无意义,则舍去。

ising: (x + 7)(x - 3) = 0. Thus x = 3 or x = -7. Since width cannot be negative, x = 3 cm.

例:一个长方形的长比宽多 4 cm,面积为 21 cm²,求宽。设宽为 x,则长为 x + 4。所以 x(x + 4) = 21,即 x² + 4x - 21 = 0。分解因式:(x + 7)(x - 3) = 0。因此 x = 3 或 x = -7。因为宽不能为负,所以 x = 3 cm。


9. Common Mistakes and Tips | 常见错误与技巧

Students often lose marks by making simple errors. Avoid these common traps:

学生常因简单错误丢分。避免以下常见陷阱:

Always check your solutions by substituting them back into the original equation.

务必把解代回原方程进行检验。


10. Practice Questions | 练习

Try these questions on your own. Write your working clearly, as Edexcel awards method marks.

请独立尝试以下题目。Edexcel 评分时会根据解题过程给方法分,请写清步骤。

  1. Solve x² - 7x + 10 = 0 by factorisation.

    用因式分解法解 x² - 7x + 10 = 0。

  2. x² + 5x - 3 = 0 using the quadratic formula. Give your answer correct to 2 decimal places.

    用二次公式解 2x² + 5x - 3 = 0,结果精确到小数点后 2 位。

  3. Find the value(s) of k for which x² + kx + 9 = 0 has exactly one real root.

    求 k 的值(或取值范围),使得 x² + kx + 9 = 0 恰有一个实根。

  4. A right-angled triangle has sides x, x + 1, and x + 2. Use Pythagoras’ theorem to form a quadratic equation and find x.

    一个直角三角形三边长分别为 x、x + 1、x + 2。用勾股定理建立二次方程并求 x。

  5. Write y = x² - 6x + 11 in the form (x + p)² + q, hence find the minimum value of y.

    将 y = x² - 6x + 11 化为 (x + p)² + q 的形式,并求 y 的最小值。

+1)² = (x+2)² → x² - 2x - 3 = 0 → x = 3 or -1, reject -1). 5. (x - 3)² + 2, minimum y = 2.

答案:1. x = 2 或 x = 5。2. x ≈ 0.50 或 x ≈ -3.00。3. k = ±6。4. x = 3(由 x² + (x+1)² = (x+2)² → x² - 2x - 3 = 0 → x = 3 或 -1,舍去 -1)。5. (x - 3)² + 2,最小值 y = 2。


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