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Solving Quadratic Equations | 解一元二次方程

2026-08-30 04:18 · 本文部分内容由AI辅助创作

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations appear frequently in IGCSE Mathematics, from simple factorisation to complex word problems. Mastering the different methods of solving them is essential for success in exams.

一元二次方程在IGCSE数学中频繁出现,从简单的因式分解到复杂的应用题。掌握不同的解法对于考试成功至关重要。


1. What is a Quadratic Equation? | 什么是一元二次方程

it can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.

一元二次方程是次数为2的多项式方程,即变量的最高次幂为2。一般来说,它可以写成 ax² + bx + c = 0 的形式,其中 a、b、c 是常数,且 a ≠ 0。

Some examples include:

一些例子包括:

The values of x that satisfy the equation are called the roots or solutions. A quadratic equation can have two distinct roots, one repeated root, or no real roots.

满足方程的 x 值称为根或解。一元二次方程可以有两个不同的根、一个重根,或没有实数根。


2. Standard Form and Key Terms | 标准形式与关键术语

t all terms are on one side and the equation is set to zero, which is necessary for most solving methods.

在解一元二次方程之前,务必将其整理为标准形式 ax² + bx + c = 0。这样可以确保所有项都在等号一侧,并使等式等于0,这是大多数解法所必需的。

the constant term. For example, in the equation 3x² - 7x + 2 = 0, a = 3, b = -7 and c = 2.

其中 a 是二次项系数,b 是一次项系数,c 是常数项。例如,在方程 3x² - 7x + 2 = 0 中,a = 3,b = -7,c = 2。

as x² - 16 = 0. If c = 0, it becomes x² - 4x = 0, which can be solved by factoring out x.

如果 b = 0,方程是纯二次方程,如 x² - 16 = 0。如果 c = 0,则变为 x² - 4x = 0,可以通过提取公因子 x 来解。


3. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method when the quadratic can be written as a product of two linear brackets. The idea is to rewrite ax² + bx + c as (px + q)(rx + s), then set each factor equal to zero.

当二次式可以写成两个一次括号的乘积时,因式分解往往是最快的方法。其思路是将 ax² + bx + c 重写为 (px + q)(rx + s),然后令每个因子等于0。

two numbers that multiply to give c and add to give b. For example, solve x² - 5x + 6 = 0:

对于首项系数为1的二次式(a = 1),寻找两个数,它们相乘得到 c,相加得到 b。例如,解 x² - 5x + 6 = 0:

x² - 5x + 6 = (x - 2)(x - 3) = 0

Therefore x - 2 = 0 or x - 3 = 0, so x = 2 or x = 3.

因此 x - 2 = 0 或 x - 3 = 0,所以 x = 2 或 x = 3。

b, then grouping. Example: 2x² + 7x + 3 = 0. Here a × c = 6, and factors 1 and 6 add to 7.

对于首项系数不为1的二次式,你需要对带有非1首项系数的二次式进行因式分解。可以通过找到 a × c 的因子,且这些因子相加等于 b,然后分组。例如:2x² + 7x + 3 = 0。这里 a × c = 6,因子1和6相加得7。

2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)

Thus x = -1/2 or x = -3. Always check your factors by expanding.

因此 x = -1/2 或 x = -3。始终通过展开来检查你的因式。


4. Solving by Completing the Square | 配方法

Completing the square transforms ax² + bx + c into a perfect square plus a constant. This is useful when factorisation is difficult or when you need to find the turning point of a quadratic graph.

配方法将 ax² + bx + c 转化为一个完全平方加一个常数。这适用于因式分解困难或需要求二次函数图像顶点的情况。

monic quadratic x² + bx + c, rewrite it as (x + b/2)² - (b/2)² + c. Solve x² + 6x - 7 = 0:

对于首项系数为1的二次式 x² + bx + c,将其重写为 (x + b/2)² - (b/2)² + c。解 x² + 6x - 7 = 0:

(x + 3)² - 9 - 7 = 0 → (x + 3)² = 16

Taking the square root of both sides gives x + 3 = ±4, so x = 1 or x = -7.

对方程两边开平方根得到 x + 3 = ±4,所以 x = 1 或 x = -7。

thod also reveals the vertex of the parabola: for y = a(x - h)² + k, the vertex is (h, k).

对于非首项系数为1的二次式,先除以首项系数 a。然后配方。此方法还揭示了抛物线的顶点:对于 y = a(x - h)² + k,顶点为 (h, k)。


5. Solving by the Quadratic Formula | 公式法

iversal method that works for any quadratic equation. It solves ax² + bx + c = 0 directly:

公式法是一种适用于任何一元二次方程的通用方法。它直接解 ax² + bx + c = 0:

x = (-b ± √b² - 4ac) / 2a

Be careful with signs, especially when b is negative. For example, solve 2x² - 4x - 3 = 0:

使用公式时,将 a、b、c 的值代入表达式。注意符号,尤其是当 b 为负数时。例如,解 2x² - 4x - 3 = 0:

x = (4 ± √((-4)² - 4 × 2 × (-3))) / (2 × 2) = (4 ± √40) / 4

Since √40 = 2√10, the exact roots are x = (4 + 2√10)/4 = (2 + √10)/2 and x = (2 - √10)/2. Remember to simplify the answer fully.

由于 √40 = 2√10,准确根为 x = (4 + 2√10)/4 = (2 + √10)/2 和 x = (2 - √10)/2。记得将答案化到最简。


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6. The Discriminant | 判别式

oot: Δ = b² - 4ac. It tells us the nature of the roots without fully solving the equation.

判别式,用 Δ 表示,是公式法中根号下的部分:Δ = b² - 4ac。它无需完全解方程就能告诉我们根的性质。

ble root at x = 2. For x² + 2x + 5 = 0, Δ = 2² - 20 = -16 < 0, so there are no real roots.

例如,对于 x² - 4x + 4 = 0,Δ = (-4)² - 4 × 1 × 4 = 0,所以在 x = 2 处有一个重根。对于 x² + 2x + 5 = 0,Δ = 2² - 20 = -16 < 0,所以没有实数根。


7. Sum and Product of Roots | 根与系数的关系

is relationship is useful for checking answers or constructing equations when roots are known.

对于具有根 α 和 β 的一元二次方程 ax² + bx + c = 0,两根之和为 α + β = -b/a,两根之积为 αβ = c/a。这一关系可用于检验答案或在已知根时构造方程。

m is 1 and the product is -6. A possible equation is x² - 1x - 6 = 0, i.e. x² - x - 6 = 0.

这些公式源于展开 a(x - α)(x - β) = 0。例如,如果根是3和-2,则和为1,积为-6。一个可能的方程是 x² - 1x - 6 = 0,即 x² - x - 6 = 0。


8. Quadratic Graphs and Roots | 二次函数图像与根

= ax² + bx + c. The graph is a parabola that opens upward if a > 0 and downward if a < 0.

一元二次方程的根是其图像 y = ax² + bx + c 的 x 轴交点。图像是抛物线,当 a > 0 时开口向上,当 a < 0 时开口向下。

= -b/(2a), and the y-coordinate can be found by substituting this value into the equation.

抛物线的顶点表示二次表达式的最值。其 x 坐标为 x = -b/(2a),将该值代入方程可求得 y 坐标。

he discriminant: two crossings for Δ > 0, one touch for Δ = 0, and no crossings for Δ < 0.

用图像法求解时,可能需要从图中估算根。抛物线与 x 轴的交点数与判别式对应:Δ > 0 时有两个交点,Δ = 0 时有一个切点,Δ < 0 时无交点。


9. Word Problems | 应用题

Many IGCSE questions model real-life situations with quadratic equations. The first step is to translate the problem into an equation, then solve it using an appropriate method, and finally check the validity of the answers in context.

许多IGCSE考题用一元二次方程模拟现实情境。第一步是将问题转化为方程,然后用合适的方法求解,最后检查答案在情境中是否合理。

he width be x cm, then length is x + 3. The area gives x(x + 3) = 40, so x² + 3x - 40 = 0.

例子:一个矩形的长比宽多3厘米,面积为40平方厘米。求其尺寸。设宽为 x 厘米,则长为 x + 3。由面积得 x(x + 3) = 40,即 x² + 3x - 40 = 0。

and the length is 8 cm. Always reject impossible negative solutions in geometry problems.

因式分解得:(x + 8)(x - 5) = 0,所以 x = -8 或 x = 5。由于宽度不能为负,x = 5。因此宽为5厘米,长为8厘米。在几何问题中务必舍去不可能的负解。


10. Common Mistakes to Avoid | 常见错误

Mistakes in solving quadratic equations often come from arithmetic errors or misunderstanding the methods. Here are some key guidelines to avoid them.

解一元二次方程时的错误通常来自算术错误或对方法的误解。以下是一些避免错误的关键要点。


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