Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear in many areas of mathematics and science. In the Edexcel IGCSE syllabus, you must be able to solve them by factorisation, by using the quadratic formula, and by completing the square. This guide covers all three methods, with worked examples and common pitfalls.

二次方程在数学和科学中广泛出现。在Edexcel IGCSE考纲中，你必须掌握因式分解法、二次公式法和配方法这三种解法。本指南将逐一讲解，并附有例题和常见错误提醒。

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1. Standard Form of a Quadratic Equation | 二次方程的标准形式

The general form of a quadratic equation is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable is 2, which is why it is called ‘quadratic’.

二次方程的一般形式为ax² + bx + c = 0，其中a、b、c为常数，且a ≠ 0。变量的最高次数是2，因此称为“二次”。

• If a = 0, the equation becomes linear, not quadratic.
• 如果a = 0，方程变为一次方程，而不是二次方程。
• The solutions of a quadratic equation are also called its roots.
• 二次方程的解也称为方程的根。

ax² + bx + c = 0

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2. Solving by Factorisation | 因式分解法

Factorisation involves writing the quadratic expression as a product of two linear factors. Then, using the fact that if the product of two numbers is zero, at least one of them must be zero, we can solve each linear equation.

因式分解法是将二次表达式写成两个一次因式的乘积。然后利用“若两个数的乘积为零，则至少有一个数为零”的性质，分别求解每个一次方程。

Example | 例题: Solve x² − 5x + 6 = 0.

Solution | 解: Factorise the left-hand side:

x² − 5x + 6 = (x − 2)(x − 3)

Set each factor equal to zero:

x − 2 = 0 or x − 3 = 0

Hence x = 2 or x = 3.

因此 x = 2 或 x = 3。

Always check your factors by expanding them. For example, (x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6.

务必通过展开来检验因式是否正确。例如，(x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6。

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3. Factorising When the Coefficient of x² is Not 1 | 当x²系数不为1时的因式分解

When the coefficient of x² is not 1, for example 2x² + 7x + 3 = 0, you need to find the correct pair of brackets by trial and error or by the ‘ac’ method.

当x²的系数不为1时，例如2x² + 7x + 3 = 0，你需要通过试错法或“ac法”找到正确的括号组合。

Example | 例题: Solve 2x² + 7x + 3 = 0.

The factors of 2x² are 2x and x. The factors of 3 are 1 and 3. Try (2x + 1)(x + 3): expanding gives 2x² + 6x + x + 3 = 2x² + 7x + 3. Correct.

2x²的因式为2x和x。3的因式为1和3。尝试(2x + 1)(x + 3)：展开得2x² + 6x + x + 3 = 2x² + 7x + 3。正确。

So (2x + 1)(x + 3) = 0, giving x = −½ or x = −3.

因此(2x + 1)(x + 3) = 0，解得x = −½或x = −3。

If you are struggling, the quadratic formula always works.

如果难以分解，二次公式始终有效。

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4. The Quadratic Formula | 二次公式

For any quadratic equation ax² + bx + c = 0, the roots are given by the quadratic formula:

对于任意二次方程ax² + bx + c = 0，求根公式为：

x = (−b ± √(b² − 4ac)) / (2a)

This formula is provided in the Edexcel IGCSE formula booklet, but you must know how to substitute values correctly.

该公式在Edexcel IGCSE公式册中会给出，但你必须知道如何正确代入数值。

Example | 例题: Solve 2x² + 3x − 4 = 0 using the formula. Give your answers correct to 2 decimal places.

Here a = 2, b = 3, c = −4.

这里a = 2, b = 3, c = −4。

x = (−3 ± √(3² − 4×2×(−4))) / (2×2) = (−3 ± √(9 + 32)) / 4

√41 ≈ 6.403

So x ≈ (−3 + 6.403)/4 ≈ 0.85 or x ≈ (−3 − 6.403)/4 ≈ −2.35.

因此x ≈ 0.85或x ≈ −2.35。

Be careful with negative signs inside the square root and when substituting c.

注意根号内的负号和代入c时的符号。

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5. The Discriminant: b² − 4ac | 判别式：b² − 4ac

The value of the discriminant, Δ = b² − 4ac, tells us how many real roots a quadratic equation has, without solving it.

判别式的值Δ = b² − 4ac能告诉我们二次方程有多少个实根，而无需解方程。

Discriminant | 判别式	Roots | 根
Δ > 0	Two distinct real roots | 两个不同的实根
Δ = 0	One repeated real root | 一个重根
Δ < 0	No real roots | 没有实根

Example | 例题: How many real roots does x² + 2x + 5 = 0 have?

Here a = 1, b = 2, c = 5. Δ = 2² − 4×1×5 = 4 − 20 = −16 < 0. So no real roots.

这里a = 1, b = 2, c = 5。Δ = 2² − 4×1×5 = 4 − 20 = −16 < 0。所以没有实根。

The graph of the quadratic does not cross the x-axis.

该二次函数的图像不与x轴相交。

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6. Solving by Completing the Square | 配方法

Completing the square rewrites the quadratic in the form a(x + p)² + q. This is useful for finding the vertex of the parabola and also for solving equations.

配方法将二次表达式改写为a(x + p)² + q的形式。这有助于求抛物线的顶点，也可用于解方程。

Example | 例题: Solve x² + 6x − 7 = 0 by completing the square.

Take half of the coefficient of x: 6/2 = 3. Write (x + 3)² − 9 − 7 = 0.

取x系数的半：6/2 = 3。写成(x + 3)² − 9 − 7 = 0。

Simplify: (x + 3)² − 16 = 0.

化简：(x + 3)² − 16 = 0。

(x + 3)² = 16

Take square roots: x + 3 = ±4. Hence x = 1 or x = −7.

开平方：x + 3 = ±4。因此x = 1或x = −7。

Check by expanding: (x + 3)² − 16 = x² + 6x + 9 − 16 = x² + 6x − 7.

展开检验：(x + 3)² − 16 = x² + 6x + 9 − 16 = x² + 6x − 7。

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7. Completing the Square When the Coefficient of x² is Not 1 | 当x²系数不为1时的配方法

If the equation is 2x² + 8x + 5 = 0, first factor out the coefficient of x² from the first two terms.

如果方程是2x² + 8x + 5 = 0，先将x²的系数从前两项中提出来。

2(x² + 4x) + 5 = 0

Complete the square inside the bracket: x² + 4x = (x + 2)² − 4.

括号内配方：x² + 4x = (x + 2)² − 4。

So 2[(x + 2)² − 4] + 5 = 0 → 2(x + 2)² − 8 + 5 = 0 → 2(x + 2)² − 3 = 0.

所以2[(x + 2)² − 4] + 5 = 0 → 2(x + 2)² − 8 + 5 = 0 → 2(x + 2)² − 3 = 0。

Now solve: (x + 2)² = 3/2. Then x = −2 ± √(3/2).

现在求解：(x + 2)² = 3/2。则x = −2 ± √(3/2)。

Rationalise if required.

如有需要可进行分母有理化。

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8. Solving Quadratic Inequalities | 解二次不等式

Solving a quadratic inequality such as x² − 5x + 6 < 0 involves finding the roots first, then using a number line or sketch of the graph to determine the range of x that satisfies the inequality.

解二次不等式如x² − 5x + 6 < 0，首先要找到根，然后利用数轴或函数图像草图确定满足不等式的x范围。

Example | 例题: Solve x² − 5x + 6 ≤ 0.

Factorise: (x − 2)(x − 3) ≤ 0. The roots are 2 and 3.

因式分解：(x − 2)(x − 3) ≤ 0。根为2和3。

The parabola y = x² − 5x + 6 opens upward. The expression is ≤ 0 between the roots (where the graph is on or below the x-axis).

抛物线y = x² − 5x + 6开口向上。表达式在两根之间≤ 0（图像位于x轴上或下方）。

2 ≤ x ≤ 3

If the inequality were > 0, the solution would be x < 2 or x > 3.

如果不等式为> 0，则解为x < 2或x > 3。

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9. Solving Equations by Substitution | 用换元法解二次方程

Some equations that are not obviously quadratic can be rewritten as quadratic by substituting a new variable. For example, x⁴ − 5x² + 4 = 0 is a quartic equation but can be treated as quadratic in x².

有些看似不是二次的方程，可以通过设新变量化为二次方程。例如x⁴ − 5x² + 4 = 0是四次方程，但可以看作关于x²的二次方程。

Let y = x². Then the equation becomes y² − 5y + 4 = 0.

令y = x²。则方程变为y² − 5y + 4 = 0。

Factorise: (y − 1)(y − 4) = 0, so y = 1 or y = 4.

因式分解：(y − 1)(y − 4) = 0，所以y = 1或y = 4。

Substitute back: x² = 1 → x = ±1; x² = 4 → x = ±2.

代回：x² = 1 → x = ±1；x² = 4 → x = ±2。

Hence the equation has four real roots: x = 1, −1, 2, −2.

因此原方程有四个实根：x = 1, −1, 2, −2。

Remember to always substitute back to find the value of the original variable.

记住一定要代回求出原变量的值。

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

Here are some common errors students make when solving quadratic equations, and tips to avoid them.

以下是学生在解二次方程时常犯的错误及避免建议。

• Forgetting to set the equation equal to zero before factorising. Always rearrange to ax² + bx + c = 0 first.
• 因式分解前忘记将方程化为等于零的形式。务必先整理成ax² + bx + c = 0。
• Misplacing negative signs when substituting into the quadratic formula. Write down the values of a, b and c with brackets.
• 代入二次公式时弄错负号。请先写出a、b、c的值，并加上括号。
• Losing solutions when solving by square roots: remember ±.
• 用开平方求解时漏解：不要忘记±。
• If the discriminant is negative, don’t try to find a real root. State ‘no real roots’ if required.
• 如果判别式为负，不要试图找实根。需要时就说明“没有实根”。

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

始终将答案代回原方程进行检验。

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