Example 1.7.1 – Solving Quadratic Equations by Factorising, Formula and Completing the Square | 示例 1.7.1 – 因式分解法、公式法与配方法解二次方程

📚 Example 1.7.1 – Solving Quadratic Equations by Factorising, Formula and Completing the Square | 示例 1.7.1 – 因式分解法、公式法与配方法解二次方程

Example 1.7.1 is a core AQA A-Level Mathematics worked example on solving a quadratic equation. It brings together the three standard algebraic methods: factorising, using the quadratic formula, and completing the square. A strong understanding of this example helps students handle many Pure Mathematics questions on quadratics, discriminants, and curve sketching.

示例 1.7.1 是 AQA A-Level 数学中关于解二次方程的核心例题。它综合了三种标准代数方法：因式分解、使用二次公式和配方法。透彻理解这个例题有助于学生解决许多关于二次函数、判别式和曲线作图的纯数学问题。

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1. The Worked Problem | 例题展示

The equation given in Example 1.7.1 is a quadratic equation with rational roots. Students are expected to solve it accurately and to verify the answer using at least two different methods. This is typical of AQA mark schemes, where method marks are awarded for a clear and logical approach.

示例 1.7.1 中给出的方程是一个具有有理根的二次方程。学生需要准确求解，并至少用两种不同的方法验证答案。这是 AQA 评分方案的典型要求，清晰且有逻辑的解题过程通常能获得方法分。

• Equation: 2x² − 7x + 6 = 0 | 方程：2x² − 7x + 6 = 0
• Find: all real solutions | 求：所有实数解
• Verify: use at least two methods | 验证：至少使用两种方法

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2. Factorisation First | 先因式分解

Factorising is often the fastest method in AQA questions when the coefficients are small and the roots are rational. For this equation, split the middle term into −4x and −3x because (−4) + (−3) = −7 and (−4)(−3) = 12, which matches 2 × 6.

当系数较小且根为有理数时，因式分解通常是 AQA 考试中最快的方法。对于这个方程，将中间项拆分为 −4x 和 −3x，因为 (−4) + (−3) = −7 且 (−4)(−3) = 12，与 2 × 6 匹配。

After grouping, the expression becomes (2x − 3)(x − 2), so setting each factor to zero gives the two roots directly.

分组后，表达式变为 (2x − 3)(x − 2)，因此令每个因式为零可直接得到两个根。

2x² − 7x + 6 = (2x − 3)(x − 2) = 0

Therefore, 2x − 3 = 0 gives x = 3/2, and x − 2 = 0 gives x = 2. Both roots are exact fractions, which is often preferred on AQA papers.

因此，2x − 3 = 0 得到 x = 3/2，x − 2 = 0 得到 x = 2。两个根都是精确分数，这在 AQA 试卷中通常是首选答案形式。

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3. Using the Quadratic Formula | 使用二次公式

The quadratic formula is a reliable second method and is especially useful when factorisation is not obvious. The formula must be applied with the correct signs for a, b, and c from the standard form ax² + bx + c = 0.

二次公式是一种可靠的第二种方法，在因式分解不明显时尤其有用。使用公式时必须从标准形式 ax² + bx + c = 0 中正确代入 a、b 和 c 的符号。

x = (−b ± √(b² − 4ac)) / (2a)

For 2x² − 7x + 6 = 0, we have a = 2, b = −7, and c = 6. Substituting these into the formula gives the discriminant first.

对于 2x² − 7x + 6 = 0，我们有 a = 2，b = −7，c = 6。将这些值代入公式，首先计算判别式。

The discriminant is b² − 4ac = (−7)² − 4(2)(6) = 49 − 48 = 1. Since the discriminant is positive, there are two distinct real roots.

判别式为 b² − 4ac = (−7)² − 4(2)(6) = 49 − 48 = 1。由于判别式为正，方程有两个不同的实根。

Substituting into the quadratic formula gives x = (7 ± √1) / 4, which simplifies to x = (7 ± 1) / 4. The two roots are therefore x = 8/4 = 2 and x = 6/4 = 3/2.

代入二次公式得到 x = (7 ± √1) / 4，化简为 x = (7 ± 1) / 4。因此两个根为 x = 8/4 = 2 和 x = 6/4 = 3/2。

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4. Completing the Square | 配方法

Completing the square is a third method that also reveals the turning point of the quadratic graph. It is often tested in AQA questions that ask for the vertex form of a quadratic function.

配方法是第三种方法，它同时也能揭示二次函数图像的顶点。AQA 考试中经常出现要求写出二次函数顶点式的题目，因此配方法非常重要。

Start by dividing the original equation by 2 so that the coefficient of x² is 1. This gives x² − (7/2)x + 3 = 0, and moving the constant term to the other side gives x² − (7/2)x = −3.

首先将原方程除以 2，使 x² 的系数为 1。得到 x² − (7/2)x + 3 = 0，将常数项移到等号右边，得到 x² − (7/2)x = −3。

Add (7/4)² to both sides because half of 7/2 is 7/4. This gives a perfect square on the left-hand side.

在等式两边同时加上 (7/4)²，因为 7/2 的一半是 7/4。这样左边就成为一个完全平方式。

x² − (7/2)x + (7/4)² = −3 + (7/4)²

Since (7/4)² = 49/16 and −3 = −48/16, the right-hand side becomes 1/16. The left-hand side is (x − 7/4)², so the equation is (x − 7/4)² = 1/16.

因为 (7/4)² = 49/16 且 −3 = −48/16，右边变为 1/16。左边为 (x − 7/4)²，因此方程为 (x − 7/4)² = 1/16。

Taking the square root of both sides gives x − 7/4 = ± 1/4. Adding 7/4 to both sides gives x = 7/4 ± 1/4, which yields x = 2 and x = 6/4 = 3/2.

对两边开平方得到 x − 7/4 = ± 1/4。两边同时加上 7/4，得到 x = 7/4 ± 1/4，从而求得 x = 2 和 x = 6/4 = 3/2。

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5. The Discriminant and Nature of Roots | 判别式与根的性质

The discriminant Δ = b² − 4ac determines how many real solutions a quadratic equation has. In Example 1.7.1, Δ = 1, so the equation has two distinct real roots. This agrees with the results from all three solving methods.

判别式 Δ = b² − 4ac 决定二次方程有多少个实数解。在示例 1.7.1 中，Δ = 1，因此方程有两个不同的实根。这与三种求解方法得到的结果一致。

The following table summarises the three possible cases. AQA questions often ask students to use the discriminant to determine the number of roots without solving the equation fully.

下表总结了三种可能的情况。AQA 考试经常要求学生利用判别式判断根的个数，而无需完整解出方程。

Discriminant | 判别式	Nature of roots | 根的性质	Graph | 图像
Δ > 0	Two distinct real roots | 两个不同实根	Crosses x-axis twice | 与 x 轴相交两次
Δ = 0	One repeated real root | 一个重根	Touches x-axis at vertex | 与 x 轴在顶点处相切
Δ < 0	No real roots | 无实根	Does not meet x-axis | 不与 x 轴相交

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6. Graphical Interpretation | 图像解释

The graph of y = 2x² − 7x + 6 is a U-shaped parabola because the coefficient of x² is positive. It crosses the x-axis at the two roots, x = 3/2 and x = 2. The vertex lies exactly halfway between the roots at x = 7/4.

y = 2x²

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