Year 11 SQA Mathematics: Common Mistakes and How to Fix Them | SQA Year 11 数学：常见误区与纠正方法

📚 Year 11 SQA Mathematics: Common Mistakes and How to Fix Them | SQA Year 11 数学：常见误区与纠正方法

Many Year 11 students sitting SQA Mathematics exams lose marks not because they don't understand the concepts, but because they fall into predictable traps. Spotting and fixing these common mistakes can dramatically boost your grade. This article highlights ten frequent errors together with clear, step‑by‑step corrections.

许多参加 SQA 数学考试的 Year 11 学生丢分，并不是因为没有理解概念，而是掉进了可预见的陷阱。发现并纠正这些常见错误，可以显著提高你的成绩。本文重点介绍十个常见错误，并提供清晰、分步的纠正方法。

1. Misapplying Laws of Indices | 错误应用指数法则

One of the most common slip‑ ing a² × a³ = a⁶. The mistake comes from multiplying the exponents instead of adding them.

最普遍的失误之一是把 a² × a³ 写成 a⁶。错误的原因是指数相乘，而不是相加。

multiplying powers with the same base, add the exponents: aᵐ × aⁿ = aᵐ⁺ⁿ, so a² × a³ = a⁵.

正确法则：同底数幂相乘，指数相加：aᵐ × aⁿ = aᵐ⁺ⁿ，因此 a² × a³ = a⁵。

reating (x³)² as x³⁺² = x⁵. The correct step is to multiply the powers: (x³)² = x³×² = x⁶.

相关错误是把 (x³)² 当作 x³⁺² = x⁵。正确的步骤是指数相乘：(x³)² = x³×² = x⁶。

ause confusion. For instance, 2⁻³ is often wrongly taken as −8. In fact, 2⁻³ = 1/2³ = 1/8.

负指数也会引起混淆。例如，2⁻³ 经常被错误地当作 −8。实际上，2⁻³ = 1/2³ = 1/8。

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2. Sign Errors When Expanding Brackets | 展开括号时的符号错误

When a negative sign sits outside a bracket, many students forget to multiply every term inside. Expanding −(3x − 4) as −3x − 4 is a classic blunder.

当括号外有负号时，许多学生忘记乘以括号内的每一项。把 −(3x − 4) 展开成 −3x − 4 是经典的错误。

s like multiplying by −1, so −(3x − 4) = −1 × (3x − 4) = −3x + 4. Every sign inside flips.

负号相当于乘以 −1，因此 −(3x − 4) = −1 × (3x − 4) = −3x + 4。括号内每一项的符号都要变。

mistake occurs with −2(x − 5). Pupils often write −2x − 10, missing that −2 × (−5) = +10.

另一个常见错误出现在 −2(x − 5) 中。学生经常写成 −2x − 10，忽视了 −2 × (−5) = +10。

ributed first. For −2(x − 5), carry out −2 × x = −2x and −2 × (−5) = +10, giving −2x + 10.

一定要先分配负号。对于 −2(x − 5)，计算 −2 × x = −2x 和 −2 × (−5) = +10，得到 −2x + 10。

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3. Misunderstanding Algebraic Fractions | 误解代数分式

When adding fractions like 3/(x+1) + 2/(x−1), students sometimes incorrectly add numerators and denominators directly, obtaining 5/(2x). This is never valid.

在计算 3/(x+1) + 2/(x−1) 这类分式加法时，学生有时错误地直接将分子和分母分别相加，得到 5/(2x)。这永远不成立。

each fraction: 3(x−1)/(x+1)(x−1) + 2(x+1)/(x+1)(x−1) = (3x−3+2x+2)/(x²−1) = (5x−1)/(x²−1).

正确方法：找到最简公分母 (x+1)(x−1)。重新写分式：3(x−1)/(x+1)(x−1) + 2(x+1)/(x+1)(x−1) = (3x−3+2x+2)/(x²−1) = (5x−1)/(x²−1)。

Simplifying algebraic fractions can also go wrong. Some students think (x²−4)/(x−2) cannot be cancelled, but the numerator factorises as (x−2)(x+2). Provided x ≠ 2, the fraction simplifies to x+2.

代数分式的化简也容易出错。有些学生认为 (x²−4)/(x−2) 不能约分，但分子可以因式分解为 (x−2)(x+2)。只要 x ≠ 2，分式就简化为 x+2。

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4. Solving Linear Equations Incorrectly | 错误解线性方程

3. Pupils sometimes multiply every term on the left by 5 but forget to multiply the right‑hand side correctly, or they misapply the operation.

含分母的方程常出现陷阱，例如 2x/5 + 1 = 3。学生有时将左边的每一项都乘以 5，却忘记正确乘右边，或者运算顺序错误。

t. Subtract 1 from both sides: 2x/5 = 2. Then multiply both sides by 5: 2x = 10, so x = 5.

稳妥的做法：先分离分数项。两边减 1：2x/5 = 2。然后两边乘 5：2x = 10，因此 x = 5。

m, including the +1: 5 × (2x/5) + 5 × 1 = 5 × 3 gives 2x + 5 = 15, again leading to x = 5.

如果你喜欢一开始就乘掉分母，必须乘以每一项，包括 +1：5 × (2x/5) + 5 × 1 = 5 × 3 得到 2x + 5 = 15，同样得出 x = 5。

orrectly write x − 2 = 4 and then x = 6. The error happens when the bracket is handled incorrectly: they might write 3x − 2 = 12. Always expand the bracket or divide the whole bracket.

另一个典型错误是忘记逆运算。对于 3(x − 2) = 12，学生有时除以 3 再加 2，可能错误地写成 x − 2 = 4 然后 x = 6，这步是对的，但错误在展开时写成 3x − 2 = 12。一定要正确展开括号或整体处理括号。

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5. Factorising Quadratics with a > 1 | 二次项系数大于1的因式分解

Factorising 2x² + 7x + 3 trips up many candidates. They often write (2x + 3)(x + 1), which expands to 2x² + 5x + 3 – wrong.

对 2x² + 7x + 3 进行因式分解难住了很多考生。他们经常写成 (2x + 3)(x + 1)，展开后却是 2x² + 5x + 3 —— 错了。

nd 1 × 3 = 3. The sum of the cross terms is 6x + 1x = 7x, matching the middle coefficient.

正确的因式分解是 (2x + 1)(x + 3)，因为 2x × x = 2x²，2x × 3 = 6x，1 × x = 1x，1 × 3 = 3。交叉项的和是 6x + 1x = 7x，与中间项系数吻合。

term: 2x² + 6x + 1x + 3, then factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

为了避免错误，使用系统方法：计算 a×c（2×3=6），找到两数，乘积为 6 和为 7：6 和 1。分解中间项：2x² + 6x + 1x + 3，然后分组分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。

, e.g. −x² + 5x − 6, many struggle. Factor out −1 first: −(x² − 5x + 6) = −(x − 2)(x − 3).

当 x² 的系数为负时，例如 −x² + 5x − 6，很多人无从下手。先提取 −1：−(x² − 5x + 6) = −(x − 2)(x − 3)。

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6. Trigonometry Confusion: SOH CAH TOA | 三角函数混淆：SOH CAH TOA

In right‑ tudents often recall SOH CAH TOA but apply it backwards, e.g. sin θ = adjacent/hypotenuse.

在直角三角形三角学中，搞混对边和邻边是丢分的头号原因。学生往往记得 SOH CAH TOA，却反向应用，比如 sin θ = 邻边/斜边。

se, tan θ = opposite/adjacent. Always label the sides relative to the angle you are using.

正确的比是：sin θ = 对边/斜边，cos θ = 邻边/斜边，tan θ = 对边/邻边。一定要根据所考虑的角度标出各边。

ght write θ = sin 0.5 instead of θ = sin⁻¹(0.5). Ensure your calculator is in degree mode.

求角度时，学生有时忘记使用反三角函数。已知 sin θ = 0.5，可能会写成 θ = sin 0.5，而不是 θ = sin⁻¹(0.5)。务必确保计算器处于角度模式。

is always the denominator, leading to tan = opposite/hypotenuse. Keep the mnemonic clear.

关于 tan θ 的常见失误也很普遍。因为 tan θ = 对边/邻边，有人总以为分母一定是斜边，从而误用 tan = 对边/斜边。要牢记口诀。

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7. Pythagoras: Finding a Shorter Side | 勾股定理：求直角边

A classic mistake occurs when students apply Pythagoras' e is 13 cm and one leg is 5 cm, they often write a² = 13² + 5², giving a² = 194, a = √194.

学生在利用勾股定理求直角边时常常犯一个经典错误。如果斜边为 13 cm，一条直角边为 5 cm，他们往往会写成 a² = 13² + 5²，得出 a² = 194，a = √194。

nged for a shorter side: a² = c² − b². So a² = 13² − 5² = 169 − 25 = 144, hence a = 12 cm.

正确的关系是 a² + b² = c²，其中 c 为斜边。求直角边时需变形：a² = c² − b²。因此 a² = 13² − 5² = 169 − 25 = 144，所以 a = 12 cm。

Using a table can help remember the rearrangement:

使用表格有助于记住变形：

Finding the hypotenuse	Finding a shorter side
c² = a² + b²	a² = c² − b²

Always start by writing the full formula and substitute carefully. If you end up with a negative under the square root, you have placed the numbers the wrong way round.

始终从写出完整公式开始，谨慎代入数值。如果根号下出现了负值，说明你代入的次序颠倒了。

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8. Misinterpreting Statistical Diagrams | 误解统计图表

Box plots often cause errors when students pick the maximum value instead of the upper quartile to find the interquartile range (IQR). The IQR is Q₃ − Q₁, not max − min.

箱线图常导致学生错误：用最大值代替上四分位数来计算四分位距 (IQR)。IQR 是 Q₃ − Q₁，而不是最大值 − 最小值。

For cumulative frequency graphs, misinterpretation of the axis is common. To find the median, draw a line from 50% of the cumulative frequency to the curve and read down to the data axis. Students sometimes read the frequency axis incorrectly, selecting the wrong percentile.

对于累积频率图，最常发生坐标轴误解。找中位数时，应从累积频率的 50% 处画水平线交曲线，再向下读取数据轴。学生有时读错频率轴，选错了百分位数。

requency ÷ class width is vital. Without this, pupils may simply compare bar heights and draw wrong conclusions.

在直方图中，频率与直条的面积成正比，而不是其高度。必须使用频率密度 = 频率 ÷ 组距。否则学生可能仅通过比较条形高度就得出错误结论。

Always annotate the diagram with the key values you extract, and double‑check whether the question asks for an estimate or an exact value.

一定要在图上标出你提取的关键值，并再次核对题目要求的是估计值还是精确值。

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9. Standard Form Slip‑ups | 科学记数法失误

m is 3.6 × 10⁻⁴ because the decimal point must move four places to the right to reach 3.6.

把 0.000 36 写成 3.6 × 10⁻³ 是一个反复出现的错误。正确的科学记数法是 3.6 × 10⁻⁴，因为小数点需要向右移动 4 位才能得到 3.6。

Rule: for a number less than 1, count the zeros before the first non‑zero digit, including the one before the decimal point? e number of moves is the negative exponent. 0.000 36 becomes 3.6 after four moves → ×10⁻⁴.

规则：对于小于 1 的数，把小数点向右移动，直到得到一个 1 到 10 之间的数；移动的位数就是负指数的绝对值。0.000 36 移动 4 次得到 3.6 → ×10⁻⁴。

y. 52 000 000 = 5.2 × 10⁷, not 52 × 10⁶. The mantissa must be at least 1 and less than 10.

转换大数如 52 000 000 时，若小数点放错位置就会

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