Mathematics · AQA 8300 §N7

Surds

AQA 8300 §N7 ⭐⭐⭐⭐⭐ 55 mins AQA · Edexcel · OCR Grade 9
  • Simplify surds and identify like surds
  • Add and subtract like surds
  • Expand brackets containing surds
  • Rationalise denominators including conjugate surds
  • Leave answers in exact surd form

🔑 Core Concepts

What is a Surd?

A surd is an irrational number expressed as the root of an integer. Irrational means it cannot be written as a fraction $\frac{p}{q}$ of two integers; its decimal expansion never terminates and never repeats. Surds arise naturally in geometry (the diagonal of a unit square has length $\sqrt{2}$) and in trigonometry (e.g. $\sin 60° = \frac{\sqrt{3}}{2}$). In GCSE examinations, exact surd form is always preferred over a decimal approximation unless explicitly instructed otherwise.

📚
DEFINITION — Surd
A surd is a root expression whose value is irrational. The expressions $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, $3\sqrt{7}$, and $2+\sqrt{11}$ are all surds. However, $\sqrt{4}=2$, $\sqrt{9}=3$, and $\sqrt{25}=5$ are not surds because they evaluate to rational integers.
🎯
EXAM TIP
If a question says “give your answer in exact form”, “leave as a surd”, or “show that”, do not convert to a decimal. An answer of $3\sqrt{2}$ is exact; $4.243$ is not. Exact form is always required at Grade 7–9.
✗
COMMON MISTAKE — Splitting Under Addition
$\sqrt{4+9} \neq \sqrt{4}+\sqrt{9}$. You cannot split a square root across an addition or subtraction. Check: $\sqrt{13} \approx 3.61$, yet $\sqrt{4}+\sqrt{9}=2+3=5$. These are completely different values.

Simplifying Surds

To simplify $\sqrt{n}$, find the largest perfect-square factor of $n$. A perfect square is a number whose square root is an integer: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, … Once you spot the largest perfect-square factor, apply the multiplication rule.

Multiplication Rule for Surds
$$\sqrt{ab} = \sqrt{a} \times \sqrt{b} \qquad (a,b \geq 0)$$
$a$ = perfect-square factor — $b$ = remaining factor (no further perfect-square factor)
Key Identity: Squaring a Surd
$$\sqrt{a} \times \sqrt{a} = a$$
Squaring any surd eliminates the root entirely
Write $n$ = (largest perfect square) × (remaining factor)
→
$n = k^2 \times m$ where $m$ has no perfect-square factor
→
$\sqrt{n} = k\sqrt{m}$
✎
WORKED EXAMPLE — Simplifying Surds
$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$
$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$
$\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$
$\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}$
🎯
EXAM TIP — Use the Largest Factor
Always identify the largest perfect-square factor so you fully simplify in one step. Using a smaller factor forces a second simplification: $\sqrt{72}=\sqrt{4\times18}=2\sqrt{18}$ is not fully simplified. You must continue: $2\sqrt{18}=2\sqrt{9\times2}=6\sqrt{2}$. One step is quicker and less error-prone.
🧠
MEMORY TRICK — Perfect Square Checklist
For any $\sqrt{n}$, test divisibility in order: 4, 9, 25, 36, 49, 64, 100, 121, 144. The first (largest) one that divides $n$ exactly is your factor. Alternatively, prime-factorise $n$ and pair up identical primes: $\sqrt{2^2 \times 3^2 \times 2} = 2\times3\times\sqrt{2}=6\sqrt{2}$.

Adding and Subtracting Like Surds

Surds behave like algebraic terms. Just as $3x + 5x = 8x$, you can only combine surd terms that share the same irrational part. These are called like surds. Terms with different surds (e.g. $\sqrt{2}$ and $\sqrt{3}$) are unlike and cannot be combined further.

📚
DEFINITION — Like Surds
Like surds share the same irrational factor. $3\sqrt{5}$ and $7\sqrt{5}$ are like surds (both contain $\sqrt{5}$). $3\sqrt{2}$ and $5\sqrt{3}$ are unlike surds and cannot be combined.
✎
WORKED EXAMPLE — Adding and Subtracting
$5\sqrt{3} + 2\sqrt{3} = 7\sqrt{3}$

$\sqrt{12} + \sqrt{27}$: simplify first → $2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$

$\sqrt{50} - \sqrt{8}$: simplify first → $5\sqrt{2} - 2\sqrt{2} = 3\sqrt{2}$

$\sqrt{75} + \sqrt{48} - \sqrt{3} = 5\sqrt{3} + 4\sqrt{3} - \sqrt{3} = 8\sqrt{3}$
✗
COMMON MISTAKE — Combining Unlike Surds
$\sqrt{3}+\sqrt{5} \neq \sqrt{8}$. Unlike surds cannot be added. Always simplify each surd fully first, then check whether the simplified forms are like surds before combining.

Multiplying Surds

Multiplication uses the same rule as simplification, applied directly. Integer coefficients multiply together separately from the surd parts.

Multiplying Surd Expressions
$$a\sqrt{b} \times c\sqrt{d} = ac\sqrt{bd}$$
Integers multiply: $a \times c$ — Surds multiply: $\sqrt{b} \times \sqrt{d} = \sqrt{bd}$
✎
WORKED EXAMPLE — Multiplying Surds
$\sqrt{3} \times \sqrt{12} = \sqrt{36} = 6$

$2\sqrt{5} \times 3\sqrt{5} = 6 \times 5 = 30$

$3\sqrt{2} \times 4\sqrt{6} = 12\sqrt{12} = 12 \times 2\sqrt{3} = 24\sqrt{3}$

$\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4$

Expanding Brackets Containing Surds

Expand exactly as in algebra using the distributive law. For double brackets, use FOIL (First, Outer, Inner, Last). The key extra rule is that $\sqrt{a}\times\sqrt{a}=a$, which eliminates the surd.

✎
WORKED EXAMPLE — Single Bracket
$\sqrt{3}(2 + \sqrt{3}) = 2\sqrt{3} + \sqrt{3}\times\sqrt{3} = 2\sqrt{3} + 3$

$\sqrt{5}(3\sqrt{5} - 2) = 3\times5 - 2\sqrt{5} = 15 - 2\sqrt{5}$
✎
WORKED EXAMPLE — Double Bracket (FOIL)
$(2+\sqrt{3})(5-\sqrt{3})$
F: $2 \times 5 = 10$
O: $2 \times (-\sqrt{3}) = -2\sqrt{3}$
I: $\sqrt{3} \times 5 = 5\sqrt{3}$
L: $\sqrt{3} \times (-\sqrt{3}) = -3$
Collect: $10 - 2\sqrt{3} + 5\sqrt{3} - 3 = 7 + 3\sqrt{3}$
Difference of Two Squares with Surds
$$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b}) = a - b$$
The surd cross-terms cancel — Result is always rational
⚠
IMPORTANT — Why the Difference of Two Squares Matters
When you multiply conjugate surds $(a+\sqrt{b})(a-\sqrt{b})$, the result $a^2-b$ contains no surds. This rational outcome is the entire basis of conjugate rationalisation. Recognise this pattern immediately: it saves working and is a favourite Grade 8–9 trick.
Grade 9: Squaring a Surd Binomial
$$(p + q\sqrt{r})^2 = p^2 + 2pq\sqrt{r} + q^2 r$$
$p^2$: rational — $2pq\sqrt{r}$: surd term (the middle term!) — $q^2 r$: rational
✗
COMMON MISTAKE — Forgetting the Middle Term
$(2+\sqrt{3})^2 \neq 4+3=7$. The correct expansion is $4 + 4\sqrt{3} + 3 = 7+4\sqrt{3}$. The middle term $2\times2\times\sqrt{3}=4\sqrt{3}$ is always present. Use FOIL or the formula above — never skip it.

Rationalising the Denominator

A fraction is not in its simplest, most presentable form when the denominator contains a surd. Rationalising removes the surd from the denominator by multiplying numerator and denominator by a carefully chosen expression. There are two distinct methods.

📚
DEFINITION — Rationalising the Denominator
To rationalise means to rewrite a fraction so that its denominator is a rational number (an integer or simple fraction) while preserving the value of the expression. You multiply by a form of 1: $\frac{\text{something}}{\text{something}}$.

Method 1 — Simple Surd Denominator ($\sqrt{n}$ only)

If the denominator is a single surd $\sqrt{n}$, multiply top and bottom by $\dfrac{\sqrt{n}}{\sqrt{n}}$. Since $\sqrt{n}\times\sqrt{n}=n$, the denominator becomes the integer $n$.

Rationalising $\dfrac{a}{\sqrt{n}}$
$$\frac{a}{\sqrt{n}} \times \frac{\sqrt{n}}{\sqrt{n}} = \frac{a\sqrt{n}}{n}$$
Denominator: $\sqrt{n}\times\sqrt{n}=n$ — Simplify $\dfrac{a}{n}$ if possible
✎
WORKED EXAMPLE — Simple Rationalisation
$\dfrac{6}{\sqrt{3}} = \dfrac{6}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$

$\dfrac{10}{\sqrt{5}} = \dfrac{10\sqrt{5}}{5} = 2\sqrt{5}$

$\dfrac{4}{\sqrt{8}} = \dfrac{4}{2\sqrt{2}} = \dfrac{2}{\sqrt{2}} = \dfrac{2\sqrt{2}}{2} = \sqrt{2}$

Method 2 — Conjugate Rationalisation (Binomial Denominator, Grade 8–9)

When the denominator is a binomial involving a surd, such as $(a+\sqrt{b})$ or $(a-\sqrt{b})$, multiply numerator and denominator by the conjugate — the same expression with the sign of the surd term reversed. This exploits the difference-of-two-squares identity to produce a rational denominator.

📚
DEFINITION — Conjugate
The conjugate of $(a+\sqrt{b})$ is $(a-\sqrt{b})$, and vice versa. Their product $(a+\sqrt{b})(a-\sqrt{b})=a^2-b$ is always rational (an integer). The surd terms cancel due to the difference-of-two-squares identity.
Rationalising $\dfrac{c}{a+\sqrt{b}}$ Using Conjugate
$$\frac{c}{a+\sqrt{b}} \times \frac{a-\sqrt{b}}{a-\sqrt{b}} = \frac{c(a-\sqrt{b})}{a^2-b}$$
Conjugate: flip sign of surd term — New denominator: $a^2-b$ (always rational)
✎
WORKED EXAMPLE — Conjugate Rationalisation
$\dfrac{5}{3+\sqrt{2}} \times \dfrac{3-\sqrt{2}}{3-\sqrt{2}} = \dfrac{5(3-\sqrt{2})}{9-2} = \dfrac{15-5\sqrt{2}}{7}$

$\dfrac{1}{2-\sqrt{3}} \times \dfrac{2+\sqrt{3}}{2+\sqrt{3}} = \dfrac{2+\sqrt{3}}{4-3} = 2+\sqrt{3}$
🎯
EXAM TIP — Check Your Conjugate
Conjugate of $(3+\sqrt{5})$ is $(3-\sqrt{5})$. The denominator product is $9-5=4$. If $a^2-b=1$, the answer is the expanded numerator — a favourite Grade 9 trap! For example: denominator $(1+\sqrt{2})(1-\sqrt{2})=1-2=-1$, so you must divide by $-1$, changing all signs.
🧠
MEMORY TRICK — Conjugate = Sign Flip
To write the conjugate: find the surd term in the denominator and flip its sign. $(5+3\sqrt{7}) \rightarrow (5-3\sqrt{7})$. Their product: $5^2-(3\sqrt{7})^2=25-63=-38$. Always compute the denominator product first.

🗺 Visual Notes

Surds
Simplifying
  • Find largest perfect-square factor
  • $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$
  • $\sqrt{72}=\sqrt{36\times2}=6\sqrt{2}$
  • Check: no perfect-square factors remain
Adding & Subtracting
  • Only like surds combine
  • Simplify each surd first
  • $3\sqrt{5}+7\sqrt{5}=10\sqrt{5}$
  • $\sqrt{12}+\sqrt{3}=2\sqrt{3}+\sqrt{3}=3\sqrt{3}$
Multiplying & Expanding
  • $\sqrt{a}\times\sqrt{a}=a$
  • FOIL for double brackets
  • $(a+\sqrt{b})(a-\sqrt{b})=a^2-b$
  • $(p+q\sqrt{r})^2=p^2+2pq\sqrt{r}+q^2r$
Rationalising
  • Simple: multiply by $\frac{\sqrt{n}}{\sqrt{n}}$
  • Conjugate: flip the surd sign
  • $(a+\sqrt{b})(a-\sqrt{b})=a^2-b$
  • Result always has rational denominator
Exact Form
  • Never convert to decimal
  • Geometry: diagonals, heights
  • $\sin 60°=\frac{\sqrt{3}}{2}$, $\sin 45°=\frac{\sqrt{2}}{2}$
  • Area and perimeter problems
Grade 9 Skills
  • Proving surd equivalences
  • Solving equations with surds
  • Conjugate with $a^2-b=1$
  • $(p+q\sqrt{r})$ form answers

Comparison: Simple vs Conjugate Rationalisation

Feature Simple Rationalisation Conjugate Rationalisation
Denominator type $\sqrt{n}$ (single surd only) $a \pm \sqrt{b}$ (binomial with surd)
Multiplier used $\dfrac{\sqrt{n}}{\sqrt{n}}$ $\dfrac{a \mp \sqrt{b}}{a \mp \sqrt{b}}$
Key identity $\sqrt{n}\times\sqrt{n}=n$ $(a+\sqrt{b})(a-\sqrt{b})=a^2-b$
New denominator $n$ $a^2-b$
Example $\dfrac{3}{\sqrt{5}}=\dfrac{3\sqrt{5}}{5}$ $\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}$
Grade level Grade 6–7 Grade 8–9

Surd or Not? Quick Classification

Expression Value Is it a surd? Reason
$\sqrt{2}$$1.4142\ldots$YesIrrational, non-terminating decimal
$\sqrt{4}$$2$NoPerfect square, rational integer
$\sqrt{9}$$3$NoPerfect square, rational integer
$\sqrt{5}$$2.2360\ldots$YesIrrational, non-terminating decimal
$\sqrt{49}$$7$NoPerfect square
$\sqrt{50}$$5\sqrt{2}$YesSimplifies to a surd form
$\sqrt{100}$$10$NoPerfect square

Decision Tree: Which Rationalisation Method?

Is there a surd in the denominator?
→
Is the denominator a single surd $\sqrt{n}$?
→
YES: multiply by $\dfrac{\sqrt{n}}{\sqrt{n}}$
→
NO: identify conjugate $a\mp\sqrt{b}$ and multiply by it
→
Expand, simplify fraction

✎ Worked Examples

Grade 4–5
Simplify $\sqrt{48}$ and $\sqrt{75}$. Hence calculate $\sqrt{48}+\sqrt{75}$, giving your answer in the form $k\sqrt{3}$.
1
Simplify $\sqrt{48}$: find the largest perfect-square factor Check divisibility by 4, 9, 16, 25… $48 \div 16 = 3$. So $48 = 16 \times 3$: $$\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16}\times\sqrt{3} = 4\sqrt{3}$$
2
Simplify $\sqrt{75}$: find the largest perfect-square factor $75 \div 25 = 3$. So $75 = 25 \times 3$: $$\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25}\times\sqrt{3} = 5\sqrt{3}$$
3
Add like surds Both terms now contain $\sqrt{3}$, so they are like surds. Collect: $$\sqrt{48}+\sqrt{75} = 4\sqrt{3}+5\sqrt{3} = 9\sqrt{3}$$
$$\sqrt{48}+\sqrt{75} = 9\sqrt{3} \qquad (k=9)$$
Grade 6–7
Expand and simplify $(3+\sqrt{5})(3-\sqrt{5})$. Hence rationalise $\dfrac{4}{3+\sqrt{5}}$, giving your answer in the form $a+b\sqrt{5}$ where $a$ and $b$ are integers.
1
Recognise the difference-of-two-squares pattern $(3+\sqrt{5})(3-\sqrt{5})$ matches $(a+b)(a-b)=a^2-b^2$ with $a=3$, $b=\sqrt{5}$: $$(3+\sqrt{5})(3-\sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4$$
2
Rationalise using conjugate $(3-\sqrt{5})$ $$\frac{4}{3+\sqrt{5}} = \frac{4}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3-\sqrt{5}} = \frac{4(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$$
3
Substitute the denominator result from Step 1 $$= \frac{4(3-\sqrt{5})}{4} = 3-\sqrt{5}$$
4
Write in the required form $a+b\sqrt{5}$ $3-\sqrt{5} = 3+(-1)\sqrt{5}$, so $a=3$ and $b=-1$.
$$\frac{4}{3+\sqrt{5}} = 3-\sqrt{5} \qquad (a=3,\; b=-1)$$
Grade 9
Show that $\dfrac{5+2\sqrt{3}}{2+\sqrt{3}} = a+b\sqrt{3}$ where $a$ and $b$ are integers. State the values of $a$ and $b$.

Hence find $\left(\dfrac{5+2\sqrt{3}}{2+\sqrt{3}}\right)^2$, leaving your answer in the form $p+q\sqrt{3}$.
1
Identify the conjugate of the denominator Denominator is $2+\sqrt{3}$. Its conjugate is $2-\sqrt{3}$ (flip the sign of the surd term).
2
Multiply numerator and denominator by the conjugate $$\frac{5+2\sqrt{3}}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{(5+2\sqrt{3})(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}$$
3
Compute the denominator (difference of two squares) $$(2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1$$
4
Expand the numerator using FOIL $$(5+2\sqrt{3})(2-\sqrt{3})$$ $$= 5\times2 + 5\times(-\sqrt{3}) + 2\sqrt{3}\times2 + 2\sqrt{3}\times(-\sqrt{3})$$ $$= 10 - 5\sqrt{3} + 4\sqrt{3} - 2\times3$$ $$= 10 - 5\sqrt{3} + 4\sqrt{3} - 6 = 4 - \sqrt{3}$$
5
Combine: divide numerator by 1 $$\frac{4-\sqrt{3}}{1} = 4-\sqrt{3}$$ Therefore $a=4$ and $b=-1$.
6
Square the result using $(p+q\sqrt{r})^2 = p^2+2pq\sqrt{r}+q^2r$ With $p=4$, $q=-1$, $r=3$: $$(4-\sqrt{3})^2 = 4^2 + 2\times4\times(-1)\times\sqrt{3} + (-1)^2\times3$$ $$= 16 - 8\sqrt{3} + 3 = 19-8\sqrt{3}$$
$$\frac{5+2\sqrt{3}}{2+\sqrt{3}} = 4-\sqrt{3} \quad (a=4,\; b=-1)$$ $$\left(\frac{5+2\sqrt{3}}{2+\sqrt{3}}\right)^2 = 19-8\sqrt{3} \quad (p=19,\; q=-8)$$

❓ Exam Questions

Q1 1 mark

Write $\sqrt{75}$ in simplified surd form.

Mark scheme:
$\sqrt{75} = \sqrt{25\times3} = 5\sqrt{3}$ ✓ (B1)

Must use perfect-square factor 25 (or equivalent method). Answer $5\sqrt{3}$ only.
Q2 2 marks

Expand and simplify $(\sqrt{5}+2)^2$, giving your answer in the form $a+b\sqrt{5}$.

Mark scheme:
$(\sqrt{5}+2)^2 = (\sqrt{5})^2 + 2\times\sqrt{5}\times2 + 2^2$ (M1 for correct three-term expansion)
$= 5 + 4\sqrt{5} + 4 = 9+4\sqrt{5}$ (A1 for correct final answer)

$a=9$, $b=4$.
Q3 3 marks

Rationalise the denominator of $\dfrac{3}{\sqrt{7}-2}$, giving your answer in the form $a+b\sqrt{7}$ where $a$ and $b$ are integers.

Mark scheme:
Multiply by $\dfrac{\sqrt{7}+2}{\sqrt{7}+2}$ (M1 for correct conjugate)

Denominator: $(\sqrt{7}-2)(\sqrt{7}+2) = 7-4 = 3$ (M1 for correct denominator)

$\dfrac{3(\sqrt{7}+2)}{3} = \sqrt{7}+2$ (A1 for correct simplified answer)

So $a=2$, $b=1$.
Q4 4 marks

Show that $\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}} = 2+\sqrt{3}$.

Mark scheme:
Multiply numerator and denominator by conjugate $(\sqrt{6}+\sqrt{2})$: (M1)

Denominator: $(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2}) = 6-2 = 4$ (M1 for correct denominator)

Numerator: $(\sqrt{6}+\sqrt{2})^2 = 6 + 2\sqrt{12} + 2 = 8+2\sqrt{12}$ (M1 for correct numerator)

Simplify: $\sqrt{12}=2\sqrt{3}$, so numerator $= 8+4\sqrt{3}$

$\dfrac{8+4\sqrt{3}}{4} = 2+\sqrt{3}$ ✓ (A1 — shown as required)
Q5 6 marks

A rectangle has length $(3+\sqrt{5})$ cm and width $(3-\sqrt{5})$ cm.

(a) Find the area of the rectangle. [2]

(b) Find the perimeter of the rectangle. [1]

(c) Show that the length of the diagonal is $2\sqrt{7}$ cm. [3]

(a) Area:
$(3+\sqrt{5})(3-\sqrt{5}) = 3^2-(\sqrt{5})^2 = 9-5 = 4$ cm² (M1 for difference of squares, A1) [2]

(b) Perimeter:
$P = 2[(3+\sqrt{5})+(3-\sqrt{5})] = 2\times6 = 12$ cm (B1) [1]
Note: $\sqrt{5}$ terms cancel when adding conjugates.

(c) Diagonal by Pythagoras:
$d^2 = (3+\sqrt{5})^2+(3-\sqrt{5})^2$ (M1 for Pythagoras)
$(3+\sqrt{5})^2 = 9+6\sqrt{5}+5 = 14+6\sqrt{5}$
$(3-\sqrt{5})^2 = 9-6\sqrt{5}+5 = 14-6\sqrt{5}$
$d^2 = 14+6\sqrt{5}+14-6\sqrt{5} = 28$ (M1 for correct sum with surds cancelling)
$d = \sqrt{28} = \sqrt{4\times7} = 2\sqrt{7}$ cm ✓ (A1 — shown) [3]

⭐ Grade 9 Model Answers

The following provides a fully annotated model answer for Question 5, which requires three different surd techniques applied in sequence in a geometrical context. This is typical of how Grade 9 surd questions are structured.

⚠
IMPORTANT — What Makes This Grade 9
Question 5 is Grade 9 because it demands: (1) immediate recognition of the difference-of-two-squares identity in a geometry context, (2) neat cancellation of conjugate surds in the perimeter, (3) careful expansion of two squared binomials using FOIL, (4) recognition that the surd terms cancel in the Pythagoras sum, and (5) correct simplification of $\sqrt{28}$. Every step tests a different surd skill.

Q5: Full Annotated Model Answer

Grade 9 Full Solution
Rectangle with length $(3+\sqrt{5})$ cm and width $(3-\sqrt{5})$ cm. Find area, perimeter, and show diagonal $= 2\sqrt{7}$ cm.
A
Part (a): Area — earns M1 + A1 Strategy: Recognise that length and width are conjugates. Apply the difference-of-two-squares identity immediately rather than expanding in full: $$\text{Area} = (3+\sqrt{5})(3-\sqrt{5}) = 3^2-(\sqrt{5})^2 = 9-5 = 4 \text{ cm}^2$$ Examiner note: M1 for correct method (either identity or full FOIL); A1 for answer 4. A student who expands fully but makes an arithmetic error may still earn M1.
B
Part (b): Perimeter — earns B1 Strategy: Add conjugates first; note the $\sqrt{5}$ terms cancel: $$l+w = (3+\sqrt{5})+(3-\sqrt{5}) = 6$$ $$P = 2\times6 = 12 \text{ cm}$$ Examiner note: B1 for 12. Surds must be seen to cancel (or their cancellation implicit in correct answer).
C
Part (c): Diagonal — earns M1 + M1 + A1 Strategy: State Pythagoras clearly, then expand each squared term individually: $$d^2 = (3+\sqrt{5})^2+(3-\sqrt{5})^2$$ Expand the first square (using $(p+q)^2=p^2+2pq+q^2$): $$(3+\sqrt{5})^2 = 9+6\sqrt{5}+5 = 14+6\sqrt{5}$$ Expand the second square: $$(3-\sqrt{5})^2 = 9-6\sqrt{5}+5 = 14-6\sqrt{5}$$ Add — observe the surd terms $+6\sqrt{5}$ and $-6\sqrt{5}$ cancel perfectly: $$d^2 = (14+6\sqrt{5})+(14-6\sqrt{5}) = 28$$ Simplify: $d = \sqrt{28} = \sqrt{4\times7} = 2\sqrt{7}$ cm ✓
Examiner note: M1 for Pythagoras set up; M1 for correctly expanding both squares; A1 for $2\sqrt{7}$ shown.
Area = 4 cm²  |  Perimeter = 12 cm  |  Diagonal = $2\sqrt{7}$ cm
All answers left in exact form — no decimals used.
🎯
EXAM TIP — Maximising Marks on ‘Show That’ Questions
When the question says “show that”, full intermediate working is compulsory. Write out $(3\pm\sqrt{5})^2$ expanded in full even if the steps feel obvious. Do not jump from $d^2=(3+\sqrt{5})^2+(3-\sqrt{5})^2$ directly to $d^2=28$ — show the intermediate sums $14\pm6\sqrt{5}$ to earn method marks. The final answer is already given; the marks reward the process.

📋 Revision Sheet

Key Definitions
TermMeaning
SurdIrrational root expression, e.g. $\sqrt{2}$, $3\sqrt{5}$
Like surdsSame irrational part, e.g. $3\sqrt{5}$ and $7\sqrt{5}$
ConjugateSame terms, opposite sign: $(a+\sqrt{b})\to(a-\sqrt{b})$
RationaliseRemove surds from the denominator
Exact formAnswer left as surd, not converted to decimal
Perfect squareInteger whose root is also an integer: 4, 9, 16, 25…
Essential Formulae

$$\sqrt{ab} = \sqrt{a}\times\sqrt{b}$$

$$\sqrt{a}\times\sqrt{a} = a$$

$$(a+\sqrt{b})(a-\sqrt{b}) = a^2-b$$

$$\frac{c}{a+\sqrt{b}} = \frac{c(a-\sqrt{b})}{a^2-b}$$

$$(p+q\sqrt{r})^2 = p^2+2pq\sqrt{r}+q^2r$$

$$a\sqrt{b}\times c\sqrt{d} = ac\sqrt{bd}$$

Memory Hooks
  • SPLIT: $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ — split at largest perfect-square factor
  • LIKE TERMS: only surds with identical irrational parts can be combined
  • FOIL: expand double surd brackets exactly as algebra
  • FLIP THE SIGN: conjugate = reverse the sign of the surd term
  • SQUARE CANCELS: $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b}) = a-b$ (always rational)
  • NEVER DECIMALISE: exact surd form is always preferred
Exam Strategy Tips
  • Always use the largest perfect-square factor to simplify in one step
  • Simplify each surd fully before attempting to add or subtract
  • Write out all FOIL steps — method marks are available for correct expansion
  • After rationalising, simplify the resulting fraction (divide by HCF)
  • Conjugate product $a^2-b$: if this equals 1, the fraction equals the expanded numerator
  • In ‘show that’ questions: show every intermediate line of working
  • Perfect squares to memorise: 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144

🔄 Flashcards

Click each card to reveal the answer. Use these for self-testing before an exam.

✗ Common Mistakes

✗
Mistake 1 — Adding Numbers Under Separate Roots
What students do: $\sqrt{4}+\sqrt{9} = \sqrt{4+9} = \sqrt{13}$
Why marks are lost: This is completely wrong. $\sqrt{4}+\sqrt{9}=2+3=5$, not $\sqrt{13}\approx3.61$. It demonstrates a fundamental misunderstanding of surds.
How to avoid: Evaluate or simplify each root separately. You can never merge two roots that are added or subtracted under a single root sign.
✗
Mistake 2 — Not Using the Largest Perfect-Square Factor
What students do: $\sqrt{72}=\sqrt{4\times18}=2\sqrt{18}$ and stop.
Why marks are lost: $2\sqrt{18}$ is not fully simplified. Examiners expect the simplest form $6\sqrt{2}$. Leaving a partially simplified answer may score no marks on the accuracy mark.
How to avoid: After simplifying, check whether the remaining number under the root has any perfect-square factors. Keep going until none remain.
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Mistake 3 — Adding Unlike Surds
What students do: $3\sqrt{2}+4\sqrt{3}=7\sqrt{5}$ or $7\sqrt{6}$
Why marks are lost: Unlike surds ($\sqrt{2}$ and $\sqrt{3}$ are different irrational values) cannot be combined, just as $3x+4y$ cannot simplify to $7xy$ in algebra. This error scores zero.
How to avoid: Ask: “Do these terms have the same number under the root?” Only if yes can they be collected. Otherwise leave as separate terms.
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Mistake 4 — Using the Same Expression Instead of the Conjugate
What students do: To rationalise $\dfrac{1}{3+\sqrt{2}}$, multiply by $\dfrac{3+\sqrt{2}}{3+\sqrt{2}}$.
Why marks are lost: The denominator becomes $(3+\sqrt{2})^2=11+6\sqrt{2}$ — still irrational, and the problem is made worse. Zero marks for the rationalisation step.
How to avoid: The conjugate always has the opposite sign in front of the surd: $(3+\sqrt{2})\to(3-\sqrt{2})$. Flip the sign, do not copy.
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Mistake 5 — Omitting the Middle Term When Squaring
What students do: $(2+\sqrt{3})^2=4+3=7$
Why marks are lost: The middle term $2\times2\times\sqrt{3}=4\sqrt{3}$ is missing. The correct answer is $7+4\sqrt{3}$. This error costs 1–2 marks and is among the most common on this topic.
How to avoid: Always write out $(a+b)^2=a^2+2ab+b^2$ or use FOIL step by step. Never skip the $2ab$ term.
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Mistake 6 — Not Simplifying After Rationalising
What students do: $\dfrac{6}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3}$ — then stop without simplifying.
Why marks are lost: The expected final answer is $2\sqrt{3}$. Leaving $\dfrac{6\sqrt{3}}{3}$ may lose the final accuracy mark, even though the method is correct.
How to avoid: After rationalising, always divide numerator and denominator by their HCF. Check whether the fraction can be simplified.

✅ Final Checklist

Click each item once you are confident with it. Your progress is saved automatically.

  • I can explain what a surd is and give three examples
  • I know which square roots are surds and which are rational integers
  • I can find the largest perfect-square factor of any given integer
  • I can simplify surds such as $\sqrt{72}$, $\sqrt{200}$, and $\sqrt{175}$ in one step
  • I can identify like surds and explain why unlike surds cannot be combined
  • I always simplify each surd fully before attempting to add or subtract
  • I can multiply surds using $\sqrt{a}\times\sqrt{b}=\sqrt{ab}$ and simplify the result
  • I can expand single brackets containing surds and simplify
  • I can expand double brackets with surds using FOIL, collecting like terms
  • I can recognise and apply the pattern $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$
  • I can rationalise a simple surd denominator by multiplying by $\dfrac{\sqrt{n}}{\sqrt{n}}$
  • I can identify the conjugate of any binomial surd expression
  • I can rationalise a denominator of the form $a\pm\sqrt{b}$ using conjugates
  • I can expand $(p+q\sqrt{r})^2$ correctly, never omitting the middle term
  • I consistently leave answers in exact surd form without converting to decimals
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