Simplifying Surds, Rationalising Denominators, and the Exact-Value Rule

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

What a surd is, and the two laws that let you simplify one

A surd is a root — almost always a square root, on this paper — that is left in exact form because it cannot be written as a whole number or a terminating fraction: $\sqrt{2}$ , $\sqrt{5}$ , $3\sqrt{7}$ are all surds. $\sqrt{9}$ is not a surd once evaluated, because it equals the rational number 3 exactly; the whole point of a surd is that no such rational number exists for it. Spec 1.2 asks for two things about them: the ability to manipulate and simplify a surd expression, and the ability to rationalise a denominator that one appears in.

Two laws do essentially all the simplifying work, and neither is a new rule to learn from scratch: $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ and $\sqrt{a}/\sqrt{b} = \sqrt{a/b}$ (for $a, b \geq 0$ , and $b > 0$ in the second case). Both are the index laws from spec 1.1 wearing a different notation. Since $\sqrt{a} = a^{1/2}$ , the first law is just $a^{m} \times b^{m} = (ab)^{m}$ with $m = \frac{1}{2}$ : $a^{1/2}b^{1/2} = (ab)^{1/2} = \sqrt{ab}$ . Nothing about square roots specifically makes this true — it is a fractional-exponent instance of a law you already have.

Simplifying a surd means pulling the LARGEST perfect-square factor out of what's under the root, in one pass — not stopping at the first square factor you happen to notice. $\sqrt{72}$ : the perfect-square factors of 72 are 1, 4, 9 and 36 — and 36 is the largest, since $72 = 36 \times 2$ . So $\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36}\sqrt{2} = 6\sqrt{2}$ . Stopping at a smaller square factor gives a technically-true but incompletely-simplified answer — $\sqrt{72} = \sqrt{4 \times 18} = 2\sqrt{18}$ is correct arithmetic, but $\sqrt{18}$ still has its own square factor (9) sitting inside it, so this is not yet the simplest form the question is asking for.

Adding and subtracting surds — like terms, with one extra step first

$a\sqrt{n} + b\sqrt{n} = (a+b)\sqrt{n}$ works for exactly the same reason $3x + 5x = 8x$ works: both are collecting a shared factor. In the algebra case the shared factor is the variable $x$ ; here it's the surd part $\sqrt{n}$ . Just as $3x$ and $5y$ cannot be combined into one term because $x$ and $y$ are different unknowns, $a\sqrt{n}$ and $b\sqrt{m}$ cannot be combined when $n \neq m$ — there is no rule that merges them, and the expression stays as two separate terms.

The extra step, and the one this topic actually tests: two surds have to be simplified to their smallest form BEFORE you can tell whether their surd parts genuinely match. $\sqrt{8}$ and $\sqrt{18}$ look unrelated on the page — different numbers under the root — but $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$ and $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$ : once simplified, both carry $\sqrt{2}$ , and $2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}$ . Trying to add before simplifying either hides a genuine match (leaving two un-combined terms that were always secretly the same surd part) or, just as commonly, tempts a student into an invalid shortcut like adding what's under the two different roots directly.

It cuts the other way too: once every surd genuinely is in simplest form, $a\sqrt{2} + b\sqrt{3}$ really is as far as it goes. $\sqrt{2}$ and $\sqrt{3}$ are not the same number wearing different disguises — they are two different irrational numbers — and no amount of further manipulation turns two unlike, fully-simplified surds into one term. Recognising when an expression is already finished is as much a part of this skill as knowing how to combine terms that do match.

Mechanism

Why multiplying by √b over itself clears a single surd from the denominator

The move for a denominator that's a single surd — $\frac{a}{\sqrt{b}}$ — is to multiply top and bottom by $\sqrt{b}$ : $\frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$ . The reason it's allowed at all is that $\frac{\sqrt{b}}{\sqrt{b}}$ equals exactly 1 (any non-zero number divided by itself is 1), so multiplying by it changes nothing about the fraction's value — only its form. And the reason it's useful is what happens on the bottom specifically: $\sqrt{b} \times \sqrt{b} = b$ , a whole, rational number, precisely because squaring is the operation a square root undoes. The surd hasn't disappeared from the calculation — it's moved to the numerator, where a surd is allowed to sit; the point of rationalising is only ever to clear it off the bottom.

Mechanism

The conjugate isn't a memorised trick — it's the one multiplication where the cross-terms cancel

A denominator like $2 + \sqrt{3}$ has two terms, and multiplying it by $\sqrt{3}$ alone — the single-surd move above — does not work: $(2+\sqrt{3}) \times \sqrt{3} = 2\sqrt{3} + 3$ , which still has a surd in it, because the '2' was never touched. What's needed is an expression that, multiplied against $2+\sqrt{3}$ , makes the surd vanish completely, not just move. Expand $(p+\sqrt{q})(p-\sqrt{q})$ in full: $p^2 - p\sqrt{q} + p\sqrt{q} - (\sqrt{q})^2 = p^2 - q$ . The two middle terms, $-p\sqrt{q}$ and $+p\sqrt{q}$ , are exact opposites and cancel outright — this is the difference-of-two-squares , $(x+y)(x-y) = x^2 - y^2$ , applied with $x=p$ and $y=\sqrt{q}$ . $p^2 - q$ is a whole number whenever $p$ and $q$ are, with no root sign left anywhere. That expression, $p - \sqrt{q}$ — same two terms as the original denominator, opposite sign in the middle — is called the conjugate, and it is the ONLY multiplier (up to an overall constant) that makes the cross-terms cancel this way. Multiplying by $p+\sqrt{q}$ again, or by $\sqrt{q}$ alone, both leave a cross-term standing, because neither one sets up the $x^2 - y^2$ cancellation the conjugate specifically produces.

Worked, in full

Rationalise $\frac{1}{3-\sqrt{5}}$ — a binomial denominator, worked from why √5/√5 alone fails to a checked final answer

01
Check whether the single-surd move even applies first. Multiplying top and bottom by $\sqrt{5}$ alone gives $(3-\sqrt{5})\times\sqrt{5} = 3\sqrt{5} - 5$ on the denominator — still a surd, because the standalone $3$ was never addressed. A two-term denominator needs the conjugate, not a single surd.
Earns: No mark of its own yet — this check is what motivates reaching for the conjugate instead of the single-surd move, not something a mark scheme scores directly. Skipping it and guessing the wrong multiplier is the single most common way to arrive at a still-irrational denominator.
02
Identify the conjugate of $3-\sqrt{5}$ : same two terms, opposite sign in the middle, so $3+\sqrt{5}$ . Multiply top and bottom by it: $\frac{1}{3-\sqrt{5}} \times \frac{3+\sqrt{5}}{3+\sqrt{5}}$ .
Earns: M1 — attempts to rationalise by multiplying numerator and denominator by the conjugate of the given denominator.
03
Expand the denominator via the difference of two squares: $(3-\sqrt{5})(3+\sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4$ . The cross terms ( $3\sqrt{5}$ and $-3\sqrt{5}$ ) cancel exactly — this is the whole reason the conjugate, and nothing else, was the right choice.
Earns: A1 — correct rational denominator, dependent on the M1 above.
04
Expand the numerator: $1 \times (3+\sqrt{5}) = 3+\sqrt{5}$ .
Earns: A1 — numerator correctly expanded.
05
State the answer: $\frac{1}{3-\sqrt{5}} = \frac{3+\sqrt{5}}{4}$ . Check by decimal substitution: $3-\sqrt{5} \approx 0.764$ , so $\frac{1}{3-\sqrt{5}} \approx 1.309$ ; and $\frac{3+\sqrt{5}}{4} \approx \frac{5.236}{4} \approx 1.309$ — matching.
Earns: A1 — final answer, correct answer only. This substitution check costs about ten seconds and catches a wrong conjugate sign every time — build it into the habit rather than trusting the algebra by eye.

In your own words

In one sentence: why does multiplying top and bottom by $p - \sqrt{q}$ clear the surd from a denominator of $p + \sqrt{q}$ , when multiplying by $p + \sqrt{q}$ again would not?

Complete it yourself

Complete the chain — rationalise $\frac{1}{\sqrt{6}-2}$

01
This denominator is a binomial with a surd, $\sqrt{6}-2$ — multiplying top and bottom by $\sqrt{6}$ alone would leave a surd behind: $\sqrt{6}\times(\sqrt{6}-2) = 6 - 2\sqrt{6}$ , still carrying a $-2\sqrt{6}$ term. A single-surd multiplier will not clear this.
02
The conjugate of $\sqrt{6}-2$ is $\sqrt{6}+2$ — same two terms, opposite sign in the middle. Multiply top and bottom by it: $\frac{1}{\sqrt{6}-2} \times \frac{\sqrt{6}+2}{\sqrt{6}+2}$ .

Same question, every valid method

Simplify $\frac{1}{\sqrt{12}}$ , giving your answer in simplest surd form. (VERIDIAN-original question — not a reproduction of any past-paper question. The per-line mark allocations are modelled on the general M1/A1 conventions for a manipulation question, verified in WMA11-verified-facts.md §3, not transcribed from a real scheme, which for an original question does not exist.)

2 valid methods · every one reaches $\frac{\sqrt{3}}{6}$ · 2 marks available

01
$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ , so $\frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}}$ .
Method mark for extracting the square factor from the denominator before attempting to rationalise it.
M1
02
$\frac{1}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{2 \times 3} = \frac{\sqrt{3}}{6}$ .
Accuracy mark for the correct final answer, in simplest form.
A1

Simplifying the denominator first keeps the rationalising multiplication small — $\sqrt{3}$ instead of $\sqrt{12}$ — which is the safer route whenever the square factor inside the denominator is easy to spot. It also often produces a fraction that's already in lowest terms, with less cancelling to do at the end.

Marked, line by line

(a) Simplify $\sqrt{75} - \sqrt{12}$ , giving your answer in the form $k\sqrt{3}$ , where $k$ is an integer. (2) (b) Rationalise the denominator of $\frac{5}{2-\sqrt{3}}$ , giving your answer in the form $p + q\sqrt{3}$ , where $p$ and $q$ are integers. (3) (c) Hence find the exact value of $(\sqrt{75} - \sqrt{12}) \times \frac{5}{2-\sqrt{3}}$ , giving your answer in the form $a + b\sqrt{3}$ , where $a$ and $b$ are integers. (3) — VERIDIAN-original question, built to combine spec 1.2's two named techniques (simplifying a surd by its largest square factor, and rationalising a binomial-conjugate denominator) in the multi-part, 'hence'-reusing-earlier-parts shape real WMA11 questions use — the same structure the pilot lesson's own marked-solution used for its discriminant/completing-the-square question. Not a reproduction of any past-paper question, and the per-line mark allocations below are modelled on verified mark-scheme conventions (M for method, A dependent on M and cao unless ft) rather than copied from a real scheme, which for an original question does not exist.

8 marks available

(a) — 2 marks

01
$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$ and $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ .
Method mark for extracting the largest square factor from at least one of the two surds. Earned for the attempt, whatever happens to the subtraction that follows.
M1
02
$\sqrt{75} - \sqrt{12} = 5\sqrt{3} - 2\sqrt{3} = 3\sqrt{3}$ .
Accuracy mark, correct answer only. Both surds had to be simplified to the SAME surd part before they could be combined at all.
A1

(b) — 3 marks

101
$\frac{5}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}}$ , multiplying by the conjugate of the denominator.
Method mark for rationalising by the conjugate of a binomial denominator. Multiplying by 2−√3 again, or by √3 alone, would not earn this — neither clears the surd.
M1
102
$(2-\sqrt{3})(2+\sqrt{3}) = 4 - 3 = 1$ ; numerator $= 5(2+\sqrt{3}) = 10+5\sqrt{3}$ .
Accuracy mark for the correct denominator (via the difference of two squares) and the correctly expanded numerator.
A1
103
$\frac{5}{2-\sqrt{3}} = 10 + 5\sqrt{3}$ , so $p=10$ , $q=5$ .
Accuracy mark, correct answer only, in the exact form the question specifies.
A1

(c) — 3 marks

201
Hence: $(\sqrt{75}-\sqrt{12}) \times \frac{5}{2-\sqrt{3}} = 3\sqrt{3} \times (10+5\sqrt{3})$ .
Method mark for multiplying their own answers to (a) and (b) together with an attempt to expand — the 'hence' instruction is a direct signal to reuse those two results rather than start the question again from scratch.
M1
202
$3\sqrt{3} \times 10 + 3\sqrt{3} \times 5\sqrt{3} = 30\sqrt{3} + 15 \times 3 = 30\sqrt{3} + 45$ .
Accuracy mark for the correct expansion. Note √3 × √3 = 3 exactly — it is this term, and only this term, that turns part of the product rational.
A1
203
Exact value $= 45 + 30\sqrt{3}$ .
Accuracy mark, correct answer only, in the exact form the question asks for. This is the line the common wrong path below diverges from — the working up to here is identical either way.
A1

Named traps

decimal-given-instead-of-exact-surd: This course's own verified mark-scheme record states the general principle plainly, checked verbatim against two real WMA11 mark schemes (Jan 2023 and Jan 2024, WMA11-verified-facts.md §3): 'Examiners' reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals.' A real, specific, per-question instance of exactly this is now on record too: Pearson WMA11/01, Summer 2024, Q2(ii) (rationalise the denominator of a surd expression, no-calculator warning printed on the question) — the examiner report's general comments state plainly, 'Question 2(ii) was a particular case where this warning was given yet candidates clearly used a calculator to simplify their fractions rather than showing the work to rationalise the denominator.' The trap is genuinely dangerous because it fights the paper's own default: WMA11's General Instructions state elsewhere that 'inexact answers should be given to three significant figures unless otherwise stated' (verified verbatim, WMA11-verified-facts.md §2) — a habit built correctly on most of this paper becomes exactly the wrong move the moment a question says 'exact value,' or a surd is left unresolved in the working.
wrong-multiplier-leaves-a-cross-term-standing: Multiplying a binomial denominator $p+\sqrt{q}$ by anything other than its true conjugate $p-\sqrt{q}$ leaves a surd behind — multiplying by $p+\sqrt{q}$ again, or by $\sqrt{q}$ alone, both leave a cross-term standing (see the mechanism block above for why). This is a reasoned consequence of the algebra itself, not a specific examiner-report finding for this topic in this course's research bank, and is flagged as such rather than dressed up with an invented citation. The check is mechanical: after multiplying out, the denominator must be a single rational number with no root sign left in it at all — if one remains, the multiplier used was not the actual conjugate.
single-term-method-applied-to-a-binomial-denominator: Multiplying a two-term denominator like $5+\sqrt{3}$ by just $\sqrt{3}$ — the way a single-term denominator like $\sqrt{3}$ alone is rationalised — does not work: it produces $5\sqrt{3}+3$ , which still carries a surd, because there was never a cross-term for that multiplier to cancel. Reasoned from the structure of the two techniques, not from a specific examiner-report quote for this topic: the single-surd move and the conjugate move solve genuinely different-shaped denominators, and reaching for the wrong one is a common way to arrive at an answer that is still irrational on the bottom.
surd-not-simplified-before-attempting-to-combine: Two surds that look unrelated on the page can share the same simplified surd part — $\sqrt{8}$ and $\sqrt{18}$ both reduce to a multiple of $\sqrt{2}$ — and the only way to see that is to simplify each one fully first. A student who tries to combine before simplifying will conclude, technically correctly but substantively wrongly, that the two terms 'don't match' and cannot be combined, leaving an answer that is algebraically valid but not in the simplest form a 'giving your answer in the form $k\sqrt{n}$ ' instruction is asking for. Reasoned from the structure of the technique itself, not from a specific examiner-report quote.
square-factor-extraction-stopped-early: Simplifying a surd means pulling out the LARGEST perfect-square factor in one pass, not just any square factor that happens to be visible. $\sqrt{72}$ correctly reduces to $6\sqrt{2}$ ( $72 = 36 \times 2$ ); stopping at a smaller factor gives $2\sqrt{18}$ ( $72 = 4 \times 18$ ) or $3\sqrt{8}$ ( $72 = 9 \times 8$ ) — both arithmetically true, and both still simplifiable, since 18 and 8 each still carry a square factor of their own. A mark scheme asking for a surd in its simplest form does not accept an intermediate step as the final answer. Reasoned from the definition of 'simplest form,' not from a specific examiner-report quote for this topic.
correct-final-surd-with-no-visible-combining-step: This trap is now backed by a real, specific, per-question mark scheme and examiner report, not just reasoning from a general convention. Pearson WMA11/01, Summer 2024, Q2(ii) ('Solve the equation x√3 − 3 = x + √3, giving your answer in the form p + q√3', 3 marks, no-calculator warning printed on the question): the real mark scheme's A1 is defined as 'For 3 + 2√3 WITH AT LEAST ONE INTERMEDIATE STEP' — and its own worked 'Example of insufficient work' shows a solution that states the correct conjugate multiplication but jumps straight from that setup to the final simplified answer with no expansion line in between; that exact solution 'Scores M1M1A0' — both method marks earned, the final accuracy mark withheld, for a completely correct final answer. The real examiner report confirms this happened in practice, not just in the scheme's wording: 'A significant number of candidates did not show the working for rationalising the denominator and hence lost the final two marks... Many lost the accuracy mark as they didn't show intermediate steps – at least one was needed.' It also follows from a genuinely verified general marking convention, WMA11-verified-facts.md §3, under 'Answers without working': 'The rubric says that these may not gain full credit... General policy is that if it could be done "in your head", detailed working would not be required.' A correct final surd on its own is not proof that the required combining or rationalising step was actually carried out rather than done silently in the candidate's head. Reaching $\frac{5}{\sqrt2}-3\sqrt2 = -\frac{\sqrt2}{2}$ in a single jump, with no intermediate line shown, and reaching that same value via a shown combining step — $\frac{5}{\sqrt2}-3\sqrt2 = \frac{5\sqrt2}{2}-3\sqrt2 = -\frac{\sqrt2}{2}$ — land on the identical final answer; only the second leaves visible evidence the method was actually followed. This is a genuinely different trap from every other one in this taxonomy: the final answer can be completely correct and a mark can still be at risk, purely because the working that reaches it skips a visible step. Always write the intermediate line, even when the final simplification feels immediate.

Retrieval — with feedback on every choice

Question 1

2 marks

Simplify $\sqrt{150}$ , leaving your answer as a surd in simplest form.

Question 2

3 marks

Rationalise the denominator of $\frac{3}{4-\sqrt{2}}$ .

Question 3

2 marks

Simplify $2\sqrt{45} + \sqrt{20}$ , giving your answer in the form $k\sqrt{5}$ .

Question 4

2 marks

A WMA11 question asks you to 'find the exact value of $x$ ', where the correct working leads to $x = 7 - 2\sqrt{5}$ . You write your final answer as $x = 2.53$ (3 s.f.). What happens to the marks for this part?

Question 5

2 marks

Simplify $\frac{10}{\sqrt{5}}$ , giving your answer in simplest surd form.

Question 6

2 marks

Which expression, multiplied against the denominator of $\frac{7}{5+\sqrt{3}}$ , gives a denominator with no surd left in it?

Reference — not a study method, a lookup

√(a×b) = √a × √b, √(a/b) = √a/√b (a,b ≥ 0) — always pull out the LARGEST square factor, not just any square factor.
a√n ± b√n = (a±b)√n — only once BOTH surds are simplified to the same surd part. Simplify first, combine second.
Single surd on the bottom: multiply top and bottom by that surd. a/√b = a√b/b.
Binomial p ± √q on the bottom: multiply top and bottom by the conjugate p ∓ √q. (p+√q)(p−√q) = p² − q — cross terms cancel, the surd doesn't.
"Exact value" or "surds required" in the question ⇒ a decimal final answer normally loses the mark, however correct the working was up to that line.

Not affiliated with or endorsed by Pearson Edexcel. A 2026 mark-scheme-bullet coverage audit located a real, specific per-question anchor for this lesson's central claim: Pearson WMA11/01, Summer 2024, Q2(ii) ('Solve the equation x√3 − 3 = x + √3, giving your answer in the form p + q√3', no-calculator warning printed on the question) — a genuine rationalising-a-denominator question whose real mark scheme gates its final accuracy mark on showing 'at least one intermediate step,' and whose real examiner report confirms candidates lost marks in practice for skipping that step or relying on a calculator instead of showing the rationalisation ('Question 2(ii) was a particular case where this warning was given yet candidates clearly used a calculator to simplify their fractions rather than showing the work to rationalise the denominator'). That citation, together with the general marking-convention principle already used elsewhere in this lesson and independently re-verified in the same audit pass — verbatim against two independent series' real WMA11 mark scheme PDFs (Jan 2023 and Jan 2024, per WMA11-verified-facts.md §3): 'Examiners' reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals' — now backs the 'decimal-given-instead-of-exact-surd' and 'correct-final-surd-with-no-visible-combining-step' trap-taxonomy items with real, per-question evidence, not general convention alone. The remaining three trap-taxonomy items ('wrong-multiplier-leaves-a-cross-term-standing', 'single-term-method-applied-to-a-binomial-denominator', 'surd-not-simplified-before-attempting-to-combine') and the 'square-factor-extraction-stopped-early' item are still reasoned from the structure of the algebra itself and from this course's own established mark-scheme conventions, not from a specific examiner-report quote about this topic — each still says so in its own text — because this audit, like the research pass before it, did not find a real per-question anchor for those specific sub-skills (combining unlike surds, choosing the wrong multiplier, stopping at a smaller-than-largest square factor). Every question in this lesson — prequestion, worked chain, chain drill, method comparison, marked solution and MCQ alike — remains VERIDIAN-original wording, not a reproduction of any real Pearson question; the per-line mark allocations attached to them are modelled on verified general mark-scheme conventions (what M, A and B marks mean; that A marks are cao unless marked ft; that a method mark can be earned by a correct method regardless of the final number) rather than transcribed from a real mark scheme, which for an original question does not exist.

Question 12 marks

Simplify $\sqrt{150}$ , leaving your answer as a surd in simplest form.

$5\sqrt{6}$
Correct. $150 = 25 \times 6$ , and 25 is the largest perfect-square factor of 150, so $\sqrt{150} = \sqrt{25}\sqrt{6} = 5\sqrt{6}$ .
B $12.2$ (3 s.f.)
This is a numerically correct approximation ( $5\sqrt{6} \approx 12.247$ ) and it still loses the mark: the instruction is to simplify, which on this topic means leave it as a surd, not evaluate it.
C $25\sqrt{6}$
The factor pair $25 \times 6$ is right and the square root of the 25 was never taken — the coefficient outside the root should be $\sqrt{25} = 5$ , not the 25 itself.
D $5\sqrt{30}$
This treats $\sqrt{5 \times 30}$ as $5\sqrt{30}$ — pulling the 5 out from under the root without taking its square root, which is only valid for a genuine SQUARE factor. $5 \times 30 = 150$ is correct, but 5 is not a perfect square, so this factor pair cannot be split this way at all.

Traps tested: Decimal given instead of exact surd · Square factor not square rooted · Non square factor pulled out unrooted

Question 23 marks

Rationalise the denominator of $\frac{3}{4-\sqrt{2}}$ .

$\frac{12+3\sqrt{2}}{14}$
Correct. Multiplying by the conjugate $\frac{4+\sqrt{2}}{4+\sqrt{2}}$ : denominator $(4-\sqrt{2})(4+\sqrt{2}) = 16-2 = 14$ ; numerator $3(4+\sqrt{2}) = 12+3\sqrt{2}$ .
B $\frac{12-3\sqrt{2}}{14}$
The denominator is right and the numerator's sign is wrong. $3 \times (4+\sqrt{2}) = 12 + 3\sqrt{2}$ — both terms of the conjugate are positive here, so both terms of the product must be, too.
C $\frac{12+3\sqrt{2}}{16}$
The numerator is right and the denominator has only used $4^2=16$ , dropping the $-(\sqrt{2})^2$ term of the difference of two squares entirely. $(4-\sqrt{2})(4+\sqrt{2}) = 4^2 - (\sqrt{2})^2 = 16-2=14$ , not just $16$ .
D $\frac{6}{7}$
The denominator, 14, is right — but the numerator has been evaluated as if $3 \times (4+\sqrt{2})$ were just $3 \times 4 = 12$ , with the $\sqrt{2}$ term dropped entirely during expansion, then $12/14$ reduced to $6/7$ . The surd term in the numerator does not disappear; only the one in the denominator does.

Traps tested: Sign error distributing into the conjugate · Difference of squares second term dropped · Surd term dropped during numerator expansion

Question 32 marks

Simplify $2\sqrt{45} + \sqrt{20}$ , giving your answer in the form $k\sqrt{5}$ .

$8\sqrt{5}$
Correct. $\sqrt{45} = 3\sqrt{5}$ , so $2\sqrt{45} = 6\sqrt{5}$ ; $\sqrt{20} = 2\sqrt{5}$ . $6\sqrt{5} + 2\sqrt{5} = 8\sqrt{5}$ .
B $6\sqrt{5} + \sqrt{20}$
This is numerically equal to the correct answer, but it isn't the answer the question asked for: the first term has been simplified correctly ( $2\sqrt{45} = 6\sqrt{5}$ ) and the second, $\sqrt{20}$ , has been left unsimplified — so the two terms never get combined into the single required form $k\sqrt{5}$ .
C $8\sqrt{65}$
This treats the two ORIGINAL numbers under the roots as if they could be added together ( $45+20=65$ ) after combining the coefficients — an invalid operation. Surds only combine when their simplified surd parts already match; the numbers under two different roots are never added directly.
D $10\sqrt{5}$
Both surds are correctly simplified to $6\sqrt{5}$ and $2\sqrt{5}$ — the error is purely arithmetic in the final addition: $6+2=8$ , not $10$ .

Traps tested: Second surd left unsimplified so never combined · Surds combined under one root · Coefficient addition slip

Question 42 marks

A WMA11 question asks you to 'find the exact value of $x$ ', where the correct working leads to $x = 7 - 2\sqrt{5}$ . You write your final answer as $x = 2.53$ (3 s.f.). What happens to the marks for this part?

The final accuracy mark is normally lost, even though $2.53$ is numerically correct to 3 s.f. — the question's own instruction to give an exact answer, and the general marking convention this course has verified, both point the same way: exact means a surd, not a rounded decimal.
Correct. $7-2\sqrt{5} \approx 2.528$ , which does round to $2.53$ — the number itself is right, and that is exactly why this trap is dangerous: the working can be flawless and the final answer still costs the mark, purely because of its form.
BNothing is lost — $2.53$ is numerically right, and mark schemes accept any correct numerical value regardless of form.
This is the assumption the whole topic is designed to correct. A mark scheme can and does distinguish between a numerically correct value and the specific exact form a question asks for — 'exact value' is itself part of the question's demand, not decoration around it.
CEvery mark in the part is lost, including the method marks for the working that produced $7-2\sqrt{5}$ in the first place.
This overstates the cost. Method marks are earned for the correct method being attempted and carried out, independent of what happens on the very last line — only the final accuracy mark, tied to the exact form, is normally at risk here.
DIt depends on whether the exam allows calculators for this question.
WMA11 is a calculator-permitted paper overall, and the exact-value convention holds regardless — it is not a no-calculator rule. The paper's own AO5 weighting (the narrowest calculator-use allocation of any unit in the qualification) is itself evidence that manual algebraic fluency is deliberately tested even where a calculator is allowed.

Traps tested: Numerically correct decimal assumed acceptable · Overestimates the cost of a late error · Calculator permission conflated with exact answer rule

Question 52 marks

Simplify $\frac{10}{\sqrt{5}}$ , giving your answer in simplest surd form.

$2\sqrt{5}$
Correct. $\frac{10}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}$ .
B $10\sqrt{5}$
The numerator has correctly become $10\sqrt{5}$ — the denominator, which should also have become 5, has simply been dropped rather than divided through.
C $\frac{2}{\sqrt{5}}$
This treats the 10 in the numerator and the 5 hidden inside $\sqrt{5}$ as though they were the same kind of quantity and cancels them directly ( $10 \div 5 = 2$ ). But $\sqrt{5} \neq 5$ , so nothing under a root can be cancelled against a whole number sitting outside one this way — and the denominator is still irrational here besides.
D $\frac{10\sqrt{5}}{5}$
This is a correctly rationalised intermediate step, and it isn't reduced to lowest terms: 10 and 5 share a common factor of 5, giving $2\sqrt{5}$ .

Traps tested: Denominator not divided through · Surd treated as its radicand for cancellation · Fraction not reduced to lowest terms

Question 62 marks

Which expression, multiplied against the denominator of $\frac{7}{5+\sqrt{3}}$ , gives a denominator with no surd left in it?

$5-\sqrt{3}$
Correct — this is the conjugate. $(5+\sqrt{3})(5-\sqrt{3}) = 25-3=22$ , a rational number, because the two middle terms cancel exactly.
B $5+\sqrt{3}$
Squaring the original denominator does not clear the surd: $(5+\sqrt{3})(5+\sqrt{3}) = 28+10\sqrt{3}$ , which still has a surd term — multiplying by the SAME expression again doesn't set up any cancellation at all.
C $\sqrt{3}$
This is the move for a single-term denominator, misapplied to a binomial one: $(5+\sqrt{3}) \times \sqrt{3} = 5\sqrt{3}+3$ , still carrying a surd, because there was no cross-term for this multiplier to cancel against in the first place.
D $5$
Multiplying by only the rational part leaves the surd term completely untouched: $(5+\sqrt{3}) \times 5 = 25+5\sqrt{3}$ — still a surd on the bottom.

Traps tested: Denominator multiplied by itself instead of conjugate · Single term method applied to binomial denominator · Only the rational part used

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA11.

Simplifying Surds, Rationalising Denominators, and the Exact-Value Rule

What a surd is, and the two laws that let you simplify one

Adding and subtracting surds — like terms, with one extra step first

Why multiplying by √b over itself clears a single surd from the denominator

The conjugate isn't a memorised trick — it's the one multiplication where the cross-terms cancel

Rationalise 13−5\frac{1}{3-\sqrt{5}}3−5​1​ — a binomial denominator, worked from why √5/√5 alone fails to a checked final answer

Complete the chain — rationalise 16−2\frac{1}{\sqrt{6}-2}6​−21​

Deep dive

Practice this for real

Rationalise $\frac{1}{3-\sqrt{5}}$ — a binomial denominator, worked from why √5/√5 alone fails to a checked final answer

Complete the chain — rationalise $\frac{1}{\sqrt{6}-2}$