Pure 2.2 - Surds and rationalising denominators | Edexcel A-Level Mathematics

Surds keep answers exact while you change their form. You will simplify and multiply square-root expressions, then rationalise single-term and two-term denominators without changing the value of a fraction.

Exact surd form

Can a fraction with an awkward square root underneath simplify to just $3-\sqrt5$ ? Try $4/(3+\sqrt5)$ . A decimal calculation suggests about $0.764$ , but an exact calculation will reveal why these two expressions are identical.

You already know

$6^2=36$ , so $\sqrt{36}=6$ . A square root means the non-negative root: $\sqrt9=3$ , whereas solving $x^2=9$ gives $x=\pm3$ .

The number $\sqrt2$ is an exact value, not a decimal waiting to be rounded. A surd is an irrational root retained in exact form; $\sqrt2$ and $\sqrt7$ are surds, while $\sqrt9=3$ is rational. A rational multiple such as $5\sqrt7$ is a surd expression.

Square-root rules

For real $u\ge0$ , $v\ge0$ ,

(\sqrt u)^2=u,\qquad \sqrt{uv}=\sqrt u\sqrt v

In particular, $\sqrt{t^2}=|t|$ for real $t$ .
Not in the formula booklet: learn it.

Extract a square factor before collecting terms: $\sqrt{72}=\sqrt{36\times2}=6\sqrt2$ . The remaining root should have no integer square factor greater than $1$ . Algebraic roots need the same care: $\sqrt{18x^2}=3|x|\sqrt2$ , not always $3x\sqrt2$ .

Worked example 1: collecting like surds

Question. Simplify $3\sqrt{18}-\sqrt{50}$ .

Plan. Extract square factors, then collect the coefficients of $\sqrt2$ .

\begin{aligned} 3\sqrt{18}-\sqrt{50}&=3(3\sqrt2)-5\sqrt2 &&\text{square factors }9,\ 25\\ &=(9-5)\sqrt2=4\sqrt2 &&\text{like surds} \end{aligned}

Why this step? The roots only become like terms after simplification.

Answer. $4\sqrt2$ .

Check. $3\sqrt{18}\approx12.728$ , $\sqrt{50}\approx7.071$ ; their difference is about $5.657=4\sqrt2$ .

Change only subtraction to addition. The square-factor work stays the same.

Worked example 2: changing the operation

Question. Simplify $3\sqrt{18}+\sqrt{50}$ .

Plan. Simplify both roots before adding their coefficients.

\begin{aligned} 3\sqrt{18}+\sqrt{50}&=9\sqrt2+5\sqrt2 &&\text{extract square factors}\\ &=14\sqrt2 &&\text{add coefficients} \end{aligned}

Why this step? The operation acts on the coefficients; the shared root remains.

Answer. $14\sqrt2$ .

Check. $(14\sqrt2)^2=392$ , and expanding $(3\sqrt{18}+\sqrt{50})^2$ also gives $162+50+180=392$ ; both expressions are positive.

The product rule does not extend to sums.

Watch out

Wrong: $\sqrt{9+16}=\sqrt9+\sqrt{16}=7$ . Right: $\sqrt{25}=5$ . Split a product under a root, not a sum.

Keep the final answer exact and collect every like term.

Multiplying and expanding surds

A surd behaves like an algebraic term in a bracket. For $(2+\sqrt3)(5-\sqrt3)$ , expand all four products: $10-2\sqrt3+5\sqrt3-3=7+3\sqrt3$ . Remember that the last product contains $(\sqrt3)^2=3$ .

Difference of two square-root terms

For real $u\ge0$ , $v\ge0$ ,

(\sqrt u+\sqrt v)(\sqrt u-\sqrt v)=u-v

The middle products cancel. Not in the formula booklet: learn it.

This is the ordinary difference of squares: $(A+B)(A-B)=A^2-B^2$ . It is especially useful when your goal is to remove roots.

Worked example 3: expanding two brackets

Question. Simplify $(3+\sqrt5)(2-\sqrt5)$ .

Plan. Expand, replace the squared root, then collect.

\begin{aligned} (3+\sqrt5)(2-\sqrt5)&=6-3\sqrt5+2\sqrt5-5 &&\text{four products}\\ &=1-\sqrt5 &&\text{collect} \end{aligned}

Why this step? The unlike constants in the brackets mean the middle terms do not cancel.

Answer. $1-\sqrt5$ .

Check. The first bracket is positive and the second negative, so the negative final answer has the correct sign.

Change only the second bracket's constant from $2$ to $3$ . Now its terms match the first bracket apart from one sign.

Worked example 4: a conjugate product

Question. Simplify $(3+\sqrt5)(3-\sqrt5)$ .

Plan. Use the difference of squares.

\begin{aligned} (3+\sqrt5)(3-\sqrt5)&=3^2-(\sqrt5)^2 &&\text{middle terms cancel}\\ &=9-5=4 &&\text{evaluate squares} \end{aligned}

Why this step? Equal constants and opposite root terms make the cross-products cancel exactly.

Answer. $4$ .

Check. Full expansion gives $9-3\sqrt5+3\sqrt5-5=4$ .

This cancellation is the engine behind the next two parts.

Watch out

Wrong: $(2\sqrt3)^2=2\times3=6$ . The coefficient is squared too: $(2\sqrt3)^2=4\times3=12$ .

Show the square calculation when the answer is supplied.

Rationalising a single-surd denominator

Rationalising means rewriting a fraction with a rational denominator. You are changing its appearance, not its value: multiply numerator and denominator by the same non-zero root. For example, $6/\sqrt3=(6\sqrt3)/3=2\sqrt3$ .

Rationalising a denominator

To rationalise a denominator is to give an equivalent fraction whose denominator is rational. A surd may remain in the numerator; $2\sqrt3$ is a valid final form.

For a denominator $k\sqrt n$ , where $k\ne0$ and $n>0$ , multiplication by $\sqrt n/\sqrt n=1$ makes the denominator $kn$ . This method is not supplied in the formula booklet.

Worked example 5: a two-term numerator

Question. Rationalise and simplify $\dfrac{4+\sqrt2}{2\sqrt2}$ .

Plan. Multiply the whole numerator and denominator by $\sqrt2$ .

\begin{aligned} \frac{4+\sqrt2}{2\sqrt2}&=\frac{(4+\sqrt2)\sqrt2}{2\sqrt2\sqrt2} &&\text{multiply by }1\\ &=\frac{4\sqrt2+2}{4}=\frac{2\sqrt2+1}{2} &&\text{expand and reduce} \end{aligned}

Why this step? Every term in the numerator must receive the multiplier.

Answer. $(2\sqrt2+1)/2$ .

Check. Multiplying the answer by $2\sqrt2$ returns $4+\sqrt2$ .

Change only the numerator's constant from $4$ to $6$ . The denominator and multiplier remain fixed.

Worked example 6: simplifying the whole numerator

Question. Rationalise and simplify $\dfrac{6+\sqrt2}{2\sqrt2}$ .

Plan. Use the same root multiplier and reduce the fraction.

\begin{aligned} \frac{6+\sqrt2}{2\sqrt2}&=\frac{6\sqrt2+2}{4} &&\text{multiply by }\frac{\sqrt2}{\sqrt2}\\ &=\frac{3\sqrt2+1}{2} &&\text{divide every term by }2 \end{aligned}

Why this step? Reduction divides all numerator terms and the denominator by the common factor.

Answer. $(3\sqrt2+1)/2$ .

Check. $((3\sqrt2+1)/2)(2\sqrt2)=6+\sqrt2$ .

It is fine for the final coefficients to be fractions.

Watch out

Wrong: $\frac{3+\sqrt5}{\sqrt5}=\frac{3\sqrt5+\sqrt5}{5}$ . The second product is $\sqrt5\sqrt5=5$ , so the correct numerator is $3\sqrt5+5$ .

Use that correction to finish the exact calculation yourself.

Rationalising a two-term denominator

A single root will not clear the denominator $3+\sqrt5$ ; it just moves the root to its other term. Instead, arrange for the middle terms to cancel by changing one sign.

Conjugate

The conjugate of $a+b\sqrt c$ is $a-b\sqrt c$ , and conversely. For example, $3-\sqrt5$ is the conjugate of $3+\sqrt5$ ; $-3-\sqrt5$ is not.

For rational $a$ , $b$ and positive integer $c$ , their product is $a^2-b^2c$ , which is rational. This result is not in the formula booklet. The original denominator and its conjugate multiplier must be non-zero.

Worked example 7: answering the opening puzzle

Question. Rationalise $\dfrac4{3+\sqrt5}$ .

Plan. Multiply top and bottom by $3-\sqrt5$ .

\begin{aligned} \frac4{3+\sqrt5}&=\frac{4(3-\sqrt5)}{(3+\sqrt5)(3-\sqrt5)} &&\text{conjugate}\\ &=\frac{4(3-\sqrt5)}{9-5}=3-\sqrt5 &&\text{difference of squares} \end{aligned}

Why this step? The opposite sign removes both middle products from the denominator.

Answer. $3-\sqrt5$ .

Check. $(3-\sqrt5)(3+\sqrt5)=4$ , the original numerator.

Change only the sign in the denominator. The conjugate's sign must change with it.

Worked example 8: a minus sign in the denominator

Question. Rationalise $\dfrac4{3-\sqrt5}$ .

Plan. Use the conjugate $3+\sqrt5$ .

\begin{aligned} \frac4{3-\sqrt5}&=\frac{4(3+\sqrt5)}{9-5} &&\text{conjugate product}\\ &=3+\sqrt5 &&\text{cancel }4 \end{aligned}

Why this step? Multiplying by the same denominator again would square it and retain a root term.

Answer. $3+\sqrt5$ .

Check. $(3+\sqrt5)(3-\sqrt5)=4$ .

A more complicated numerator simply needs expansion: $(1+\sqrt2)/(3-\sqrt2)=(5+4\sqrt2)/7$ . The denominator is still $9-2$ , and the numerator comes from all four products. There is a pleasant exact surprise here: the two opening fractions multiply to $4$ , since their simplified forms are conjugates.

Exam tip

Write the chosen multiplier explicitly. A correct conjugate method can earn method credit even when a later expansion contains a slip.

Your turn keeps the same conjugate idea and adds a numerator to expand.

Choosing a clean final form

Rationalising may leave another step: collect like roots, use a common denominator and match the requested form. “Exact” means retaining roots and fractions; it does not mean leaving an expression unsimplified. This habit is useful later for exact trigonometric values and quadratic roots.

Worked example 9: combining after rationalising

Question. Show that $\dfrac2{\sqrt5-1}-\sqrt5=\dfrac{1-\sqrt5}{2}$ .

Plan. Rationalise the fraction, then subtract using denominator $2$ .

\begin{aligned} \frac2{\sqrt5-1}-\sqrt5&=\frac{2(\sqrt5+1)}{5-1}-\sqrt5 &&\text{conjugate}\\ &=\frac{\sqrt5+1-2\sqrt5}{2} &&\text{common denominator}\\ &=\frac{1-\sqrt5}{2} &&\text{collect} \end{aligned}

Why this step? Rewrite the separate $\sqrt5$ as $2\sqrt5/2$ before combining.

Answer. The required identity follows with exact working.

Check. Both sides are approximately $-0.618$ .

Change only the subtraction to addition. The rationalisation stays exactly the same.

Worked example 10: matching a requested form

Question. Express $\dfrac2{\sqrt5-1}+\sqrt5$ as $p+q\sqrt5$ , where $p$ , $q$ are rational.

Plan. Rationalise, add, then split the final fraction.

\begin{aligned} \frac2{\sqrt5-1}+\sqrt5&=\frac{\sqrt5+1}{2}+\sqrt5 &&\text{rationalise}\\ &=\frac{1+3\sqrt5}{2}=\frac12+\frac32\sqrt5 &&\text{collect and split} \end{aligned}

Why this step? The requested form identifies the rational coefficients separately.

Answer. $p=\frac12$ , $q=\frac32$ .

Check. Multiplying the answer minus $\sqrt5$ by $\sqrt5-1$ gives $2$ .

The final form must describe the same exact number throughout.

Watch out

Wrong: $(1+3\sqrt5)/2=1+(3/2)\sqrt5$ . The denominator divides both terms, so the rational term is $1/2$ .

For the last calculation, change the coefficient and denominator sign while keeping the same rationalise-then-collect strategy.

Quick Recap

Simplify roots by extracting square factors; collect only like surds.
Expand every bracket product and use $(\sqrt u)^2=u$ for $u\ge0$ .
A conjugate changes one sign and turns a denominator product into a difference of squares.
Rationalising preserves value; finish by collecting terms and matching the requested exact form.

The opening puzzle is resolved: $4/(3+\sqrt5)=3-\sqrt5$ , since their denominator and conjugate multiply to $4$ .

Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 2.2 - Surds and rationalising denominators

Exact surd form

Multiplying and expanding surds

Rationalising a single-surd denominator

Rationalising a two-term denominator

Choosing a clean final form

Quick Recap

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