Ma10, Ma11 - Surds and rationalising denominators | OCR A-Level Mathematics B (MEI)

Use Ma10 to simplify and calculate with exact surds, then Ma11 to rationalise their denominators. You will see why conjugates work, keep the domain of a variable expression and make short exact arguments without replacing roots by rounded decimals.

Exact roots and indices

How could three awkward fractions add to just $\sqrt7-2$ ? Consider

\frac1{\sqrt5+2}+\frac1{\sqrt6+\sqrt5}+\frac1{\sqrt7+\sqrt6}.

The denominators do not look alike, so a common denominator looks uninviting. We will find a much shorter route after sorting out the roots themselves.

You already know

Fractional indices and roots describe the same operation. For a positive $u$ , $u^{m/n}=(\sqrt[n]{u})^m$ . Multiply the top and bottom of a fraction by the same nonzero expression to keep its value.

Exact form means keeping the number itself, not merely many decimal places. For example, $\sqrt2$ is exact, but $1.414$ is rounded.

Surd

A surd is an irrational root of a rational number. Examples are $\sqrt2$ and $\sqrt[3]{5}$ . A root need not be a surd: $\sqrt9=3$ is rational.

For an even root, the real radicand must be non-negative. The principal square root is also non-negative: $\sqrt9=3$ , whereas the two solutions of $z^2=9$ are $z=\pm3$ . Thus $\sqrt{u^2}=|u|$ , and at $u=-2$ , $\sqrt{9u^2}=6=3|u|$ . This principal-root identity is not in the formula sheet: learn it.

Working with square roots

For $a,b\ge0$ ,

\sqrt{ab}=\sqrt a\sqrt b.

For $a\ge0,b>0$ ,

\sqrt{a/b}=\frac{\sqrt a}{\sqrt b}.

The conditions keep all roots real and the divisor nonzero.

Not in the formula sheet: learn it.

The product rule follows because $\sqrt a\sqrt b$ is non-negative and squares to $ab$ . Extract square factors with that rule; do not assume a similar rule for addition.

Worked example 1: simplifying a root

Question. Express $\sqrt{72}$ in simplest surd form.

Plan. Pull out the square factor $36$ .

\begin{aligned} \sqrt{72}&=\sqrt{36\times2} && \text{separate a square factor}\\ &=6\sqrt2 && \text{product rule} \end{aligned}

Why this step? Only a factor with an exact rational square root can leave the radical as a rational multiplier.

Answer. $6\sqrt2$ .

Check. $(6\sqrt2)^2=72$ , and the expression is positive.

Change only the square factor, keeping the same remaining root. This is why unlike-looking roots can become like terms later.

Worked example 2: recognising the same root

Question. Express $\sqrt{32}$ in simplest surd form.

Plan. Separate the square factor $16$ .

\begin{aligned} \sqrt{32}&=\sqrt{16\times2} && \text{factor the radicand}\\ &=4\sqrt2 && \text{take the exact root} \end{aligned}

Why this step? The simplified root is again $\sqrt2$ , so a sum involving $\sqrt{72}$ and $\sqrt{32}$ can now be collected.

Answer. $4\sqrt2$ .

Check. $(4\sqrt2)^2=32$ , with a non-negative value.

Root and index notation must agree: $\sqrt u=u^{1/2}$ , $\sqrt[4]{u^3}=u^{3/4}$ for $u\ge0$ , and $81^{3/4}=3^3=27$ . These root/index identities are not in the formula sheet: learn them. Odd roots can accept negative values, such as $\sqrt[3]{-8}=-2$ .

Watch out

Wrong: $\sqrt{25t^2}=5t$ for unrestricted real $t$ . Right: $\sqrt{25t^2}=5|t|$ . A principal square root never supplies a negative answer.

Use a new square factor, then check the distinction between roots and powers.

Surd arithmetic

Simplification turns surd arithmetic back into familiar algebra: collect matching roots and multiply brackets term by term.

For example,

2\sqrt{12}-\sqrt{27}+\sqrt3 =4\sqrt3-3\sqrt3+\sqrt3=2\sqrt3.

The coefficient arithmetic is ordinary arithmetic once each root is simplified. Terms such as $\sqrt2+\sqrt3$ do not collect. In particular, $\sqrt{9+16}=5$ , while $\sqrt9+\sqrt{16}=7$ .

Surd bracket identities

For real $a,b$ ,

(a+b)^2=a^2+2ab+b^2,\qquad (a+b)(a-b)=a^2-b^2.

These identities apply when $a$ or $b$ is a real surd. The mixed terms add in the square and cancel in the conjugate product.

Not in the formula sheet: learn it.

The signs of the mixed products decide whether a root remains. Make each product visible before simplifying.

Worked example 3: four products

Question. Expand and simplify $(\sqrt5+2)(\sqrt5-3)$ .

Plan. Expand each term into the other bracket, then collect like terms.

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

Why this step? Squaring the root gives the rational contribution $5$ ; each mixed product still has one root.

Answer. $-1-\sqrt5$ .

Check. Since $2<\sqrt5<3$ , the first bracket is positive and the second negative, agreeing with the answer's sign.

Change only the constant in the second bracket, from minus three to minus two. The brackets become conjugates and the mixed terms now cancel.

Worked example 4: a rational product

Question. Expand and simplify $(\sqrt5+2)(\sqrt5-2)$ .

Plan. Show the cancelling mixed terms, or use the difference of squares.

\begin{aligned} (\sqrt5+2)(\sqrt5-2) &=5-2\sqrt5+2\sqrt5-4 && \text{expand}\\ &=1 && \text{cancel and subtract} \end{aligned}

Why this step? Changing that one constant makes the mixed products equal and opposite.

Answer. $1$ .

Check. The two positive factors are reciprocal: dividing one by their product gives $1/(\sqrt5+2)=\sqrt5-2$ .

That small rational answer is the key to rationalising. In a square, however, the mixed terms have the same sign: $(2+\sqrt3)^2=7+4\sqrt3$ .

Watch out

Wrong: $(2+\sqrt3)^2=7$ . This drops the cross term. Right: expand both brackets, giving $4+2\sqrt3+2\sqrt3+3=7+4\sqrt3$ . A product of conjugates is a different expression.

Use the same conjugate-product method with a new root.

Rationalising denominators

Choose a multiplier that cancels the root in the denominator, rather than multiplying several awkward fractions together first.

For a single root, $3/\sqrt5=3\sqrt5/5$ : multiply the top and bottom by $\sqrt5$ . For two terms, multiply by the conjugate. The conjugate of $a+\sqrt b$ is $a-\sqrt b$ ; only the middle sign changes. Both the starting denominator and the multiplier must be nonzero.

Worked example 5: a rational term and a root

Question. Rationalise the denominator of $\dfrac2{3-\sqrt5}$ .

Plan. Multiply top and bottom by $3+\sqrt5$ , then use the difference of squares.

\begin{aligned} \frac2{3-\sqrt5} &=\frac{2(3+\sqrt5)}{(3-\sqrt5)(3+\sqrt5)} && \text{multiply by one}\\ &=\frac{2(3+\sqrt5)}{9-5} && \text{conjugate product}\\ &=\frac{3+\sqrt5}{2} && \text{cancel a factor }2 \end{aligned}

Why this step? The sign change makes the mixed products cancel, leaving the rational denominator $4$ .

Answer. $(3+\sqrt5)/2$ .

Check. Multiplying by $3-\sqrt5$ gives $(9-5)/2=2$ .

Change only the first denominator term from $3$ to $\sqrt7$ . A difference of squares still works when both terms are roots.

Worked example 6: two roots in the denominator

Question. Rationalise the denominator of $\dfrac2{\sqrt7-\sqrt5}$ .

Plan. Multiply by the conjugate $\sqrt7+\sqrt5$ .

\begin{aligned} \frac2{\sqrt7-\sqrt5} &=\frac{2(\sqrt7+\sqrt5)}{(\sqrt7-\sqrt5)(\sqrt7+\sqrt5)} && \text{conjugate multiplier}\\ &=\frac{2(\sqrt7+\sqrt5)}{7-5} && \text{difference of squares}\\ &=\sqrt7+\sqrt5 && \text{cancel }2 \end{aligned}

Why this step? Squaring each root gives a rational number, so their difference is rational too.

Answer. $\sqrt7+\sqrt5$ .

Check. $(\sqrt7+\sqrt5)(\sqrt7-\sqrt5)=7-5=2$ .

Now read the opening fractions as a connected argument. For $n\ge0$ , rationalising gives

\frac1{\sqrt{n+1}+\sqrt n} =\frac{\sqrt{n+1}-\sqrt n}{(n+1)-n} =\sqrt{n+1}-\sqrt n.

The denominator becomes one. Use this result on each fraction separately:

\begin{aligned} \frac1{\sqrt5+2}+\frac1{\sqrt6+\sqrt5}+\frac1{\sqrt7+\sqrt6} &=(\sqrt5-2)+(\sqrt6-\sqrt5)+(\sqrt7-\sqrt6)\\ &=\sqrt7-2 \end{aligned}

That is the surprise: all the middle roots disappear. The index $n$ names a radicand, not a number of terms; the three substitutions are $n=4,5,6$ . Reading the symbols before manipulating them keeps the cancellation clear.

A variable denominator needs its restrictions too:

\frac1{\sqrt x-2}=\frac{\sqrt x+2}{x-4},\qquad x\ge0,\quad x\ne4.

At $x=0$ , both forms give $-1/2$ ; $x=4$ is excluded. Also simplify perfect roots first: $1/(2+\sqrt4)=1/4$ is already rational, whereas its proposed conjugate multiplier is zero.

Watch out

Wrong: $1/(2+\sqrt4)$ can be rationalised by multiplying by $(2-\sqrt4)/(2-\sqrt4)$ . That proposed multiplier is $0/0$ , not one. Right: simplify the perfect root first to get $1/4$ .

Exact algebra needs only a few clear lines when each equality is justified.

Exam tip

For MEI's detailed-reasoning questions, show at least one actual multiplication by the conjugate over itself and the rational denominator it produces. Writing the rationalised answer alone, even correctly, does not demonstrate that step. Keep the exact surd; extra decimal places do not make a rounded answer exact.

Use a different first radicand in the second worked example. The rational denominator no longer cancels completely.

Quick Recap

Ma10 uses exact roots: simplify square factors, collect matching surds and expand every product.
Principal square roots are non-negative, so $\sqrt{x^2}=|x|$ .
Ma11 rationalises with a root or conjugate; show the same nonzero multiplier on top and bottom.
Rationalise separate terms before combining them when that exposes a simple cancellation, and keep variable restrictions.

The opening sum is $\sqrt7-2$ : rationalisation cancels every middle root.

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Ma10, Ma11 - Surds and rationalising denominators

Exact roots and indices

Surd arithmetic

Rationalising denominators

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