1.02b - Surds and rationalising denominators | OCR A-Level Mathematics A

Keep roots exact while you simplify, expand and divide expressions. You will connect surd and index notation, use conjugates to clear denominators and preserve the restrictions that make each step valid.

Exact roots and indices

Can $1/(\sqrt2-1)$ really equal $\sqrt2+1$ ? One looks like division by a small number; the other is a little more than two. Before explaining why they agree, we need to be precise about what a root means.

You already know

For a positive base, $u^{m/n}=(\sqrt[n]{u})^m$ . A fraction retains its value when you multiply its numerator and denominator by the same nonzero number.

A finite decimal often hides a rounding decision. Writing $\sqrt2$ gives the exact number, while $1.414$ is only an approximation.

Surd

A surd is an irrational root of a rational number, such as $\sqrt2$ or $\sqrt[3]{5}$ . A root that simplifies to a rational number is not a surd: $\sqrt9=3$ .

The square-root sign means the principal, non-negative root. Thus $\sqrt9=3$ , while solving $z^2=9$ gives $z=\pm3$ . For real $u$ , $\sqrt{u^2}=|u|$ ; if $u=-2$ , then $\sqrt{9u^2}=6=3|u|$ . This principal-root identity is not in the formula sheet: learn it.

Root products and quotients

For $a,b\ge0$ ,

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

For $a\ge0,b>0$ ,

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

The conditions ensure that the real roots exist and the divisor is nonzero.

Not in the formula sheet: learn it.

Why does the product rule work? The product $\sqrt a\sqrt b$ is non-negative, and its square is $ab$ , so it must be the principal square root of $ab$ . Use this to extract square factors.

Worked example 1: extracting a square factor

Question. Simplify $\sqrt{72}$ .

Plan. Separate the largest square factor from the remaining factor.

\begin{aligned} \sqrt{72}&=\sqrt{36\times2} && \text{factor the radicand}\\ &=6\sqrt2 && \text{take the square root of }36 \end{aligned}

Why this step? A square factor can leave the root as an integer; the factor $2$ has no further square factor.

Answer. $6\sqrt2$ .

Check. $(6\sqrt2)^2=36\times2=72$ , and $6\sqrt2>0$ .

Change only the square factor from $36$ to $25$ . The unsquared factor stays $2$ .

Worked example 2: the same surd in another disguise

Question. Simplify $\sqrt{50}$ .

Plan. Separate $25$ and apply the product rule.

\begin{aligned} \sqrt{50}&=\sqrt{25\times2} && \text{find a square factor}\\ &=5\sqrt2 && \text{simplify the root} \end{aligned}

Why this step? Simplest surd form leaves no square factor greater than one inside the root.

Answer. $5\sqrt2$ .

Check. $(5\sqrt2)^2=25\times2=50$ , with a non-negative result.

The same root may be written with an index: $\sqrt u=u^{1/2}$ and $\sqrt[4]{u^3}=u^{3/4}$ for $u\ge0$ . For example, $81^{3/4}=3^3=27$ . These equivalent root/index forms are not in the formula sheet: learn them. Even roots require non-negative radicands; odd roots can accept negative ones, such as $\sqrt[3]{-8}=-2$ .

Watch out

Wrong: $\sqrt{25t^2}=5t$ for all real $t$ . Right: $\sqrt{25t^2}=5|t|$ . A principal square root cannot be negative.

Now change the square factor once more, and keep the index and sign conventions in mind.

Surd arithmetic

Surds obey ordinary algebra: simplify their roots before collecting terms, and expand every product before combining like terms.

For example, $3\sqrt5+2\sqrt5=5\sqrt5$ . The common $\sqrt5$ plays the same role as a common variable. Terms that initially look unlike can match after simplification:

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

There is no addition law for roots: $\sqrt{9+16}=5$ , but $\sqrt9+\sqrt{16}=7$ . The product rule cannot be borrowed for a sum.

Expanding a square

For real $a,b$ , including real surd values,

(a+b)^2=a^2+2ab+b^2.

The cross term comes from the two products $ab$ and $ba$ .

Not in the formula sheet: learn it.

A product is exact when every contribution is retained. Keep the rational terms and surd terms separate until the final line.

Worked example 3: multiplying two brackets

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

Plan. Form all four products, then collect the rational and surd terms.

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

Why this step? $\sqrt5\times\sqrt5=5$ , while the mixed products still contain one root.

Answer. $-1-\sqrt5$ .

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

Change only the second bracket so that it equals the first. This makes a square, but it does not remove the mixed products.

Worked example 4: squaring a surd expression

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

Plan. Expand as two equal brackets and collect.

\begin{aligned} (\sqrt5+2)^2 &=5+2\sqrt5+2\sqrt5+4 && \text{four products}\\ &=9+4\sqrt5 && \text{collect} \end{aligned}

Why this step? The two mixed products add; they do not cancel just because the brackets match.

Answer. $9+4\sqrt5$ .

Check. $\sqrt5>2$ , so the bracket exceeds $4$ and its square exceeds $16$ . The answer exceeds $17$ , whereas omitting the mixed term would give only $9$ .

You can use approximate decimals as a check, but exact questions need exact working and an exact final form.

Watch out

Wrong: $(2+\sqrt3)^2=7$ . Right: $(2+\sqrt3)^2=7+4\sqrt3$ . Squaring a sum always includes its two mixed products.

Replace the root in the second worked example and carry all four products through.

Rationalising denominators

Rationalising changes the form of a fraction while leaving its value intact: multiply by a carefully chosen version of one.

For a single square root in the denominator, that root is the useful multiplier. When there are two terms, the conjugate makes the mixed products cancel.

Conjugate surd expressions

The expressions $a+\sqrt b$ and $a-\sqrt b$ , with real $a$ and $b\ge0$ , are conjugates: their middle sign differs. For example, the conjugate of $3-\sqrt5$ is $3+\sqrt5$ , not $-3+\sqrt5$ .

Their product is a difference of squares, so a rational denominator appears when the squared terms are rational.

Difference of squares

For real $u,v$ ,

(u+v)(u-v)=u^2-v^2.

To rationalise a fraction this way, the starting denominator and the conjugate multiplier must both be nonzero where used.

Not in the formula sheet: learn it.

Keep the multiplier visible on top and bottom. That one line explains why the fraction still has its starting value.

Worked example 5: a single root denominator

Question. Rationalise the denominator of $\dfrac3{\sqrt5}$ .

Plan. Multiply numerator and denominator by $\sqrt5$ .

\begin{aligned} \frac3{\sqrt5} &=\frac{3\sqrt5}{\sqrt5\sqrt5} && \text{multiply by one}\\ &=\frac{3\sqrt5}{5} && \text{square the root} \end{aligned}

Why this step? A root multiplied by itself becomes its rational radicand.

Answer. $3\sqrt5/5$ .

Check. Multiplying the answer by $\sqrt5$ gives $3$ , recovering the numerator.

Change only the denominator by adding a rational term. Multiplying by the root alone would leave a mixed term, so use the conjugate instead.

Worked example 6: a two-term denominator

Question. Rationalise the denominator of $\dfrac3{2+\sqrt5}$ .

Plan. Multiply numerator and denominator by $2-\sqrt5$ , then simplify the sign.

\begin{aligned} \frac3{2+\sqrt5} &=\frac{3(2-\sqrt5)}{(2+\sqrt5)(2-\sqrt5)} && \text{conjugate multiplier}\\ &=\frac{3(2-\sqrt5)}{4-5} && \text{difference of squares}\\ &=3\sqrt5-6 && \text{divide by }-1 \end{aligned}

Why this step? The conjugate cancels both mixed terms, leaving a rational denominator even when that denominator is negative.

Answer. $3\sqrt5-6$ .

Check. $(3\sqrt5-6)(2+\sqrt5)=15-12=3$ .

The same cancellation works with two roots: $1/(\sqrt7+\sqrt3)=(\sqrt7-\sqrt3)/4$ , since the denominator becomes $7-3$ . It also settles our opening puzzle:

\frac1{\sqrt2-1} =\frac{\sqrt2+1}{(\sqrt2-1)(\sqrt2+1)} =\sqrt2+1.

A surprising consequence is that $(\sqrt2+1)^2+(\sqrt2-1)^2=6$ : the irrational mixed terms cancel when the two squares are added.

For a variable denominator, retain its domain:

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

Here $x=0$ is allowed and both forms give $-1/2$ ; $x=4$ remains excluded. Simplify perfect roots before choosing a multiplier: $1/(2+\sqrt4)=1/4$ is already rational. Multiplying by $(2-\sqrt4)/(2-\sqrt4)$ would be multiplying by an undefined fraction.

Watch out

Wrong: change only the denominator to its conjugate. Right: multiply both numerator and denominator by the same nonzero conjugate. A zero conjugate cannot form a valid multiplier by one.

In an exact OCR question, keep the surd rather than rounding it. For detailed reasoning, display the fraction multiplication and rational denominator. Try one new rational term now.

Quick Recap

Surds preserve exact irrational root values; extract square factors before collecting like terms.
Principal square roots are non-negative: $\sqrt{x^2}=|x|$ .
Expand all four products, including the mixed terms when squaring a sum.
Rationalise with a root or conjugate, multiplying both numerator and denominator and retaining restrictions.

The opening equality holds because $(\sqrt2-1)(\sqrt2+1)=1$ , so $1/(\sqrt2-1)=\sqrt2+1$ .

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1.02b - Surds and rationalising denominators

Exact roots and indices

Surd arithmetic

Rationalising denominators

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