Surds: Simplification, Manipulation and Rationalising Denominators

Surds allow irrational numbers to be written exactly. For example, $\sqrt{3}$ cannot be represented by a terminating or recurring decimal, but the expression $\sqrt{3}$ records its exact value without approximation.

In the TMUA, the challenge is not simply knowing how to simplify a square root. Students must recognise useful factorisations, manipulate expressions efficiently, rationalise denominators and keep track of signs when square roots are involved.

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Surds may appear within:

• algebraic expressions;
• fractions requiring rationalisation;
• equations and inequalities;
• exact comparisons;
• expressions that are disguised perfect squares;
• multiple-choice options written in different but equivalent forms.

Efficient work with surds requires students to:

• simplify roots before combining terms;
• recognise conjugate expressions;
• avoid invalid square-root rules;
• preserve exact values rather than using decimals;
• check the sign of an expression after taking a square root.

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A surd is an irrational root written in exact form.

Examples include

$$\sqrt{2},\qquad \sqrt{7},\qquad \sqrt[3]{5}.$$

More generally, an expression such as

$$3+2\sqrt{5}$$

is called a surd expression because it contains the irrational number $\sqrt{5}$.

An expression containing a root is not necessarily a surd. For example,

$$\sqrt{81}=9,$$

so $\sqrt{81}$ simplifies to a rational number.

The principal square root

For $a\geq 0$, the notation $\sqrt{a}$ means the non-negative square root of $a$.

Therefore,

$$\sqrt{49}=7,$$

not $\pm7$.

However, the equation

$$x^2=49$$

has two solutions:

$$x=\pm7.$$

The square-root symbol represents one non-negative value, whereas solving a quadratic equation may produce both a positive and a negative solution.

Introductory example

Simplify

$$\sqrt{72}.$$

Since

$$72=36\times2,$$

we have

$$\sqrt{72} = \sqrt{36\times2} = \sqrt{36}\sqrt{2} = 6\sqrt{2}.$$

Therefore,

$$\boxed{\sqrt{72}=6\sqrt{2}}.$$

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For $a\geq0$ and $b\geq0$,

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

For $a\geq0$ and $b>0$,

$$\boxed{\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}}.$$

For $a\geq0$,

$$\boxed{\left(\sqrt{a}\right)^2=a}.$$

Also,

$$\boxed{\sqrt{a}\sqrt{a}=a}.$$

For example,

$$\sqrt{7}\sqrt{7}=7.$$

It does not equal $49$.

Combining like surds

Surds can be combined only when their irrational parts are the same.

For example,

$$3\sqrt{5}+7\sqrt{5}=10\sqrt{5}.$$

Similarly,

$$8\sqrt{3}-2\sqrt{3}=6\sqrt{3}.$$

However,

$$\sqrt{2}+\sqrt{3}$$

cannot be simplified further because the two surds are unlike terms.

A rule that is not valid

In general,

$$\sqrt{a+b}\neq\sqrt{a}+\sqrt{b}.$$

For example,

$$\sqrt{9+16}=5,$$

but

$$\sqrt{9}+\sqrt{16}=3+4=7.$$

The product rule for square roots does not extend to addition.

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To simplify $\sqrt{n}$, identify the largest square number that divides $n$.

Example

Simplify

$$\sqrt{180}.$$

Since

$$180=36\times5,$$

we have

$$\sqrt{180} = \sqrt{36}\sqrt{5} = 6\sqrt{5}.$$

Therefore,

$$\boxed{\sqrt{180}=6\sqrt{5}}.$$

Using the largest square factor usually produces the simplified form immediately.

Simplifying before combining

Consider

$$3\sqrt{45}-2\sqrt{20}.$$

Simplify each surd:

$$\sqrt{45} = \sqrt{9\times5} = 3\sqrt{5},$$

and

$$\sqrt{20} = \sqrt{4\times5} = 2\sqrt{5}.$$

Therefore,

$$3\sqrt{45}-2\sqrt{20} = 9\sqrt{5}-4\sqrt{5} = \boxed{5\sqrt{5}}.$$

The original surds looked different, but both reduced to multiples of $\sqrt{5}$.

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Surds can be treated like algebraic terms during expansion.

For example,

$$(a+b\sqrt{c})(d+e\sqrt{c})$$

can be expanded using the usual distributive law.

Example

Expand and simplify

$$(3-2\sqrt{2})(1+\sqrt{2}).$$

Expanding gives

$$3+3\sqrt{2}-2\sqrt{2}-2\sqrt{2}\sqrt{2}.$$

Since

$$\sqrt{2}\sqrt{2}=2,$$

we obtain

$$3+\sqrt{2}-4 = \boxed{\sqrt{2}-1}.$$

Squaring a surd expression

For real numbers $a$, $b$ and $c\geq0$,

$$(a+b\sqrt{c})^2 = a^2+2ab\sqrt{c}+b^2c.$$

For example,

$$(2+\sqrt{3})^2 = 4+4\sqrt{3}+3 = \boxed{7+4\sqrt{3}}.$$

The same algebraic identities used for ordinary expressions continue to apply.

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A surd may involve a cube root or another higher-order root.

For example,

$$\sqrt[3]{54} = \sqrt[3]{27\times2} = 3\sqrt[3]{2}.$$

The same principle applies: extract factors that are perfect powers of the relevant degree.

Rationalising a simple cube-root denominator

Consider

$$\frac{1}{\sqrt[3]{2}}.$$

Multiply the numerator and denominator by $\sqrt[3]{4}$:

$$\frac{1}{\sqrt[3]{2}} \times \frac{\sqrt[3]{4}}{\sqrt[3]{4}} = \frac{\sqrt[3]{4}}{\sqrt[3]{8}} = \boxed{\frac{\sqrt[3]{4}}{2}}.$$

For expressions involving sums or differences of cube roots, the identities

$$x^3-y^3=(x-y)(x^2+xy+y^2)$$

and

$$x^3+y^3=(x+y)(x^2-xy+y^2)$$

may be useful.

For example,

$$\frac{1}{\sqrt[3]{2}-1}$$

can be rationalised by multiplying by

$$\sqrt[3]{4}+\sqrt[3]{2}+1,$$

because

$$(\sqrt[3]{2}-1) \left(\sqrt[3]{4}+\sqrt[3]{2}+1\right) = 2-1 = 1.$$

Therefore,

$$\boxed{ \frac{1}{\sqrt[3]{2}-1} = \sqrt[3]{4}+\sqrt[3]{2}+1 }.$$

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The most important rationalisation technique is the use of a conjugate.

The conjugate of

$$a+b\sqrt{c}$$

is

$$a-b\sqrt{c}.$$

Their product contains no surd:

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

This is an application of the difference-of-two-squares identity:

$$(x+y)(x-y)=x^2-y^2.$$

When to use a conjugate

Use a conjugate when the denominator contains:

• a rational term and a surd;
• two square-root surds;
• an expression of the form $a\pm b\sqrt{c}$.

Example

Rationalise and simplify

$$\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}.$$

The conjugate of the denominator is

$$\sqrt{3}-\sqrt{2}.$$

Multiply the numerator and denominator by this conjugate:

$$\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}.$$

The denominator becomes

$$(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2}) = 3-2 = 1.$$

The numerator becomes

$$\sqrt{6}\sqrt{3}-\sqrt{6}\sqrt{2} = \sqrt{18}-\sqrt{12}.$$

Simplifying the surds,

$$\sqrt{18}=3\sqrt{2}$$

and

$$\sqrt{12}=2\sqrt{3}.$$

Therefore,

$$\boxed{ \frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}} = 3\sqrt{2}-2\sqrt{3} }.$$

The conjugate was chosen because it converted the denominator into a rational number.

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Trick 1: Extract the Largest Square Factor

When simplifying $\sqrt{n}$, look first for the greatest square factor of $n$.

For example,

$$\sqrt{288} = \sqrt{144\times2} = 12\sqrt{2}.$$

Using $288=4\times72$ would also work, but it would require additional steps.

ExamAlly insight: Search for the largest square factor before beginning a chain of smaller simplifications.

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Trick 2: Simplify Before Combining

Expressions that initially contain unlike surds may become like terms after simplification.

For example,

$$\sqrt{48}+\sqrt{75} = 4\sqrt{3}+5\sqrt{3} = 9\sqrt{3}.$$

ExamAlly insight: Do not decide that surds are unlike until each one has been fully simplified.

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Trick 3: Change the Sign to Find the Conjugate

The conjugate is obtained by changing the sign between the two terms:

$$a+\sqrt{b} \quad\longleftrightarrow\quad a-\sqrt{b}.$$

For example,

$$(4+\sqrt{7})(4-\sqrt{7}) = 16-7 = 9.$$

ExamAlly insight: Keep both terms unchanged and reverse only the sign connecting them.

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Trick 4: Look for a Denominator That Becomes One

Some denominators are deliberately chosen so that multiplying by the conjugate gives $1$.

For example,

$$(\sqrt{5}+2)(\sqrt{5}-2) = 5-4 = 1.$$

Therefore,

$$\frac{1}{\sqrt{5}+2} = \sqrt{5}-2.$$

ExamAlly insight: Before expanding fully, check whether the conjugate product is a particularly simple integer.

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Trick 5: Recognise Disguised Perfect Squares

An expression of the form

$$A+B\sqrt{c}$$

may be the square of

$$a+b\sqrt{c}.$$

Since

$$(a+b\sqrt{c})^2 = a^2+b^2c+2ab\sqrt{c},$$

we compare

$$a^2+b^2c=A$$

and

$$2ab=B.$$

For example,

$$11+6\sqrt{2} = (3+\sqrt{2})^2.$$

Therefore,

$$\sqrt{11+6\sqrt{2}} = 3+\sqrt{2}.$$

ExamAlly insight: Use the coefficient of the surd to identify the product $2ab$ before testing possible values of $a$ and $b$.

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Trick 6: Check the Sign After Taking a Square Root

Even if

$$N=(a-b\sqrt{c})^2,$$

it does not always follow that

$$\sqrt{N}=a-b\sqrt{c}.$$

The correct rule is

$$\sqrt{x^2}=|x|.$$

For example,

$$13-4\sqrt{3} = (1-2\sqrt{3})^2.$$

But

$$1-2\sqrt{3}<0.$$

Therefore,

$$\sqrt{13-4\sqrt{3}} = |1-2\sqrt{3}| = \boxed{2\sqrt{3}-1}.$$

ExamAlly insight: After factorising a nested surd as a square, determine whether the expression inside the square is positive or negative.

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Trick 7: Compare Positive Surds by Squaring

To compare two positive quantities containing square roots, it may be easier to compare their squares.

For example, compare

$$3\sqrt{5} \quad\text{and}\quad 2\sqrt{11}.$$

Both are positive. Squaring gives

$$(3\sqrt{5})^2=45$$

and

$$(2\sqrt{11})^2=44.$$

Therefore,

$$\boxed{3\sqrt{5}>2\sqrt{11}}.$$

ExamAlly insight: Squaring is safe for comparison only when both quantities are known to be non-negative.

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Trick 8: Exploit Symmetric Conjugate Expressions

Expressions containing both

$$\frac{1}{a+b} \quad\text{and}\quad \frac{1}{a-b}$$

often simplify efficiently when combined before rationalising separately.

For example,

$$\frac{1}{\sqrt{7}-\sqrt{2}} + \frac{1}{\sqrt{7}+\sqrt{2}}.$$

Combining the fractions gives

$$\frac{(\sqrt{7}+\sqrt{2})+(\sqrt{7}-\sqrt{2})} {7-2} = \frac{2\sqrt{7}}{5}.$$

ExamAlly insight: When conjugate denominators appear together, combine them structurally before carrying out two separate rationalisations.

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Simplify

$$\frac{2+\sqrt{3}}{2-\sqrt{3}} + \frac{2-\sqrt{3}}{2+\sqrt{3}}.$$

Since

$$(2+\sqrt{3})(2-\sqrt{3}) = 4-3 = 1,$$

the first fraction becomes

$$\frac{2+\sqrt{3}}{2-\sqrt{3}} = \frac{(2+\sqrt{3})^2}{1} = (2+\sqrt{3})^2.$$

Similarly,

$$\frac{2-\sqrt{3}}{2+\sqrt{3}} = (2-\sqrt{3})^2.$$

Therefore, the expression is

$$(2+\sqrt{3})^2+(2-\sqrt{3})^2.$$

Expanding,

$$(7+4\sqrt{3})+(7-4\sqrt{3}) = 14.$$

Hence,

$$\boxed{14}.$$

The irrational terms cancel because the two expanded expressions are conjugates.

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Compare

$$4\sqrt{3} \quad\text{and}\quad 7.$$

Both quantities are positive, so squaring preserves their order.

We have

$$(4\sqrt{3})^2 = 16\times3 = 48,$$

while

$$7^2=49.$$

Since

$$48<49,$$

it follows that

$$\boxed{4\sqrt{3}<7}.$$

No decimal approximation of $\sqrt{3}$ is required.

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Mistake 1: Splitting a Sum Inside a Square Root

The statement

$$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$$

is generally false.

For example,

$$\sqrt{4+5}=3,$$

but

$$\sqrt{4}+\sqrt{5}=2+\sqrt{5}.$$

Square-root rules apply to products and quotients under suitable conditions, not to sums.

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Mistake 2: Writing $\sqrt{x^2}=x$

The correct identity is

$$\boxed{\sqrt{x^2}=|x|}.$$

For example, if $x=-4$,

$$\sqrt{x^2} = \sqrt{16} = 4,$$

not $-4$.

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Mistake 3: Combining Unlike Surds

In general,

$$\sqrt{2}+\sqrt{3}$$

cannot be simplified.

However, students should simplify each surd first. For example,

$$\sqrt{12}+\sqrt{27} = 2\sqrt{3}+3\sqrt{3} = 5\sqrt{3}.$$

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Mistake 4: Squaring a Binomial Incorrectly

The expression

$$(a+b)^2$$

is not equal to

$$a^2+b^2.$$

The correct expansion is

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

Therefore,

$$(2+\sqrt{5})^2 = 4+4\sqrt{5}+5 = 9+4\sqrt{5}.$$

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Mistake 5: Rationalising Only Part of the Denominator

For

$$\frac{1}{3+\sqrt{2}},$$

multiplying by $\sqrt{2}$ does not rationalise the denominator:

$$(3+\sqrt{2})\sqrt{2} = 3\sqrt{2}+2,$$

which still contains a surd.

The correct multiplier is the conjugate:

$$3-\sqrt{2}.$$

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Mistake 6: Miscalculating the Product of Equal Surds

The correct result is

$$\sqrt{a}\sqrt{a}=a,$$

not $a^2$.

For example,

$$\sqrt{5}\sqrt{5}=5.$$

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Mistake 7: Ignoring the Principal Square Root

If

$$N=(a-b\sqrt{c})^2,$$

then

$$\sqrt{N}=|a-b\sqrt{c}|.$$

The sign must be checked before removing the square and square root.

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Real square roots

For $\sqrt{x}$ to be real,

$$x\geq0.$$

For example, $\sqrt{-3}$ is not a real number.

Denominators

A denominator must never equal zero.

For example,

$$\frac{1}{\sqrt{x}-2}$$

is undefined when

$$\sqrt{x}=2,$$

which occurs when

$$x=4.$$

Rationalising a denominator does not remove this restriction.

Product rules

The rule

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

is valid over the real numbers when $a\geq0$ and $b\geq0$.

It should not be applied blindly to negative values.

Principal square roots

The value of $\sqrt{x}$ is always non-negative.

Therefore,

$$\sqrt{x^2}=|x|.$$

Rationalisation preserves value

When rationalising a fraction, multiply the numerator and denominator by the same non-zero expression. This is multiplication by $1$, so the value of the fraction is unchanged.

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1. Simplify every surd by extracting perfect square or perfect cube factors.
2. Combine only surds with the same simplified irrational part.
3. Identify whether a denominator requires a simple surd multiplier or a conjugate.
4. Change only the central sign when forming a conjugate.
5. Use difference of two squares before expanding unnecessarily.
6. Look for expressions that may be disguised perfect squares.
7. Check the sign before simplifying $\sqrt{x^2}$.
8. Compare positive surds by squaring when this removes the roots.
9. Preserve restrictions from the original denominator.
10. Verify the final expression by estimating its sign or approximate size mentally where useful.

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Simplify exactly

$$E= \frac{1}{\sqrt{7}-\sqrt{3}} + \frac{1}{\sqrt{7}+\sqrt{3}} - \frac12\sqrt{10+2\sqrt{21}}.$$

Step 1: Combine the conjugate fractions

The first two terms have conjugate denominators:

$$\frac{1}{\sqrt{7}-\sqrt{3}} + \frac{1}{\sqrt{7}+\sqrt{3}}.$$

Combining them gives

$$\frac{(\sqrt{7}+\sqrt{3})+(\sqrt{7}-\sqrt{3})} {(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})}.$$

The numerator simplifies to

$$2\sqrt{7},$$

and the denominator simplifies to

$$7-3=4.$$

Therefore,

$$\frac{1}{\sqrt{7}-\sqrt{3}} + \frac{1}{\sqrt{7}+\sqrt{3}} = \frac{\sqrt{7}}{2}.$$

Step 2: Simplify the nested surd

Consider

$$\sqrt{10+2\sqrt{21}}.$$

Since

$$(\sqrt{7}+\sqrt{3})^2 = 7+3+2\sqrt{21} = 10+2\sqrt{21},$$

we have

$$\sqrt{10+2\sqrt{21}} = \sqrt{7}+\sqrt{3}.$$

Both terms are positive, so there is no sign ambiguity.

Step 3: Substitute

Therefore,

$$E = \frac{\sqrt{7}}{2} - \frac{\sqrt{7}+\sqrt{3}}{2}.$$

Hence,

$$E = \frac{\sqrt{7}-\sqrt{7}-\sqrt{3}}{2} = \boxed{-\frac{\sqrt{3}}{2}}.$$

This example combines:

• conjugate denominators;
• difference of two squares;
• cancellation of surd terms;
• recognition of a disguised perfect square;
• the principal square-root convention.

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For $a\geq0$ and $b\geq0$,

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

For $a\geq0$ and $b>0$,

$$\boxed{\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}}$$

For real $x$,

$$\boxed{\sqrt{x^2}=|x|}$$

For $a,b\geq0$,

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

For real $a$, $b$ and $c\geq0$,

$$\boxed{(a+b\sqrt{c})^2=a^2+2ab\sqrt{c}+b^2c}$$

For real $a$, $b$ and $c\geq0$,

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

For cube expressions,

$$\boxed{x^3-y^3=(x-y)(x^2+xy+y^2)}$$

$$\boxed{x^3+y^3=(x+y)(x^2-xy+y^2)}$$

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