# Surds

## Surds

Surds are special types of irrational number, such as $\sqrt { 2 }$ and $3\sqrt { 5 }$. Some surds give rational values, such as $\sqrt { 9 } = 3$, however others do not have an exact decimal value.

Think: What is the number between $\sqrt{4}$ and $\sqrt{16}$. Is that number $\sqrt{10}$? Why not?

Basic Properties of Surds

$\sqrt { a } \times \sqrt { b } =\sqrt { ab } \\ \\ \sqrt { a } \div \sqrt { b } =\sqrt { \frac { a }{ b } } \\ \\ { (\sqrt { x } ) }^{ 2 }=\sqrt { { x }^{ 2 } } = x \\ \\ a\sqrt { b } \times c\sqrt { d } =ac\sqrt { bd } \\ \\ \sqrt { a } \times \sqrt { a } =\sqrt { { a }^{ 2 } } =a\\ \\ \frac { \sqrt { a } }{ \sqrt { b } } =\sqrt { \frac { a }{ b } }$

### Examples:

Treat surds in the same way as pronumerals in algebra. Only add or subtract 'like surds' the same way 'like terms' are calculated.

1. Simplify $3\sqrt { 2 } +4\sqrt { 2 }$

$3\sqrt { 2 } +4\sqrt { 2 } =7\sqrt { 2 }$

2. Simplify $\sqrt{3} - \sqrt{12}$

$\sqrt{3} - \sqrt{12} \\ =\sqrt{3} - \sqrt{4} \times \sqrt{3} \\ = \sqrt{3} - 2 \sqrt{3} \\ = -\sqrt{3}$

Multiplication and Division

3. Simplify $2\sqrt { 2 } \times -5\sqrt { 8 }$

$2\sqrt { 2 } \times -5\sqrt { 8 } \\ = -10\sqrt { 16 } \\= -10\times 4 = -40$

4. Simplify $4\sqrt { 2 } \times \sqrt { 18 }$

$4\sqrt { 2 } \times \sqrt { 18 } \\ =\quad 4\sqrt { 2 } \times \sqrt { 9\times 2 } \\ =\quad 4\sqrt { 2 } \times \sqrt { 9 } \times \sqrt { 2 } \\ =\quad 4\sqrt { 2 } \times 3\sqrt { 2 } \\ =\quad 12\sqrt { 4 } \\ =\quad 12\times 2\\ =\quad 24$

5. Simplify $\cfrac { 2\sqrt { 14 } }{ 4\sqrt { 2 } }$

$\cfrac { 2\sqrt { 14 } }{ 4\sqrt { 2 } } \\ =\quad \cfrac { \not2 \times \sqrt { 2 } \times \sqrt { 7 } }{ \not2 \times 2\times \sqrt { 2 } } \\ =\quad \cfrac { \sqrt { 7 } }{ 2 }$

6. Simplify ${ \left( \cfrac { \sqrt { 20 } }{ \sqrt { 7 } } \right) }^{ 2 }$

${ \left( \cfrac { \sqrt { 20 } }{ \sqrt { 7 } } \right) }^{ 2 }\\ =\quad \cfrac { { (\sqrt { 20 } ) }^{ 2 } }{ { (\sqrt { 7 } ) }^{ 2 } } \\ =\quad \cfrac { 20 }{ 7 } \\ =\quad 2\cfrac { 6 }{ 7 }$

## Expanding Brackets

The same rules for expanding brackets and binomial products that you use in algebra also applies to surds. Go to Group Symbols and Binomials to find out more.

Properties of Surds as Binomials

$\sqrt { a } (\sqrt { b } +\sqrt { c } )=\sqrt { ab } +\sqrt { ac } \\ \\ (\sqrt { a } +\sqrt { b } )(\sqrt { c } +\sqrt { d } )=\sqrt { ac } +\sqrt { ad } +\sqrt { bc } +\sqrt { bc } \\ \\ { (\sqrt { a } +\sqrt { b } ) }^{ 2 }=a+2\sqrt { ab } +b\\ \\ { (\sqrt { a } -\sqrt { b } ) }^{ 2 }=a-2\sqrt { ab } +b\\ \\ (\sqrt { a } +\sqrt { b } )(\sqrt { a } -\sqrt { b } )=a-b$

### Examples:

Expand and simplify:

1. $2\sqrt{3}(\sqrt { 12 } -4\sqrt { 5 })$

$2\sqrt { 3 } (\sqrt { 12 } -4\sqrt { 5 }) \\ =\quad 2\sqrt { 36 } -8\sqrt { 15 } \\ =\quad 2\times 6-8\sqrt { 15 } \\ =\quad 12-8\sqrt { 15 }$

2. $(\sqrt { 5 } +2\sqrt { 3 } )(\sqrt { 5 } -2\sqrt { 3 } )$

$(\sqrt { 5 } +2\sqrt { 3 } )(\sqrt { 5 } -2\sqrt { 3 } )\\ =\quad 5-2\sqrt { 15 } +2\sqrt { 15 } -4\sqrt { 9 } \\ =\quad 5\quad -\quad 4\times 3\\ =\quad 5\quad -\quad 12\quad =\quad -7$

(This example is called the difference of two squares.)

## Rationalising the Denominator

Rationalising the denominator of a fractional surd means writing it with a rational number (not a surd) in the denominator. For example, after rationalising the denominator, $\cfrac { 3 }{ \sqrt { 5 } }$ becomes $\cfrac { 3\sqrt { 5 } }{ 5 }$.

Back when calculators were available, rationalising denominator is used to make it easier to divide a fraction by a rational number than an irrational one.

$\cfrac { a }{ \sqrt { b } } \times \cfrac { \sqrt { b } }{ \sqrt { b } } =\cfrac { a\sqrt { b } }{ b }$

When there is a binomial denominator, we use the difference of two squares to rationalise it, as the result is always a rational number.

The sum and difference of two simple quadratic surds are said to be conjugate surds to each other.

Conjugate surds are also known as complementary surds.

Two surds $(2\sqrt{3}+\sqrt { 5 } )$ and $(2\sqrt{3}-\sqrt { 5 } )$ are conjugate to each other.

To rationalise the denominator of $\cfrac { \sqrt { a } +\sqrt { b } }{ \sqrt { c } +\sqrt { d } }$, multiply by $\cfrac { \sqrt { c } -\sqrt { d } }{ \sqrt { c } -\sqrt { d } }$.

### Examples:

1. Rationalise the denominator of $\cfrac { 4 }{ \sqrt { 10 } }$

$\cfrac { 4 }{ \sqrt { 10 } } \times \cfrac { \sqrt { 10 } }{ \sqrt { 10 } } \\ \\=\quad \cfrac { 4\sqrt { 10 } }{ 10 } =\quad \cfrac { 2\sqrt { 10 } }{ 5 }$

2. Rationalise the denominator of $\cfrac { 4 }{ 2\sqrt { 10 } }$

$\cfrac { 4 }{ 2\sqrt { 10 } } \times \cfrac { \sqrt { 10 } }{ \sqrt { 10 } } \\ \\=\quad \cfrac { 4\sqrt { 10 } }{ 2\times 10 } \\ \\ =\quad \cfrac { 4\sqrt { 10 } }{ 20 } =\quad \cfrac { 1\sqrt { 10 } }{ 5 }$

3. Write with a rational denominator: $\cfrac { 2\sqrt { 3 } +\sqrt { 5 } }{ \sqrt { 3 } +4\sqrt { 2 } }$

$\cfrac { 2\sqrt { 3 } +\sqrt { 5 } }{ \sqrt { 3 } +4\sqrt { 2 } } \\ \\ =\quad \cfrac { 2\sqrt { 3 } +\sqrt { 5 } }{ \sqrt { 3 } +4\sqrt { 2 } } \times \cfrac { \sqrt { 3 } -4\sqrt { 2 } }{ \sqrt { 3 } -4\sqrt { 2 } } \\ \\ =\quad \cfrac { (2\sqrt { 3 } +\sqrt { 5 } )(\sqrt { 3 } -4\sqrt { 2 } ) }{ { (\sqrt { 3 } ) }^{ 2 }-{ (4\sqrt { 2 } ) }^{ 2 } } \\ \\ =\quad \cfrac { 2\times 3-8\sqrt { 6 } +\sqrt { 15 } -4\sqrt { 10 } }{ 3-16\times 2 } \\ \\=\quad \cfrac { 6-8\sqrt { 6 } +\sqrt { 15 } -4\sqrt { 10 } }{ -29 } \\ \\ =\quad \cfrac { -6+8\sqrt { 6 } -\sqrt { 15 } +4\sqrt { 10 } }{ 29 }$

4. Evaluate as a cfraction with rational denominator: $\cfrac { 2 }{ \sqrt { 3 } +2 } +\cfrac { \sqrt { 5 } }{ \sqrt { 3 } -2 }$

$\cfrac { 2 }{ \sqrt { 3 } +2 } +\cfrac { \sqrt { 5 } }{ \sqrt { 3 } -2 } \\ \\ =\quad \cfrac { 2(\sqrt { 3 } -2)+\sqrt { 5 } (\sqrt { 3 } +2) }{ (\sqrt { 3 } +2)(\sqrt { 3 } -2) } \\ \\ =\quad \cfrac { 2\sqrt { 3 } -4+\sqrt { 15 } +2\sqrt { 5 } }{ { (\sqrt { 3 } ) }^{ 2 }-{ 2 }^{ 2 } } \\ \\ =\quad \cfrac { 2\sqrt { 3 } -4+\sqrt { 15 } +2\sqrt { 5 } }{ 3-4 } \\ \\ =\quad \cfrac { 2\sqrt { 3 } -4+\sqrt { 15 } +2\sqrt { 5 } }{ -1 } \\ \\ =\quad -2\sqrt { 3 } +4-\sqrt { 15 } -2\sqrt { 5 }$

### Challenge

Can you simplify:

$\sqrt { 5-\sqrt { 13+\sqrt { 48 } } }$

Solution:

$\sqrt { 5-\sqrt { 13+\sqrt { 48 } } } \\ \\ =\sqrt { 5-\sqrt { 13+4\sqrt { 3 } } } \\ \\ But,\quad 13+4\sqrt { 3 } \quad =\quad 1+4\sqrt { 3 } +12\\ \qquad \qquad \qquad \qquad =\quad { \left( 1+2\sqrt { 3 } \right) }^{ 2 }\\ \\ Hence,\quad \\ \\ =\sqrt { 5-\sqrt { { \left( 1+2\sqrt { 3 } \right) }^{ 2 } } } \\ \\ =\sqrt { 5-(1+2\sqrt { 3 } ) } \\ \\ =\sqrt { 5-1-2\sqrt { 3 } } \\ \\ =\sqrt { 4-2\sqrt { 3 } } \\ \\ But,\quad 4-2\sqrt { 3 } \quad =\quad 1-2\sqrt { 3 } +3\\ \qquad \qquad \qquad \qquad =\quad { \left( 1-\sqrt { 3 } \right) }^{ 2 }\\ \\ Hence,\\ \\ =\sqrt { { \left( 1-\sqrt { 3 } \right) }^{ 2 } } \\ \\ =1-\sqrt { 3 }$