Surds

Surds are roots (e.g. √2, √3, ∛5) whose values are irrational, so they can’t be written as a ratio of two integers. Their decimals go on forever without repeating.

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When working with surds, it’s important to simplify expressions by factoring out perfect squares and rationalising denominators when necessary. By doing this, surds can be presented in a simpler and more manageable form, making them easier to work with in mathematical problems.

Let’s look at how to simplify and rationalise surds.

Simplifying Surds

To simplify a surd, we use the rule \sqrt{ab}=\sqrt{a} \times \sqrt{b} (for a,b\ge0).

Consider the following examples:

  • \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}
  • \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}

Now, let’s simplify an expression involving surds:

\sqrt{12} + \sqrt{300} - \sqrt{48} = \sqrt{4 \times 3} + \sqrt{100 \times 3} - \sqrt{16 \times 3}

= 2\sqrt{3} + 10\sqrt{3} - 4\sqrt{3} = 8\sqrt{3}

Ok, let’s look at some more examples.

Example 1:

Simplify \frac{2}{\sqrt{2}}

Solution

\frac { 2 }{ \sqrt { 2 } } =\frac { 2 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } }

=\frac { 2\sqrt { 2 } }{ 2 }

=\sqrt { 2 }

Example 2:

Simplify (\sqrt { 2 } +\sqrt { 5 } )^2

Solution

Start by expanding the expression:

(\sqrt { 2 } +\sqrt { 5 } )^2 = (\sqrt { 2 } +\sqrt { 5 } )(\sqrt { 2 } +\sqrt { 5 } )

Now, multiply the terms (use (a + b)^{2} = a^{2} + 2ab + b^{2}):

=(\sqrt{2})^2 + 2(\sqrt{2})(\sqrt{5}) + (\sqrt{5})^2

Simplify:

= 2 + 2\sqrt{10} + 5

Then combine like terms:

= 7 + 2\sqrt{10}

Example 3:

Simplify (\sqrt { 3 } +\sqrt { 2 } )^2 - (\sqrt { 3 } -\sqrt { 2 } )^2

Solution

We can use the difference of squares formula a^2 - b^2 = (a + b)(a - b) with:

a = \sqrt { 3 } +\sqrt { 2 } and b = \sqrt { 3 } -\sqrt { 2 }

Compute a + b and a - b:

a + b = \sqrt { 3 } +\sqrt { 2 } +\sqrt { 3 } -\sqrt { 2 } = 2\sqrt { 3 }

a - b = \sqrt { 3 } +\sqrt { 2 } - (\sqrt { 3 } -\sqrt { 2 }) = 2\sqrt { 2 }

Remember, \sqrt { A } - \sqrt { A } = 0 and \sqrt { A } + \sqrt { A } = 2\sqrt { A }

Now apply the difference of squares formula:

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

Multiply the terms:

= 4\sqrt { 6 }

The simplified expression is:

= 4\sqrt { 6 }

Rationalising Surds

When working with surds (square roots of non-perfect squares), sometimes we need to rationalise the denominator, which means removing the surds from the denominator. Let’s look at how to rationalise fractions containing surds in their denominators.

Conjugates

The conjugate of \sqrt{a} +\sqrt{b} is \sqrt{a} -\sqrt{b}, and the conjugate of \sqrt{a} -\sqrt{b} is \sqrt{a} +\sqrt{b}. Conjugates are a useful tool for rationalising surds. To illustrate the process, let’s look at some examples.

Keep in mind this key formula when working with conjugates:

(a+b)(a-b)=(a-b)(a+b)=a^{2}-b^{2}

It’s also important to remember this formula:

(a-b)(a-b)=a^{2}+b^{2}-2ab

Example 1:

Rationalise \frac{5}{\sqrt{3}}

Solution

\frac{5}{\sqrt{3}} = \frac{5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}

Example 2:

Rationalise the following:

1. \frac{8}{\sqrt{2}}

2. \frac{12}{\sqrt{3}}

Solution

1. \frac{8}{\sqrt{2}} = \frac{8}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{8\sqrt{2}}{2}

= 4\sqrt{2}

2. \frac{12}{\sqrt{3}} = \frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{12\sqrt{3}}{3}

= 4\sqrt{3}

Example 3:

Rationalise \dfrac{1}{\left( \sqrt{2}-1\right) }

Solution

To rationalise this fraction, multiply the numerator and denominator by the conjugate of \sqrt{2}-1, which is \sqrt{2}+1:

\dfrac{1}{\sqrt{2}-1}=\dfrac{1}{\sqrt{2}-1}\times \dfrac{\sqrt{2}+1}{\sqrt{2}+1}

Now apply the formula (a+b)(a-b)=a^{2}-b^{2}:

=\dfrac{\sqrt{2}+1}{(\sqrt{2})^{2}-(1)^{2}}

Simplify the expression:

=\dfrac{\sqrt{2}+1}{2-1}

=\sqrt{2}+1

Example 4:

Rationalise: \frac { 1 }{ \sqrt { 2 } +1 }

Solution

To rationalise this fraction, multiply the numerator and denominator by the conjugate of \sqrt { 2 } +1, which is \sqrt { 2 } -1:

\frac { 1 }{ \sqrt { 2 } +1 } =\frac { 1 }{ \sqrt { 2 } +1 } \times \frac { \sqrt { 2 } -1 }{ \sqrt { 2 } -1 }

Now apply the formula (a+b)(a-b)=a^{2}-b^{2}:

=\frac { \sqrt { 2 } -1 }{ (\sqrt { 2 })^{ 2 } -(1)^{ 2 } }

Simplify the expression:

=\frac { \sqrt { 2 } -1 }{ 2-1 }

=\sqrt{2}-1

Example 5:

Rationalise \frac { 1+\sqrt { 2 } }{ 1-\sqrt { 2 } }

Solution

To rationalise the denominator, multiply the numerator and denominator by the conjugate of the denominator.

\frac { 1+\sqrt { 2 } }{ 1-\sqrt { 2 } } =\frac { 1+\sqrt { 2 } }{ 1-\sqrt { 2 } } \times \frac { 1+\sqrt { 2 } }{ 1+\sqrt { 2 } }

This results in:

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

=\frac { (1+\sqrt { 2 } ) (1+\sqrt { 2 } )}{ 1-2 }

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=\frac { 1+2\sqrt { 2 } +2 }{ -1 }

=-(3+2\sqrt { 2 } )

Simplified, we get:

=-3-2\sqrt { 2 }

Example 6:

Rationalise \frac { 2-\sqrt { 3 } }{ \sqrt { 2 } -1 }

Solution

Multiply by the conjugate of the denominator:

\frac { 2-\sqrt { 3 } }{ \sqrt { 2 } -1 } =\frac { 2-\sqrt { 3 } }{ \sqrt {2 } -1 } \times \frac { \sqrt { 2 } +1 }{ \sqrt { 2 } +1 }

This results in:

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

=\frac { 2\sqrt { 2 } +2-\sqrt { 6 } -\sqrt { 3 } }{ 2-1 }

Simplified, we get:

=2\sqrt { 2 } +2-\sqrt { 6 } -\sqrt { 3 }

Example 7:

Rationalise \frac { 27 }{ 9-\sqrt { 3 } }

Solution

Multiply by the conjugate of the denominator:

\frac { 27 }{ 9-\sqrt { 3 } } =\frac { 27 }{ 9-\sqrt { 3 } } \times \frac { 9+\sqrt { 3 } }{ 9+\sqrt { 3 } }

This results in:

=\frac { 27(9+\sqrt { 3 }) }{ 9^2-(\sqrt { 3 })^2 }

=\frac { 27(9+\sqrt { 3 }) }{ 81-3 }

=\frac { 27(9+\sqrt { 3 }) }{ 78 }

Simplify the expression:

=\frac { 27\times9 + 27\times\sqrt { 3 } }{ 78 }

=\frac { 243 + 27\sqrt { 3 } }{ 78 }

Finally, we can simplify further by dividing both the numerator and the denominator by their highest common factor, which is 3:

=\frac { 81 + 9\sqrt { 3 } }{ 26 }

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