### Surds Part I

Surds are terms that contain the square root sign e.g. $\sqrt{3} $

It is useful to know that $\sqrt{x} $ is the same as $\ x^{\frac{1}{2}} $ . Hence the rules used in the evaluation of surds is similar to the rules of indices.

Commonly used rules

Rule 1

$\ 3\sqrt{x} + \sqrt{x} = 4\sqrt{x} $

Note the similarity to to $\ 3y + y = 4y $

Rule 2

$\ \sqrt{x} \times \sqrt{x} = x $

Note the similarity to

$\begin {aligned}

x^{\frac{1}{2}} \times x^{\frac{1}{2}} &= x^{\frac{1}{2} + \frac{1}{2}} \\

&= x^1 \\

\end{aligned} $

Related to rule 2 is the rationalization of surds

Rationalization of surds

Example 1

Simplify $\ \frac{2}{\sqrt{3}} $

In order to rationalize this fraction, we need to get rid of the square root in the denominator. The way to get rid of the square root is to make use of rule 2

$\begin {aligned}

\frac{2}{\sqrt{3}} &= \frac{2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3} } \\

&= \frac{2\sqrt{3}}{3} \\

\end{aligned} $

Example 2

Simplify $\ \frac{2}{\sqrt{3} - 1} $

Again we want to get rid of the square root in the denominator.

To do so, we make use of an algebraic identity $\ (a+b)(a-b) = a^2 - b^2 $

Hence we multiply the denominator by $\ \sqrt{3} +1 $

$\begin {aligned}

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

&= 3 -1 \qquad \text{Note that:}\sqrt{3} \times \sqrt{3} = 3 \\

&= 2

\end{aligned} $

Going back to the question

$\begin {aligned}

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

&= \frac {2\sqrt{3} + 2}{2} \\

&= \sqrt{3} +1 \\

\end{aligned} $

Example 3 (a more challenging question)

Simplify $ \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{2} + \sqrt{3} - \sqrt{5} } $

$\begin{aligned}

\frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{2} + \sqrt{3} - \sqrt{5} } &= \frac{ (\sqrt{5} - \sqrt{2}) \times (\sqrt{2} + \sqrt{3} + \sqrt{5}) }{ (\sqrt{2} + \sqrt{3} - \sqrt{5}) \times (\sqrt{2} + \sqrt{3} + \sqrt{5}) } \\

&= \frac{ \sqrt{10} + \sqrt{15} + 5 - 2 - \sqrt{6} - \sqrt{10} }{ (\sqrt{2} + \sqrt{3})^2 - 5 } \\

&= \frac{ \sqrt{15} - \sqrt{6} - 3 }{ 2 + 2\sqrt{6} + 3 - 5 } \\

&= \frac{ \sqrt{15} - \sqrt{6} - 3 }{ 2\sqrt{6} } \\

&= \frac{ \sqrt{90} - 6 - 3 \sqrt{6} }{ 12 } \\

&= \frac{ 3\sqrt{10} - 6 - 3\sqrt{6} }{ 12 } \\

&= \frac{ \sqrt{10} - 2 - \sqrt{6} }{ 4 } \\

\end{aligned} $

Rule 3

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

This rule is useful when we are simplifying surds

For example $ \sqrt{72} = \sqrt{36 \times 2} = \sqrt{36}\sqrt{2} = 6 \sqrt{2} $

We know that 72 is not a perfect square but 36 is a perfect square

Example 4

Simplify $ \sqrt{27} + \sqrt{12} - 2 \sqrt{75} + \sqrt{48} $

$\begin{aligned}

\sqrt{27} + \sqrt{12} - 2 \sqrt{75} + \sqrt{48} &= \sqrt{9 \times 3} + \sqrt{4 \times 3} - 2 \sqrt{5 \times 25} + \sqrt{16 \times 3} \\

&= \sqrt{9}\sqrt{3} +\sqrt{4}\sqrt{3} -2\sqrt{5}\sqrt{25} + \sqrt{16}\sqrt{3} \\

&= 3\sqrt{3} + 2\sqrt{3} -10\sqrt{5} + 4\sqrt{3} \\

&= 9\sqrt{3} -10\sqrt{5} \\

\end{aligned} $

It is useful to know that $\sqrt{x} $ is the same as $\ x^{\frac{1}{2}} $ . Hence the rules used in the evaluation of surds is similar to the rules of indices.

Commonly used rules

Rule 1

$\ 3\sqrt{x} + \sqrt{x} = 4\sqrt{x} $

Note the similarity to to $\ 3y + y = 4y $

Rule 2

$\ \sqrt{x} \times \sqrt{x} = x $

Note the similarity to

$\begin {aligned}

x^{\frac{1}{2}} \times x^{\frac{1}{2}} &= x^{\frac{1}{2} + \frac{1}{2}} \\

&= x^1 \\

\end{aligned} $

Related to rule 2 is the rationalization of surds

Rationalization of surds

Example 1

Simplify $\ \frac{2}{\sqrt{3}} $

In order to rationalize this fraction, we need to get rid of the square root in the denominator. The way to get rid of the square root is to make use of rule 2

$\begin {aligned}

\frac{2}{\sqrt{3}} &= \frac{2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3} } \\

&= \frac{2\sqrt{3}}{3} \\

\end{aligned} $

Example 2

Simplify $\ \frac{2}{\sqrt{3} - 1} $

Again we want to get rid of the square root in the denominator.

To do so, we make use of an algebraic identity $\ (a+b)(a-b) = a^2 - b^2 $

Hence we multiply the denominator by $\ \sqrt{3} +1 $

$\begin {aligned}

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

&= 3 -1 \qquad \text{Note that:}\sqrt{3} \times \sqrt{3} = 3 \\

&= 2

\end{aligned} $

Going back to the question

$\begin {aligned}

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

&= \frac {2\sqrt{3} + 2}{2} \\

&= \sqrt{3} +1 \\

\end{aligned} $

Example 3 (a more challenging question)

Simplify $ \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{2} + \sqrt{3} - \sqrt{5} } $

$\begin{aligned}

\frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{2} + \sqrt{3} - \sqrt{5} } &= \frac{ (\sqrt{5} - \sqrt{2}) \times (\sqrt{2} + \sqrt{3} + \sqrt{5}) }{ (\sqrt{2} + \sqrt{3} - \sqrt{5}) \times (\sqrt{2} + \sqrt{3} + \sqrt{5}) } \\

&= \frac{ \sqrt{10} + \sqrt{15} + 5 - 2 - \sqrt{6} - \sqrt{10} }{ (\sqrt{2} + \sqrt{3})^2 - 5 } \\

&= \frac{ \sqrt{15} - \sqrt{6} - 3 }{ 2 + 2\sqrt{6} + 3 - 5 } \\

&= \frac{ \sqrt{15} - \sqrt{6} - 3 }{ 2\sqrt{6} } \\

&= \frac{ \sqrt{90} - 6 - 3 \sqrt{6} }{ 12 } \\

&= \frac{ 3\sqrt{10} - 6 - 3\sqrt{6} }{ 12 } \\

&= \frac{ \sqrt{10} - 2 - \sqrt{6} }{ 4 } \\

\end{aligned} $

Rule 3

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

This rule is useful when we are simplifying surds

For example $ \sqrt{72} = \sqrt{36 \times 2} = \sqrt{36}\sqrt{2} = 6 \sqrt{2} $

We know that 72 is not a perfect square but 36 is a perfect square

Example 4

Simplify $ \sqrt{27} + \sqrt{12} - 2 \sqrt{75} + \sqrt{48} $

$\begin{aligned}

\sqrt{27} + \sqrt{12} - 2 \sqrt{75} + \sqrt{48} &= \sqrt{9 \times 3} + \sqrt{4 \times 3} - 2 \sqrt{5 \times 25} + \sqrt{16 \times 3} \\

&= \sqrt{9}\sqrt{3} +\sqrt{4}\sqrt{3} -2\sqrt{5}\sqrt{25} + \sqrt{16}\sqrt{3} \\

&= 3\sqrt{3} + 2\sqrt{3} -10\sqrt{5} + 4\sqrt{3} \\

&= 9\sqrt{3} -10\sqrt{5} \\

\end{aligned} $

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