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Revision:OCR Core 1 - Surds

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 1 - Surds

1 2. Surds
2 Also See
3 Comments

2. Surds

Basic Principles

Consider the set of real numbers ( $\mathbb{R}$ ) and all the different categories that these numbers can fit into. Some of the numbers are odd, some are even, some are negative, some are positive, some are integers, some are not, etc. these are all ways in which one can further categorise the real numbers. What about rational numbers? Rational numbers (as the name might suggest) are merely those numbers which can be expressed in ratio. For instance:

$\frac{1}{1},\ \frac{1}{2},\ \frac{2}{3},\ \frac{1729}{1728}$

There are numbers which cannot be expressed as a fraction, such as:

$\pi,\ e,\ \phi$

These numbers are irrational (they cannot be expressed as a ratio of two numbers).

Numbers such as $\sqrt{2}$ are referred to as surds. Often surds can be obtained when applying Pythagoras' Theorem, or using the Quadratic Equation formula. It is useful to be able to simplify surds, however there are rules to follow when attempting to simplify them.

x, y > 0

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

$\sqrt{ \frac{x}{y}} = \frac{ \sqrt{x}}{\sqrt{y}}$

To explain this, one can simply state the following:

$\right( \sqrt{x} \times \sqrt{y} \left) ^{2} = \sqrt{x} \times \sqrt{y} \times \sqrt{x} \times \sqrt{y} = x \times y$

Hence:

$\sqrt{x} \times \sqrt{y} = \sqrt{\right( \sqrt{x} \times \sqrt{y} \left) ^{2}} = \sqrt{x \times y} = \sqrt{xy}$

For the second rule:

$\left( \frac{ \sqrt{x}}{\sqrt{y}} \right) ^{2} = \frac{x}{y}$

Hence:

$\frac{ \sqrt{x}}{\sqrt{y}} = \sqrt{ \frac{x}{y} }$

These rules can be used to simplify surds.

Example

1. Simplify fully $\sqrt{125}$ .

First one must look for factors of 125:

$\sqrt{125} = \sqrt{5 \times 25} = \sqrt{5} \times \sqrt{25}$

Notice that 25 is a perfect square, hence:

$\sqrt{125} = 5 \sqrt{5}$

2. Simplify fully $\sqrt{175} + \sqrt{343}$ .

One might notice that $\sqrt{343} = \sqrt{7 \times 7 \times 7} = \sqrt{7 \times 49} = \sqrt{7} \times \sqrt{49} = 7 \sqrt{7}$

Now one can search for factors. It is a good idea to assume that there is going to be a factor of 7 in the surd that one is adding (in this example) as it is likely that the question will have an answer expressed in the form $a \sqrt{b}$ .

Hence:

$\sqrt{175} = \sqrt{7 \times 25} = \sqrt{7} \times \sqrt{25} = 5 \sqrt{7}$

Hence:

$\sqrt{175} + \sqrt{343} = 5 \sqrt{7} + 7 \sqrt{7} = 12 \sqrt{7}$

Do not make the common mistake of adding the square roots:

$\sqrt{x} + \sqrt{y} \not= \sqrt{x + y}$

This is a common mistake.

One can apply these principles to roots of greater power.

Example

1. Simplify fully $\sqrt[5]{12500}$ .

One uses the usual technique:

$\sqrt[5]{12500} = \sqrt[5]{3125 \times 4} = \sqrt[5]{4} \times \sqrt[5]{3125} = \sqrt[5]{4} \times 5 = 5 \times \sqrt[5]{4}$

The difference of two squares

It is known that:

$x^{2} - y^{2} = (x + y)(x - y)$

Now consider:

$\left( \sqrt{x} \right) ^{2} - \left( \sqrt{y} \right) ^{2} = \left( \sqrt{x} + \sqrt{y} \right) \left( \sqrt{x} - \sqrt{y} \right)$

This can be a helpful piece of knowledge when rationalising denominators.

Rationalising Denominators

It is conventional to leave the denominator in a rational state, thus changing a fraction such that there is no surd on the denominator. The technique used is the same as one would use to change the denominator if it was rational.

Example

1. Rationalise the denominator of the following expression:

$\frac{3}{ \sqrt{5} }$ .

This is a simple example. The basic rule is to ensure that one only multiples by 1 (as this will produce an identical expression), and this means that when one multiplies the expression by a fraction, it is one whose numerator is identical to the denominator (as a number divided by itself is 1).

So, one would do the following:

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

It is merely a matter of multiplying by a surd that will cause the denominator to become rational (this can be the surd that is on the denominator, as with the previous example).

Generalisation

Rationalise the denominator of:

$\frac{m}{n \sqrt{x} }$

It is simple to see that multiplying by $\frac{ \sqrt{x} }{ \sqrt{x} }$ will simplify this.

Hence:

$\frac{m}{n \sqrt{x} } \times \frac{ \sqrt{x} }{ \sqrt{x} } = \frac{m \sqrt{x} }{nx}$

Questions can involve a slightly different format (and hence the difference of two squares is required).

Example

1. Rationalise the denominator of the following expression:

$\frac{1}{3 + \sqrt{5}}$

One cannot use the same method as before, as when one multiplied the denominator by $\sqrt{5}$ one would merely obtain a term with a surd in it (namely $3 \sqrt{5}$ .

One must use the difference of two squares.

Consider the denominator: $3 + \sqrt{5}$ . This is the same as one of the brackets of the difference of two squares, and therefore one must multiply by the other:

$\frac{1}{3 + \sqrt{5}} = \frac{1}{3 + \sqrt{5}} \times \frac{3 - \sqrt{5}}{3 - \sqrt{5}} = \frac{3 - \sqrt{5}}{ \left( 3 + \sqrt{5} \right) \left( 3 - \sqrt{5} \right)} = \frac{3 - \sqrt{5}}{9 - 5} = \frac{3 - \sqrt{5}}{4}$

This can be generalised.

Generalisation

Rationalise the denominator of the following expression:

$\frac{n}{m + \sqrt{x}}$

First one would begin as above (in the numerical example):

$\frac{n}{m + \sqrt{x}} = \frac{n}{m + \sqrt{x}} \times \frac{m - \sqrt{x}}{m - \sqrt{x}} = \frac{nm + n \sqrt{x}}{\left( m + \sqrt{x} \right) \left( m - \sqrt{x} \right)} = \frac{nm + n \sqrt{x}}{m^{2} - x}$

It is noteworthy that the question could involve roots other than the square root, but similar methods can be applied. The general principle is to ensure that one multiples the whole denominator and the whole numerator. It is a common error to multiply only the surd and forget about the rational term.