In this article, we look at indices and surds and teach you how to use index notation and how to deal with surds.

In this article, we’re going to give you a run down of what indices and surds are and how to evaluate their operations, such as binomial products.

These are important subjects for Year 9 students. It is important that you have a thorough understanding of these concepts or you will really struggle in higher levels of Maths.

This article addresses the following syllabus outcomes:

Surds and Indices:

- Uses and interprets formal definitions and generalisations when explaining solutions and/or conjectures. MA5.3-1WM
- Generalises mathematical ideas and techniques to analyse and solve problems efficiently. MA5.3-2WM
- Uses deductive reasoning in presenting arguments and formal proofs. MA5.3-3WM
- Performs operations with surds and indices. MA5.3-6NA

This may seem like a long and convoluted set of statements. But what it is saying is really quite straightforward.

Essentially, these Outcomes state that in this unit you will learn that we have index laws for numbers and we have to adhere to these laws to solve problems.

This topic introduces you to operations involving indices and surds.

You are expected to evaluate more complex expressions involving negative, zero and fractional indices, including simplification and binomial products.

Often, students have trouble simplifying surds and using index laws correctly. They tend to become confused and forget the fundamentals of the indices and surds topic when they get overwhelmed with a combination of different techniques.

Common mistakes include:

- Mixing up addition and multiplication of indices
- Confusing what to do when there are different bases as opposed to different indices
- Uncertainty around the role of brackets in expressions
- Incorrect simplification of more complex expressions

Manipulating surds and indices can be challenging, especially when you have to combine multiple ideas from both topics to get the correct answer – however, not to worry!

This subject guide will point out common errors and help you better understand these new concepts.

There is some prior knowledge needed before starting this topic. You should have a basic understanding of surds.

For example, the root function – what is \(\sqrt{3}\)? How do you find the square root of a number?

And basic indices – what does \(2^{3}\)** **mean? What does \(3^{2}\) mean? How are they different?

Additionally, you should be familiar with elementary BODMAS (**Bracket**, **Of**, **Division**, **Multiplication**, **Addition** and **Subtraction**) operations and simple algebraic expressions.

If you aren’t already comfortable with the concepts above, it’s a good idea to revise them before starting this topic – it will be extremely useful.

Index notation is a concise way of writing the repeated multiplication of the same factor. The factor is called the ‘base’ and the number of times it is repeated it called the ‘index’ or ‘power’.

eg. \(x^{4} \rightarrow x\) is the base, and 4 is the index. It expands to \(x × x × x × x\)** **

There are special index laws that allow us to combine these sorts of expressions.

When multiplying terms with the same base, indices are added.

\(x^m × x^n=x^{(m+n)}\)

You **cannot** multiply different bases in the same way to create a single base with the sum of indices.

So, \(x^{m} × y^{n} ≠xy^{(m+n)}\)

However, if they are different bases with the same power, the bases can then be multiplied. The power (or index) stays the same.

\(x^{m} × y^{m}= (xy)^{m}\)

Examples

Let’s have a look at a few examples.

1. \(y^6 × y^2\)

This question is quite simple; since they have the same base, \(y\), the indices can be added to each other to get the answer \(y^8\).

2. \(a^2 × b × a^3 × b^4 × a^4\)

We are looking for terms with the same base. We see that there are 3 terms with base \(a\), and 2 terms with base \(b\).

This can essentially be rewritten as \(a^{2} \times a^{3} \times a^{4} \times b \times b^{4}\) to get \(a^{2+3+4} \times b^{1+4}\), which equals \(a^{9} \times b^{5}\) (can also be written as \(a^9b^5\))

3. \(m^{x-5} \times 5m^{8-x}\)

The index here only applies to \(m\), so the 5 (coefficient) is treated separately.

In most cases, coefficients are added to the front of the other expressions.

\(5 \times m^{x-5} × m^{8-x} =5 \times m^{x-5+8-x}=5m^{3}\)

Division works quite similarly to multiplication. When dividing terms with the same base, the indices are subtracted from one another.

\(x^{m} \div x^{n} = x^{m-n}\)

When different bases with the same indices are divided, the bases are divided by each other.

\(x^{m} \div y^{m}=(\frac{x}{y})^m\)

Have a go at some examples:

1. \(10y^{9} \div 2y\)

As mentioned before, we treat the coefficients separately. This expression becomes

\(

10 \div 2 \times y^{9} \div y \\

=5 \times y^{9-1} \\

=5y^{8} \\

\)

2. \(40h^{20} \div 5h^{3} \div 2h^{2} \)

Same method as above, but repeated once again.

Remember that we do not have to work on everything at the same time.

You should aim to work on each question step by step. It will be correct – as long as you follow the BODMAS operations.

\(8h^{20-3} \div 2h^{2} \\

=4h^{17} \div 2h^{2} \\

=2h^{15} \\

\)

3. \(30x^{11} y^{2} ÷2x^{3} × 4xy\)

Same method as above, but repeated once again.

Remember that we do not have to work on everything at the same time.

You should aim to work on each question step by step. It will be correct – as long as you follow the BODMAS operations.

\(

30x^{11} y^{2} ÷2x^{3} × 4xy \\

= 15x^{8} y^{2} × 4xy \\

=60x^{9} y^{3} \\

\)

When you have a power of a power, you multiply the two indices together as such:

\((x^{m})^{n}=x^{mn}\)

This is a common point of confusion among many students.

When you have an expression like this, you should remember to *multiply *the two powers together, instead of doing this:

Other important points to note:

\((ax^{m})^{n}=a^{n} x^{mn}\) \(a(x^{m} )^{n}=ax^{mn}\) \( (\frac{x}{y})^{n}= \frac{x^{n}}{y^{n}}\)

Let’s have a look at some examples.

1. \((2xy^{3})^{4}\)

All terms inside the bracket are raised to the power of 4;

\((2xy^{3} )^{4}

=2^{4}×x^{1×4}×y^{3×4}

=2x^{4} y^{12}

\)

2. \(5(m^{2} n^{5} )^{3}\)

Here, only the terms inside the bracket are raised to the power of 3. The 5 stays as it is.

Hence the answer will be:

\(5×m^{2×3}×n^{5×3}=5m^{6} n^{15}\)

3. \( (4x^{2} )^{3} × 2(x^{4} )^{2} \)

Every term in the first part is cubed, while the 2 is not squared in the second part. Therefore:

\((4x^2 )^3 × 2(x^4 )^2 \\

=4^{3} x^{6} × 2×x^{8} (\text{combining numbers and pronumerals}) \\

=8x^{14}

\)

When any expression is raised to the power of zero, the answer will always be 1.

\(x^{0}=1

(5x)^{0}=1

(6x+2)^{0}=1

\)

But how about \(5x^{0}\)? This is different to the second example above. In this case, only \(x\) is raised to the power of \(0\).

So, using our knowledge of index operations, we know that \(5x^{0}=5×1=5\)

Try this one:

\(7^{0} × (2^{2} )^{0}\)

Negative indices essentially give you the reciprocal of the expression. Many questions will ask you to write expressions with positive indices. This means that you would have to use the properties below:

\(x^{-m}=\frac{1}{x^{m}} \\

ax^{-m}=\frac{a}{x^{m}} \\

\)

If you have \(5x^{-2}\), remember that the negative is only applied to x here. You would have to treat the number and the index expression separately \( \rightarrow 5 × x^{-2}\), and combine them into one after reciprocating \(x \rightarrow 5× \frac{1}{x^{2}} = \frac{5}{x^{2}}\)

Try these:

1. \(\Big( \frac{2}{3} \Big) ^{-1}\)

2. \(5x^{-7}\)

You should get \(\frac{3}{2}\) and \(\frac{5}{x^{7}}\).

\(x^{\frac{1}{n}}= \sqrt[n]{x}\)

\(x^{\frac{m}{n}}= \sqrt[n]{x^{m}}\)

For fractional indices, the base is raised to the power of the numerator, and then rooted by the denominator. That is, if you have \(x^{\frac{2}{3}}\), \(x\) is first squared then put to the cube root.

\(x^{\frac{2}{3}}= \sqrt[3]{x^{2}}\)Take another example:

\(y^{\frac{-4}{5}}\)This combines both negative and fractional indices together. If the question asks you to write it with positive integer indices, you would first deal with the negative:

\(y^{\frac{-4}{5}}=\frac{1}{y^{\frac{4}{5}}}\)then:

\( \frac{1}{y^{\frac{4}{5}}} =\frac{1}{\sqrt[5]{y^{4}}}\)If you were to convert from surd form to index form,

\(\sqrt[4]{x^{5}}=x^{\frac{5}{4}}\)

Real numbers are any numbers that can be represented on a number line. There is also something called an “imaginary” number, but you don’t need to worry about it. This is learned in more advanced maths later on!

The real number system can be divided into two subgroups: rational and irrational.

Rational numbers are those that can be written as a fraction, \(\frac{p}{q}\), where \(p\) and \(q\) are whole numbers.

For example:

- Integers
- Fractions (proper, improper, mixed)
- Decimals (terminating and recurring)

Irrational numbers are, therefore, those that cannot be written as a fraction.

This includes:

- Roots (surds)
- Decimals that continue without terminating or repeating

As we’ve explored in the definitions above, most surds are irrational numbers and cannot be written as fractions. There are special ways of treating surds in operations:

\sqrt{a} × \sqrt{b}= \sqrt{ab} \\

\sqrt{a^2}=(\sqrt{a})^{2}=a \\

a\sqrt{b} × c\sqrt{d}= ac\sqrt{bd}\\

\)

When multiplying two surds, you first multiply the values under them, then put it under the same root.

\(\sqrt{3} × \sqrt{5}= \sqrt{3 × 5}=\sqrt{15}\)When multiplying surds that have an integer outside of it, multiply the integers and surds separately.

\(5\sqrt{2} × 3\sqrt{7}= 15\sqrt{14}\)

1. \(\sqrt{3} × \sqrt{12}\)

2. \(2\sqrt{5} × \sqrt{2}\)

3. \((4\sqrt{3})^{3}\)

Using the first rule, we multiply the two values under the surds first:

1. \(\sqrt{3×12}=\sqrt{36}\), and using the second rule, \(\sqrt{36}=\sqrt{6^{2}}=6\)

2.Using the third rule, the integers and surds are multiplied separately to get \(2\sqrt{10}\)

3.This can be treated as \((4\sqrt{3}) × (4\sqrt{3}) × (4\sqrt{3})\), which equals \(64 \sqrt{27}\).

For many of these operations, you will have to simplify the surd first. A surd is in its simplest form when the number under the square root sign is not a factor of a perfect square.

For example, \(\sqrt{8}\) is not a simplified surd as \(4 = 2^2\) is a factor of 8. However, \(\sqrt{10}\), is a simplified surd as it does not have any factors that are perfect squares.

Students often have trouble understanding this.

Here is a breakdown of surd simplification. Let’s work with the example \(\sqrt{128}\).

Look for a square number that is a factor of the integer inside the root. If you can’t spot one immediately, start with small numbers such as \(4\) or \(9\) – the simplification process can be done multiple times after this.

For \(\sqrt{128}\), we can see that \(128\) is divisible by \(64\) – a perfect square!

Write the inside of the surd as the product of the perfect square and its other factor.

\(\sqrt{128} = \sqrt{64×2}\)Using Rule 1, we know that the reverse is true as well. Hence we are able to split the number into two different surds.

\(\sqrt{128}= \sqrt{64×2}= \sqrt{64}×\sqrt{2}\)Here is where the perfect square is used. We know that the root of a perfect square simplifies down to an integer. Therefore, we can just eliminate the square root and multiply the resulting integer with our other root.

\(\sqrt{64} × \sqrt{2}=8×\sqrt{2} =8\sqrt{2}\)It is useful to become familiar with this process – it allows you to deal with a much wider range of surds.

When dividing two surds, you divide the values inside them first, then put them under the same root.

\(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\)

\(\frac{a\sqrt{b}}{c\sqrt{d}}=\frac{a}{c}×\sqrt{\frac{b}{d}}\)

Have a look at the following:

1. \(\sqrt{\frac{24}{8}}\)

The fraction inside the surd can be simplified first to \(3\). The answer is then just \(\sqrt{3}\).

2. \(\frac{3\sqrt{8}}{\sqrt{2}}\)

At first glance, it seems that we aren’t able to solve this. But – remember we are able to simplify! Using what we know, \(\sqrt{8}\) can be written as \(2\sqrt{2}\).

Then the fraction becomes \(\frac{3×2\sqrt{2}}{\sqrt{2}}\). We can cancel out the \(\sqrt{2}\) to give \(6\).

Using the same method as above, simplify:

\(\frac{\sqrt{99}}{4\sqrt{3}}\).

You’ve been introduced to how addition and subtraction of algebraic terms works; a similar idea applies to addition and subtraction of surds. Like \(x\)’s and \(y\)’s, surds can only be added or subtracted from each other if they are ‘like’ – that is, if they have the same value under the surd.

\(\sqrt{3}+\sqrt{5}\) is NOT equal to \(8\), as \(3\) and \(5\) are not “like” terms – they are completely different surds.

\(2\sqrt{3}\) essentially means there are “\(2\) of \(\sqrt{3}\)‘s” – in other words \( \sqrt{3}+\sqrt{3}\).

That’s why, if you were to evaluate \(2\sqrt{3}+ \sqrt{3}\), it would be \(3\sqrt{3}\); this is similar to saying \(2x+x=3x\)

Similarly, \(6\sqrt{5}+2\sqrt{5}=8\sqrt{5}\) (think: \(6x+2x=8x\))

A similar concept applies to subtraction; only “like” terms can be subtracted.

What is \(8\sqrt{5}-12\sqrt{5}-7\sqrt{5}+ 3\sqrt{10}\)?

Here we see that there are two different terms; \(\sqrt{5}\) and \(\sqrt{10}\), so these have to be treated separately to each other. We can think of the expression as something like \(8x-12x-7x+3y\) – how would you solve this?

Now solve the above in the same way – it just becomes \(-11\sqrt{5}+3\sqrt{10}\).

Recall our definition of a rational number – it is one that can be written as a fraction. Now, when we have a denominator in the form of a root (eg. \(\sqrt{a}\), where a is a rational number, and hence irrational), we want to make it a rational number (hence the term ‘rationalise’).

Think about what we can multiply \(\sqrt{a}\) by to make it rational: \(\sqrt{a}\) itself! Doing this turns it into \(a\), which we know is a rational number.

Hence, to rationalise a single term denominator of a fraction, simply multiply both numerator and denominator by the surd.

Rationalise the denominator for \(\frac{2}{\sqrt{3}}\)

Multiply both numerator and denominator by the surd in the denominator

\(\frac{6}{\sqrt{3}} ×\frac{\sqrt{3}}{\sqrt{3}}\)Since \(\frac{\sqrt{3}}{\sqrt{3}}=1\), we’re not changing the fraction in any way; we are essentially just multiplying it by \(1\).

Simplify the fraction if possible.

\(\frac{6}{\sqrt{3}} ×\frac{\sqrt{3}}{\sqrt{3}}=\frac{6\sqrt{3}}{3}\)The \(6\) in the numerator and \(3\) in the denominator can cancel out to give our result, \(2\sqrt{3}\).

Another example:

\(\frac{2\sqrt{5}-3\sqrt{2}}{\sqrt{5}}\)

\(= \frac{\sqrt{5} (2\sqrt{5}-3\sqrt{2})}{\sqrt{5}×\sqrt{5}}\)

\(=\frac{10-3\sqrt{10}}{5}\)

These are the concepts we’ve covered in this guide:

\(x^m × x^n=x^{m+n}\)

\(x^m × y^m= (xy)^m\)

\(x^m ÷ x^n=x^{(m-n)}\)

\(x^m ÷ y^m= \Big( \frac{x}{y} \Big) ^{m}\)

\(\sqrt{a} × \sqrt{b}= \sqrt{ab}\)

\(\sqrt{a^{2}}=(\sqrt{a})^{2}=a\)

\(a\sqrt{b} × c\sqrt{d} = ac\sqrt{bd}\)

You should be able to remember all of these. This sounds daunting. However, with enough practice, you will be able to know them off by heart and be readily able to apply them in your calculations. Just practise by doing as many questions as you can!

If you encounter a question that looks especially complex and difficult, remember to recall your basic understanding of the surds and indices rules and tackle the question step-by-step. If you are able to break down the question into chunks, it’s a lot easier to think clearly and to manage each chunk individually. It’ll eventually boil down to a simple expression!

Take note of the common mistakes that students make and remember this while doing questions.

In our next article, we discuss surface are to prepare you for working out volumes.

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