GCSE Tutoring Programme

"Our chosen students improved 1.19 of a grade on average - 0.45 more than those who didn't have the tutoring."

In order to access this I need to be confident with:

Square numbers and square roots Cube numbers and cube roots Standard form / scientific notation BIDMAS Fractions Negative numbers DecimalsThis topic is relevant for:

Here we will learn about fractional powers, including what they are, how to simplify and evaluate them, and how to combine them with other laws of exponents.

There are also powers and roots worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if youβre still stuck.

**Fractional powers** are a type of index that represents the nth root of a number.

In general, the rule for fractional exponents is x^{\frac{m}{n}}=\sqrt[n]{x^{m}} .

Where m and n are constants and n β 0 .

For example,

9^{\frac{1}{2}}=\sqrt{9}=3 m=1, n=2, and x=9 , so we are calculating the second root, or the **square root** of 9 which is equal to 3 .

In order to use fractional powers:

**Simplify any powers using laws of indices.****Evaluate or solve (if required).**

Get your free fractional powers worksheet of 20+ powers and roots questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free fractional powers worksheet of 20+ powers and roots questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**Fractional powers**Β is part of our series of lessons to support revision onΒ **powers and roots**. You may find it helpful to start with the mainΒ powers and rootsΒ lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

Simplify and evaluate the expression

9^{\frac{1}{2}} \times\ 9 .

**Simplify any powers using laws of indices.**

As x^{a} \times x^{b}=x^{a+b} , and 9=9^{1} , we have

9^{\frac{1}{2}}=9^{\frac{1}{2}+1}=9^{1.5}=9^{\frac{3}{2}} .

2**Evaluate or solve (if required).**

Evaluating the expression, we get

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

So 9^{\frac{1}{2}}\times\ {9}=27 .

Write the following expression in its simplest form

(xy)^{a}\div\ {(x)^\frac{a}{2}} .

**Simplify any powers using laws of indices.**

As we have multiple variables in the first expression raised to the power of a , letβs remove the brackets so we know the power of each variable separately,

(xy)^{a}=x^{a}\times y^{a} .

We can then divide x^{a} \times y^{a} by (x)^\frac{a}{2} to get

x^{a}\times\ {y^{a}}\div{(x)^\frac{a}{2}}=x^{a-\frac{a}{2}}\times\ {y^{a}}=x^{\frac{a}{2}}\times\ {y^{a}} .

(Remember we can only simplify powers if the base is the same.)

So we now have x^{\frac{a}{2}}\times\ {y^{a}} .

**Evaluate or solve (if required).**

The question asks us to simplify the expression and so the solution is,

x^{\frac{a}{2}}y^{a} .

Solve the inequality

64^{\frac{1}{3}}\geq x^2 .

**Simplify any powers using laws of indices.**

Looking at the left hand side of the inequality sign, we have 64 being raised to the power of \frac{1}{3} and so we can write this as

64^{\frac{1}{3}}=\sqrt[3]{64}=4 .

So we now have the inequality 4\geq x^2 .

**Evaluate or solve (if required).**

We now need to solve this quadratic inequality. We can do this by rewriting it as

x^2-4\leq 0This can be factorised to (x-2)(x+2)\leq0 ,

Giving

-2\leq x\leq 2 .

This is the solution.

Evaluate the following expression,

216^{-\frac{2}{3}} .

**Simplify any powers using laws of indices.**

As the power is a fraction and a negative number, we are going to simplify the base by looking at the denominator of the power first.

As the denominator of the fraction is the nth root of the base number we can start to simplify the expression,

216^{-\frac{2}{3}}=(216^{\frac{1}{3}})^{-2}=(\sqrt[3]{216})^{-2}=6^{-2} .

(The power of \frac{1}{3} is exactly the same as the cube root.)

6^{-2}=\frac{1}{6^{2}}**Evaluate or solve (if required).**

The expression \frac{1}{6^{2}} is the simplified form of the expression, but we can now evaluate the expression as we know that 6^{2}=36 so \frac{1}{6^{2}}=\frac{1}{36} .

So the solution is \frac{1}{36} .

Simplify fully,

(2\frac{113}{256})^{\frac{1}{4}} .

Write your answer as a mixed number.

**Simplify any powers using laws of indices.**

As the base number is a mixed number, letβs convert it to an improper fraction first,

2\frac{113}{256}=\frac{2\times{256}+113}{256}=\frac{625}{256} .

So we now have (\frac{625}{256})^{\frac{1}{4}} .

As the fraction is raised to the power of \frac{1}{4} , both the numerator and the denominator can be raised to the power of \frac{1}{4} .

(\frac{625}{256})^{\frac{1}{4}}=\frac{625^{\frac{1}{4}}}{256^{\frac{1}{4}}}**Evaluate or solve (if required).**

As the index is equal to \frac{1}{4} , we can calculate the 4^{th} root of each part of the fraction.

The fourth root of 625 is 5 .

The fourth root of 256 is 4 .

So, \frac{\sqrt[4]{625}}{\sqrt[4]{256}}=\frac{5}{4}=1\frac{1}{4} .

So, (2\frac{113}{256})^{\frac{1}{4}}=1\frac{1}{4} .

Simplify fully,

(8^{-10})^{\frac{1}{5}} .

Write your answer in the form \frac{1}{2^n} .

**Simplify any powers using laws of indices.**

The law of indices we are going to use involves parentheses, namely (x^{a})^{b}=x^{a\times b} , we get

(8^{-10})^{\frac{1}{5}}=8^{-10\times{\frac{1}{5}}}=8^{-2} .

8^{-2}=\frac{1}{8^{2}} .

**Evaluate or solve (if required).**

The question does not want us to state the value of \frac{1}{8^{2}} .

Instead, the question asks for the solution to be written form \frac{1}{2^n} , so we need to change the expression so that the base number is 2 , instead of 8 .

As 8=2^{3} , we have \frac{1}{8^{2}}=\frac{1}{(2^{3})^{2}}=\frac{1}{2^{3\times{2}}}=\frac{1}{2^{6}}

So the solution is \frac{1}{2^{6}} .

**Fractional power and fraction of amounts**

A common error is to mistake a fractional power with multiplying by a fraction.

For example,

27^{\frac{1}{3}}=27\times{\frac{1}{3}}=9 is incorrect.

This is because the denominator of the fraction is the nth root of the base.

The correct answer is

27^{\frac{1}{3}}=\sqrt[3]{27}=3 .

**BIDMAS**

When raising a base to a power, the power is associated with a variable or an expression.

For example, letβs look at 3x^{-2} .

Using BIDMAS we can see that x is being raised to the power of -2 , and then is multiplied by 3 to get the answer \frac{3}{x^{2}}

However a common error is to calculate 3x raised to the power of -2 , giving the answer \frac{1}{9x^{2}} which is incorrect. If this was the case, the question would be written as (3x)^{-2}.

**Incorrect application of the exponent**

A similar circumstance is applied to fractions that are raised to a power.

For example, letβs look at (\frac{5}{2})^{-3} .

The correct application of the laws of indices would give a **correct **answer of (\frac{5}{2})^{-3}=(\frac{2}{5})^{3}=\frac{8}{125} .

However a common error would be to apply the -3 only to the numerator, leaving the denominator unchanged. This would give the **incorrect **answer of (\frac{5}{2})^{-3}=\frac{1}{125}\times\frac{1}{2}=\frac{1}{250} .

1. Evaluate

4^{\frac{1}{2}}\times\ {4^{\frac{3}{2}}} .

4^{\frac{3}{4}}

12

8

16

4^{\frac{1}{2}}\times\ {4^{\frac{3}{2}}}=4^{\frac{1}{2}+\frac{3}{2}}=4^{\frac{4}{2}}=4^{2}=16

2. Simplify fully,

(x^{a}y)^{\frac{1}{a}} .

\frac{x^{a}}{y^{a}}

(xy)^{a}

xy^{\frac{1}{a}}

\frac{1}{ax^{a}y}

(x^{a}y)^{\frac{1}{a}}=x^{a\times{\frac{1}{a}}}\times\ {y^{\frac{1}{a}}}=x\times\ {y^{\frac{1}{a}}}=xy^{\frac{1}{a}}

3. State the range of values for the following inequality,

8^{\frac{2}{3}}<\frac{x}{5} .

x>20

x<20

x>30

x<30

8^{\frac{2}{3}}=(\sqrt[3]{8})^{2}=2^{2}=4

4<\frac{x}{5}

20<x so x>20

4. Evaluate

625^{-\frac{3}{4}} .

-125

125

\frac{1}{125}

-\frac{1}{125}

625^{-\frac{3}{4}}=\frac{1}{625^{\frac{3}{4}}}=\frac{1}{(\sqrt[4]{625})^{3}}=\frac{1}{5^{3}}=\frac{1}{125}

5. Evaluate

(\frac{1024}{32})^{\frac{1}{5}} .

\frac{1}{8}

2

6.4

160

(\frac{1024}{32})^{\frac{1}{5}}=\frac{\sqrt[5]{1024}}{\sqrt[5]{32}}=\frac{4}{2}=2

6. Simplify the expression below. Write your answer in the form 3^{n}.

(9^{-3})^{\frac{1}{2}}

-\frac{3}{2}

3^{-3}

\frac{1}{27}

27

(9^{-3})^{\frac{1}{2}}=(9^{\frac{1}{2}})^{-3}=3^{-3}

1. Solve the inequality 3\div \sqrt[3]{x}<x^{\frac{2}{3}} .

**(4 marks)**

Show answer

2. (a) Evaluate 16^{-1\frac{1}{2}} .

(b) Simplify a^{\frac{1}{2}}\times{a^{\frac{1}{2}}} .

**(5 marks)**

Show answer

(a)

-1\frac{1}{2}=-\frac{3}{2}**(1)**

**(1)**

**(1)**

(b)

a^{\frac{1}{2}}\times{a^{\frac{1}{2}}}=a^{\frac{1}{2}+\frac{1}{2}}**(1)**

**(1)**

3. Express the following as a single power of 2 ,

4^{4}\times{\sqrt{8}} .

**(6 marks)**

Show answer

4^{4}=(2^{2})^{4}

**(1)**

**(1)**

**(1)**

**(1)**

**(1)**

**(1)**

You have now learned how to:

- Consolidate your numerical and mathematical capability from key stage 3 and extend your understanding of the number system to fractional indices
- Calculate with roots, and fractional indices

Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.

Find out more about our GCSE maths tuition programme.