Rationalising the denominator - surds - Numbas at mathcentre.ac.uk

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Christian Lawson-Perfect	said	Ready to use	7 years, 8 months ago
Lauren Richards	said	Needs to be tested	7 years, 8 months ago
Aiden McCall	said	Has some problems	7 years, 9 months ago

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Elliott Fletcher 7 years, 8 months ago

Published this.

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Christian Lawson-Perfect 7 years, 8 months ago

Gave some feedback: Ready to use

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Christian Lawson-Perfect 7 years, 8 months ago

Saved a checkpoint:

Good question!

It seems you didn't know about display mode maths while writing the advice: enclose maths in \[ and \] instead of dollar signs to place it at the centre of its own line with a good margin.

Restore this checkpoint

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Lauren Richards 7 years, 8 months ago

Gave some feedback: Needs to be tested

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Vicky Hall commented 7 years, 8 months ago

I've fixed the variables so that part e) will simplify if it needs to in the advice. I also changed the variables slightly in part b) to avoid ever having square numbers under the roots.

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Lauren Richards commented 7 years, 9 months ago

I've randomised parts e) and f) now too but having problems with getting e) to finally simplify. Vicky would you mind having a look at this please? I've written THIS NEEDS TO SIMPLIFY next to the issue!

As a result of not getting it simplified, I haven't set the correct answers yet.

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Lauren Richards 7 years, 9 months ago

Gave some feedback: Has some problems

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Lauren Richards commented 7 years, 9 months ago

I've randomised parts a) to d).

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Lauren Richards 7 years, 9 months ago

Gave some feedback: Needs to be tested

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Aiden McCall commented 7 years, 9 months ago

Advice:

a) The advice for part a does not match the question. You could also possibly show another method of multiplying by the denominator to realise the bottom; I could be overcomplicating it with this alternative method, so it is up to you.
b) I would put the 12 in '12 is the denominator' in math mode. Not sure what the last line means I think it is suppose to read $\frac{\sqrt{ab}}{1}$ but reads as $\frac{\sqrt{ab}}{[1,1,1,1,1]}$ .
c) Is good
d) There could be some confusing with the wording, "this is the denominator with a changed sign." It could be interpretted as the whole denominator changing signs; I would possibly change it to the rational term sign changed, or something to that effect.

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Aiden McCall commented 7 years, 9 months ago

a) On my first attempt of the question, I had received the same question for parts i) & ii). This can be fixed with except in your random variables so the same two values do not occur.

b) Everything fine.

c) I'm not sure if you need a semi-colon if you have an 'and' as this does not create two main clauses. When testing this question it always came up as the same values also. It would be more beneficial for someone practising to have a randomising example everytime.

d) Personally I think the question could be presented as "Express [equation] in the form $a \sqrt{b} + m$ where a, b and m are integers or something of that nature. I also think the answer is incorrect and not displayed as one of the options. $\frac{4}{(\sqrt{3}-2)}=-8-4\sqrt{3}&. The question is also not randomised so comes up the same each time.

e),f) both not randomised so the student cannot try another like this and get different values.

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Aiden McCall 7 years, 9 months ago

Gave some feedback: Has some problems

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Lauren Richards 7 years, 9 months ago

Gave some feedback: Needs to be tested

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Lauren Richards 7 years, 9 months ago

Gave some feedback: Has some problems

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Lauren Richards commented 7 years, 9 months ago

I have put all of the questions about rationalising the denominator together.

I didn't know how to enact your feedback for part e) in the previous question (which is now part b) in this question). It was "I got $\frac{\sqrt{6}}{\sqrt{3}}$ which it expected me to write as $\frac{\sqrt{18}}{3}$ . I'd write that as just $\sqrt{2}$ . Can you set it up so there's only one way of writing it? Making sure the top and bottom of the original fraction are coprime might do it."

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Lauren Richards 7 years, 9 months ago

Gave some feedback: Needs to be tested

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Lauren Richards 7 years, 9 months ago

Created this.

Name	Status	Author	Last Modified
Rationalising the denominator - surds	Ready to use	Lauren Richards	18/06/2024 11:49
Simon's copy of Rationalising the denominator - surds	draft	Simon Thomas	06/06/2019 11:00
Irrationale Zahlen - Nenner rational machen	Ready to use	Ulrich Görtz	20/10/2020 21:27

There are 8 other versions that do you not have access to.

Question statement

Rationalising the denominator means removing any surds on the denominator.

Try the following questions to practise rationalising with expressions in different forms.

Give any introductory information the student needs.

			Name	Type	Generated Value

			square_nums	list	[ 4, 9, 25, 16 ]
			prime_nums	list	[ 11, 7, 5, 3, 13 ]

			Name	Type	Generated Value

n	integer	2
m	integer	4
n_lcm	number	44
m_lcm	number	20
a	list	[ 11, 7, 11, 11, 11 ]
b	list	[ 5, 3, 3, 3, 3 ]
num	number	880

			Name	Type	Generated Value

c	integer	5
d	integer	8
f	integer	5
df	integer	40
gcd_cdf	number	5
c_coprime	number	1
df_coprime	number	8

			Name	Type	Generated Value

g	number	-4
h	number	15
j	integer	4
l	number	16
k	integer	4
o	number	4

			Name	Type	Generated Value

p	integer	3
r	integer	1
s	integer	4
r_coprime	number	1
s_coprime	number	4
p_p1	integer	1
qwerty	number	0
one	number	1
two	number	4

			Name	Type	Generated Value

t	integer	3
u	integer	7
ghi	number	2
denom_simp	number	1
gcd_num	number	2
gcd_frac	number	2
num_simp_1	number	8
num_simp_2	number	3

			Name	Type	Generated Value

sqrt

Definition of variable sqrt is empty.

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
shuffle([4,9,16,25])

JME function reference

Description

List of square numbers.

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `list`

[ 4, 9, 25, 16 ]

→ Used by:

This variable doesn't seem to be used anywhere.

Parts

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Part b) Gap-fill
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Part c) Gap-fill
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Part e) Gap-fill
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Part f) Choose one from a list

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Rationalise the denominator of the following surds to simplify them down to their integer value.

$\displaystyle\frac{\sqrt{\simplify{{square_nums[0]}{prime_nums[0]}}}}{\sqrt{\var{prime_nums[0]}}}=$ Gap 0.

ii)

$\displaystyle\frac{\sqrt{\simplify{{square_nums[1]}{prime_nums[1]}}}}{\sqrt{\var{prime_nums[1]}}}=$ Gap 1.

Advice

a)

The rule that should be used is $\displaystyle\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ .

$\displaystyle\frac{\sqrt{\simplify{{square_nums[0]}{prime_nums[0]}}}}{\sqrt{\var{prime_nums[0]}}} = \sqrt{\frac{{\simplify{{square_nums[0]}{prime_nums[0]}}}}{{\var{prime_nums[0]}}}} = \sqrt{\var{square_nums[0]}}= \simplify{{sqrt(square_nums[0])}}$ ; and

ii)

$\displaystyle\frac{\sqrt{\simplify{{square_nums[1]}{prime_nums[1]}}}}{\sqrt{\var{prime_nums[1]}}} = \sqrt{\frac{{\simplify{{square_nums[1]}{prime_nums[1]}}}}{{\var{prime_nums[1]}}}} = \sqrt{\var{square_nums[1]}} = \simplify{{sqrt(square_nums[1])}}$ .

b)

To rationalise the denominator, you have to use the rule $\displaystyle\sqrt{a}\times\sqrt{a}=a$ .

You are asked to rationalise the denominator of the expression $\displaystyle\frac{\sqrt{\var{n_lcm}}}{\sqrt{\var{m_lcm}}}$ .

As ${\sqrt{\var{m_lcm}}}$ is the denominator, we must multiply the whole fraction by ${\sqrt{\var{m_lcm}}}$ .

$\frac{\sqrt{\var{n_lcm}}}{\sqrt{\var{m_lcm}}}\times\frac{\sqrt{\var{m_lcm}}}{\sqrt{\var{m_lcm}}}$

By using the rule $\sqrt{a}\times\sqrt{b}$ = $\sqrt{ab}$ , we can get the numerator as $\sqrt{\var{num}}$ .

The denominator is ${\sqrt{\var{m_lcm}}}\times{\sqrt{\var{m_lcm}}} = {\var{m_lcm}}$ .

So the whole expression is

$\frac{\sqrt{\var{num}}}{\var{m_lcm}}$

Finally, we can cancel down by finding the greatest common divisor of the top and bottom to obtain

$\frac{\sqrt{\var{a[0]*b[0]}}}{\var{b[0]}}$

c)

The first thing to do in this question is multiply the whole fraction by the surd on the denominator, $\sqrt{\var{f}}$ . Using the rule $\displaystyle\sqrt{a}\times\sqrt{a}=a$ , the new value of the denominator is $\var{d}\times\var{f}=\var{df}$ .

$\displaystyle\frac{\var{c}}{\var{d}\sqrt{\var{f}}}\times\simplify[all,!simplifyFractions,!sqrtDivision]{sqrt({f})/(sqrt({f}))}={\frac{\simplify{{c}*sqrt({f})}}{{(\var{d}\times\var{f})}}}=\frac{\simplify{{c}*sqrt({f})}}{\var{d*f}}$ .

Make sure to simplify your fractions down to their simplest form. The fraction can be cancelled down using the highest common divisor, $\var{gcd_cdf}$ , to give a final answer of

$\frac{\simplify{{c}*sqrt({f})}}{\var{d*f}}=\frac{\simplify{{c_coprime}*sqrt({f})}}{\var{df_coprime}}$

We are asked to find an answer in the form $\frac{a \sqrt{\var{f}}}{b}$ , so $a = \var{c_coprime}$ and $b = \var{df_coprime}$ .

d)

When rationalising an expression with a compound denominator, like $\displaystyle\frac{\var{g}}{\sqrt{\var{h}}+\var{j}}$ , we must multiply the fraction by the conjugate of the denominator.

The conjugate of $(\sqrt{a}+b)$ is $(\sqrt{a}-b)$ , and therefore the conjugate of $(\sqrt{\var{h}}+\var{j})$ is $(\sqrt{\var{h}}-\var{j})$ .

$\frac{\var{g}}{\sqrt{\var{h}}+\var{j}}\times\frac{(\sqrt{\var{h}}-\var{j})}{(\sqrt{\var{h}}-\var{j})}=\frac{\var{g}(\sqrt{\var{h}}-\var{j})}{(\sqrt{\var{h}}+\var{j})(\sqrt{\var{h}}-\var{j})}\text{.}$

From this point, we need to multiply out the brackets on the numerator and denominator, using the rule $\displaystyle\sqrt{a}\times\sqrt{a}=a$ .

$\displaystyle\frac{\var{g}(\sqrt{\var{h}}-\var{j})}{(\sqrt{\var{h}}+\var{j})(\sqrt{\var{h}}-\var{j})}==\frac{\simplify[all,!collectNumbers,!noLeadingMinus]{({-{g}*{j}})+{g}}\sqrt{\var{h}}}{\var{h}{-\simplify{{j}*sqrt({h})}}+\simplify{{j}*sqrt({h})}-{{\var{j^2}}}}$ .

The two $\simplify{{j}*sqrt({h})}$ terms will cancel on the denominator, so that you are left with

$\frac{ \simplify{ {g*-j} + {g}*sqrt({h})} }{\var{h-j^2}}\text{,}$

which simplifies to

$\var{o}\sqrt{\var{h}}-{\var{l}}\text{.}$

e)

$\displaystyle\frac{\var{t}-\sqrt{\var{u}}}{\var{t}+\sqrt{\var{u}}}$ =

To rationalise the denominator of this fraction, multiply the whole fraction by the conjugate of the denominator, multiply out the brackets and collect like terms.

$\displaystyle\frac{\var{t}-\sqrt{\var{u}}}{\var{t}+\sqrt{\var{u}}}\times\frac{\var{t}-\sqrt{\var{u}}}{\var{t}-\sqrt{\var{u}}}=\frac{(\var{t}-\sqrt{\var{u}})(\var{t}-\sqrt{\var{u}})}{(\var{t}+\sqrt{\var{u}})(\var{t}-\sqrt{\var{u}})}=\frac{\var{t^2}-\var{2t}\sqrt{\var{u}}+\var{u}}{\var{t^2}-\var{u}}$ .

$\displaystyle\frac{\var{t^2}-\var{2t}\sqrt{\var{u}}+\var{u}}{\var{t^2}-\var{u}}=\frac{\var{t^2+u}-\var{2t}{\sqrt{\var{u}}}}{\var{t^2-u}}$ .

Dividing all terms by their highest common divisor, ${\var{gcd_frac}}$ , gives

$\displaystyle\frac{\var{num_simp_1}-\simplify{{num_simp_2}*sqrt({u})}}{\var{denom_simp}}$ .

f)

To add $\displaystyle\frac{\var{p}}{\sqrt{\var{p}}+\sqrt{\var{p^3}}}+\frac{\var{r_coprime}}{\var{s_coprime}}$ you must put both terms over the same denominator.

It is good to notice that $\sqrt{\var{p^3}}$ can be rewritten as $\var{p}\sqrt{\var{p}}$ , meaning we can rewrite the first fraction as

$\frac{\var{p}}{\sqrt{\var{p}}+\var{p}\sqrt{\var{p}}} \text{.}$

This can be simplified further by collecting like terms to obtain

$\frac{\var{p}}{\var{p+1}\sqrt{\var{p}}} \text{.}$

Therefore, the expression is now $\displaystyle\frac{\var{p}}{{\var{p+1}}\sqrt{\var{p}}}+\frac{\var{r_coprime}}{\var{s_coprime}}$ .

In order to complete the addition, the second fraction also has to have the same denominator as the first fraction. We have to multiply it by $\simplify{{p_p1}sqrt({p})}$ in order to get common denominators across both fractions.

$\displaystyle\frac{\var{r_coprime}}{\var{s_coprime}}\times\simplify[all,!simplifyFractions,!sqrtDivision]{{p_p1}*sqrt({p})/({p_p1}*sqrt({p}))}=\frac{\simplify{{one}*sqrt({p})}}{\var{two}\sqrt{\var{p}}}$ ,

Both parts of the expression now have the same denominator and we can add them.

$\frac{\var{p}}{\var{p+1}\sqrt{\var{p}}}+\frac{\simplify{{one}*sqrt({p})}}{\var{two}\sqrt{\var{p}}}= \frac{\var{p}+\simplify{{one}*sqrt({p})}}{\var{p+1}\sqrt{\var{p}}}$

When simplifying, if we note that $\displaystyle\sqrt{a}\times\sqrt{a}=a$ , we can pull $\sqrt{\var{p}}$ out of the numerator as a common factor:

$\frac{\sqrt{\var{p}}(\sqrt{\var{p}}+\var{r_coprime*p_p1})}{\var{p+1}\sqrt{\var{p}}}\text{.}$

We can then cancel the common $\sqrt{\var{p}}$ term on the numerator and denominator to simplify the expression further and get a final answer of

$\frac{\sqrt{\var{p}}+\var{one}}{\var{p+1}} \text{.}$

Give a worked solution to the whole question.

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This question is used in the following exams:

Surds by Christian Lawson-Perfect in Transition to university.
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Test Homework by Connor Fenton in Connor's workspace.
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Rationalising the denominator - surds

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Checkpoint description

Elliott Fletcher 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Lauren Richards 7 years, 8 months ago

Vicky Hall commented 7 years, 8 months ago

Lauren Richards commented 7 years, 9 months ago

Lauren Richards 7 years, 9 months ago

Lauren Richards commented 7 years, 9 months ago

Lauren Richards 7 years, 9 months ago

Aiden McCall commented 7 years, 9 months ago

Aiden McCall commented 7 years, 9 months ago

Aiden McCall 7 years, 9 months ago

Lauren Richards 7 years, 9 months ago

Lauren Richards 7 years, 9 months ago

Lauren Richards commented 7 years, 9 months ago

Lauren Richards 7 years, 9 months ago

Lauren Richards 7 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: list

Parts

Gap-fill

a)

b)

c)

d)

e)

f)

Generated value: `list`