Using Surds, Rationalising the Denominator

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From users who are members of Transition to university :

Christian Lawson-Perfect	said	Ready to use	7 years, 9 months ago
Elliott Fletcher	said	Needs to be tested	7 years, 10 months ago
Lauren Richards	said	Has some problems	7 years, 10 months ago
Chris Graham	said	Has some problems	7 years, 10 months ago

From users who are not members of Transition to university :

francisco Glezortiz

said

Ready to use

4 years, 3 months ago

History

○

Christian Lawson-Perfect 1 year, 7 months ago

Saved a checkpoint:

Added a `+- to the pattern for part e, to allow negative coefficients.

Restore this checkpoint

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francisco Glezortiz 4 years, 3 months ago

Gave some feedback: Ready to use

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francisco Glezortiz 4 years, 3 months ago

Gave some feedback: Needs to be tested

○

Christian Lawson-Perfect 5 years, 11 months ago

Saved a checkpoint:

Replaced some javascript with pattern restrictions. There are still some string restrictions that could be replaced with pattern restrictions.

Restore this checkpoint

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Elliott Fletcher 7 years, 9 months ago

Published this.

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Christian Lawson-Perfect 7 years, 9 months ago

Gave some feedback: Ready to use

○

Christian Lawson-Perfect 7 years, 9 months ago

Saved a checkpoint:

I changed the denominators for parts f,g and h from number entry to mathematical expression. While the answer is always a number, a student who's misunderstood might want to write a surd. You should let them fail!

I don't know why each line of the derivations ended with a comma, but this isn't conventional.

With those things fixed, this is a good question.

Restore this checkpoint

○

Elliott Fletcher 7 years, 10 months ago

Gave some feedback: Needs to be tested

○

Lauren Richards 7 years, 10 months ago

Gave some feedback: Has some problems

○

Lauren Richards commented 7 years, 10 months ago

MAIN PARTS

Part a) should have more numbers rather than just two.

I think you will have a problem with people not knowing how to insert a square root sign into their answers, meaning they will be scuppered for all of the questions. I think you will need to tell them how to in the parts by giving an example. Say like "3\sqrt{5} can be written as 3*sqrt(5)". Alternatively, you could set it up via gap fill so that all of the answers can be given in a pre-prepared square root sign.

Part b)'s steps could be a little off-putting for someone. They only need the first rule to do part b) but you have given them an extra rule that isn't relevant. It might confuse them.

You don't need colons after "Simplify" in each part.

Part h), your model answer included an unsimplified numerator, which might not be what the user gives. You didn't multiply out 3(\sqrt{3}+18) but writing 3\sqrt{13}+54 is not wrong.

ADVICE

I think you should restate the question in part a) after stating the rule and before the actual advice for part a).

The examples listed for part a) and b) were not the ones I was asked in the test run. It was talking about \sqrt{405} both times when I had been asked about \sqrt{1377} in part a) and \sqrt{1053} in part b).

In part c), I think you should show how the fraction cancels down a bit more by stating that the two \sqrt{17} cancel on the numerator and denominator. Someone might get lost if they followed the first method.

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Elliott Fletcher commented 7 years, 10 months ago

Thanks for your feedback, i have corrected the mistakes

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Elliott Fletcher 7 years, 10 months ago

Gave some feedback: Needs to be tested

○

Chris Graham commented 7 years, 10 months ago

Rather than having "Answer:" before the gap, I think I'd rather that the surd was repeated, e.g. " $\frac{1}{\sqrt{11}}=$ "

In (g) and (h) your answers are displayed with a decimal on the numerator, which I presume is not what you intended. The reason for this is that the braces evaluate {{n}-sqrt({a})} to a decimal number. Try this instead to retain the integer values of n and a:

({n}-sqrt({a}))/{{n}^2-{a}}.

Good work with the advice, which is well laid out and punctuated.

○

Chris Graham 7 years, 10 months ago

Gave some feedback: Has some problems

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Elliott Fletcher 7 years, 10 months ago

Gave some feedback: Needs to be tested

○

Elliott Fletcher 7 years, 10 months ago

Created this.

Name	Status	Author	Last Modified
Using Surds, Rationalising the Denominator	Ready to use	Elliott Fletcher	27/09/2023 09:14
Katherine's copy of Using Surds, Rationalising the Denominator	draft	Katherine Tomlinson	31/07/2017 11:50
Using Surds, Rationalising the Denominator	draft	Adrian Jannetta	16/08/2023 14:56
Surds Rationalisation	draft	Johnny Yi	24/01/2019 04:29
Surds Rationalisation (Diagnostic Test)	draft	Johnny Yi	24/01/2019 04:30
Simon's copy of Using Surds, Rationalising the Denominator	draft	Simon Thomas	06/06/2019 09:40
John's copy of Using Surds, Rationalising the Denominator	draft	John Wick	17/06/2019 11:00
John's copy of John's copy of John's copy of John's copy of John's copy of John's copy of John's copy of John's copy of John's copy of John's copy of John's Using Surds, Rationalising the Denominator	draft	John Wick	10/07/2019 16:00
Ausdrücke mit Quadratwurzeln vereinfachen	Ready to use	Ulrich Görtz	26/11/2020 10:28

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Question statement

To include a square root sign in your answer use sqrt(). For example, to write $\sqrt{3}$ , type sqrt(3) into the answer box. If you are entering a number multiplied by the square root of some other number, for example $3\sqrt{5}$ , type 3*sqrt(5) into the answer box.

Give any introductory information the student needs.

			Name	Type	Generated Value

p	integer	19
d	integer	3
n	integer	9
m	integer	5
b	integer	5
a	integer	17
v	integer	3
t	integer	2
c	rational	1
s	integer	7

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Name

Data type

Value

xxxxxxxxxx
 
random(1..20 #2 except 1 except 9 except 15 except a)

JME function reference

Description

prime number for parts a,b and h

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Generated value: `integer`

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Parts

Part a) Match choices with answers
Part b) Gap-fill
1. Gap Mathematical expression
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1. Step Information only
Part c) Gap-fill
1. Gap Mathematical expression
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1. Step Information only
Part d) Gap-fill
1. Gap Mathematical expression
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Part e) Gap-fill
1. Gap Mathematical expression
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Part f) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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Part g) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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1. Step Information only
Part h) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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1. Step Information only

Match choices with answers

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

Which of the following can be simplified further?

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Can be simplified further
Cannot be simplified further

$\sqrt{\var{p}}$
$\sqrt{\simplify{{a}*{n}^2}}$
$\sqrt{\var{a}}$

For each combination of answer and choice, specify the number of marks to add or subtract when the student picks it.

	$\sqrt{\var{p}}$	$\sqrt{\simplify{{a}*{n}^2}}$	$\sqrt{\var{a}}$
Can be simplified further
Cannot be simplified further

Advice

a)

Surds can be manipulated using the rule

$\sqrt{(ab)} = \sqrt{a} \times \sqrt{b}.$

We are asked to state which of $\sqrt{\var{p}}$ , $\sqrt{\simplify{{a}*{n}^2}}$ , and $\sqrt{\var{a}}$ can be simplified further. Commonly, surds can be simplified if the number inside of the square root has a square number as a factor.

Here, $\var{p}$ is a prime number which means that its only divisors are $\var{p}$ and $1$ .

Therefore, $\sqrt{\var{p}}$ cannot be simplified any further.

Similarly, $\var{a}$ is also a prime number, so $\sqrt{\var{a}}$ also cannot be simplified any further.

On the other hand, $\simplify{{a}*{n}^2}$ is not a prime number and we can use the previous rule to simplify $\sqrt{\simplify{{a}*{n}^2}}$ as

$\begin{align} \sqrt{\simplify{{a}*{n}^2}} &= \sqrt{\simplify{{n}^2}} \times \sqrt{\var{a}}\\ &= \simplify{{n}*sqrt({a})}. \end{align}$

b)

Using the same rule of manipulation as in part a), we can simplify $\sqrt{\simplify{{n}^2*{p}}}$ as

$\begin{align} \sqrt{\simplify{{n}^2*{p}}} &= \sqrt{\simplify{{n}^2}} \times \sqrt{\var{p}}\\ &= \simplify{{n}*sqrt({p})}. \end{align}$

c)

Here, we can use both of the rules for manipulating surds:

$\sqrt{(ab)} = \sqrt{a} \times \sqrt{b} \text{.}$

$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \text{.}$

We can simplify $\displaystyle\frac{ \sqrt{\simplify{{a}*{v}}} }{ \sqrt{\var{a}} }$ as follows.

$\begin{align} \frac{\sqrt{\simplify{{a}*{v}}}}{\sqrt{\var{a}}} &= \frac{\sqrt{\var{a}} \times \sqrt{\var{v}}}{\sqrt{\var{a}}} \\[0.5em] &= \frac{\sqrt{\var{a}}}{\sqrt{\var{a}}} \times \sqrt{\var{v}} \\[0.5em] &= \simplify{{sqrt(a)/sqrt(a)}} \times \sqrt{\var{v}} \\[0.5em] &= \sqrt{\var{v}} \text{.} \end{align}$

Or,

$\begin{align} \frac{\sqrt{\simplify{{a}*{v}}}}{\sqrt{\var{a}}} &= \sqrt{\frac{\simplify{{a}*{v}}}{\var{a}}} \\[0.5em] &= \sqrt{\var{v}} \text{.} \end{align}$

d)

We can simplify the fraction as

$\begin{align} \frac{\sqrt{\simplify{({b}{m})^2*{s}}}}{\var{m}} &= \frac{\sqrt{\simplify{({b*m})^2}} \times \sqrt{\var{s}}}{\var{m}} \\[0.5em] &= \frac{\simplify{{b*m}} \times \sqrt{\var{s}}}{\var{m}} \\[0.5em] &= \simplify{{b}*sqrt({s})} \text{.} \end{align}$

e)

$\begin{align} \simplify{{d}sqrt({a}) - {b}sqrt({v}^2{a})+{n}sqrt({b}^2*{a})} &= \var{d}\sqrt{\var{a}} - \var{b}(\sqrt{\simplify{{v}^2}} \times \sqrt{\var{a}})+\var{n}(\sqrt{\simplify{{b}^2}} \times \sqrt{\var{a}}) \\ &= \var{d}\sqrt{\var{a}} -\var{b}(\simplify{{v}*sqrt({a})})+\var{n}(\simplify{{b}*sqrt({a})}) \\ &= \simplify{{d}sqrt({a})}-\simplify{{b}*{v}sqrt({a})}+\simplify{{n}*{b}sqrt({a})} \\ &= \simplify{({d}-{b}*{v}+{n}*{b})sqrt({a})} \text{.} \end{align}$

f)

We rationalise the denominator of fractions of the form $\displaystyle\frac{1}{\sqrt{a}}$ , by multiplying the top and bottom by $\sqrt{a}$ .

Therefore, to rationalise the denominator of the fraction $\displaystyle\frac{1}{\sqrt{\var{a}}}$ , we multiply top and bottom by $\sqrt{\var{a}}$ .

$\begin{align} \frac{1}{\sqrt{\var{a}}} &= \frac{1}{\sqrt{\var{a}}} \times \frac{\sqrt{\var{a}}}{\sqrt{\var{a}}} \\[0.5em] &= \frac{\sqrt{\var{a}}}{\var{a}} \text{.} \end{align}$

g)

We rationalise the denominator of fractions of the form $\displaystyle\frac{1}{a+\sqrt{b}}$ by multiplying the top and bottom by $a-\sqrt{b}$ .

Therefore, to rationalise the denominator of the fraction $\displaystyle\frac{1}{\var{n}+\sqrt{\var{a}}}$ , we multiply the top and bottom by $\var{n} - \sqrt{\var{a}}$ .

$\begin{align} \frac{1}{\var{n}+\sqrt{\var{a}}} &= \frac{1}{\var{n}+\sqrt{\var{a}}} \times \frac{\var{n}-\sqrt{\var{a}}}{\var{n}-\sqrt{\var{a}}} \\[0.5em] &=\frac{\var{n}-\sqrt{\var{a}}}{(\var{n}+\sqrt{\var{a}})(\var{n}-\sqrt{\var{a}})} \\[0.5em] &=\frac{\var{n}-\sqrt{\var{a}}}{\simplify{{n}^2}-\var{a}} \\[0.5em] &=\frac{\var{n}-\sqrt{\var{a}}}{\simplify{{n}^2-{a}}} \text{.} \end{align}$

h)

We rationalise the denominator of fractions of the form $\displaystyle\frac{1}{a-\sqrt{b}}$ by multiplying the top and bottom by $a+\sqrt{b}$ .

Therefore, to rationalise the denominator of the fraction $\displaystyle\frac{\var{t}}{\var{d+p}-\sqrt{\var{p}}}$ , we multiply the top and bottom by $\var{d+p}+\sqrt{\var{p}}$ .

$\begin{align} \frac{\var{t}}{\var{d+p}-\sqrt{\var{p}}} &= \frac{\var{t}}{\var{d+p}-\sqrt{\var{p}}} \times \frac{\var{d+p}+\sqrt{\var{p}}}{\var{d+p}+\sqrt{\var{p}}} \\[0.5em] &=\frac{\var{t}(\var{d+p}+\sqrt{\var{p}})}{(\var{d+p}-\sqrt{\var{p}})(\var{d+p}+\sqrt{\var{p}})} \\[0.5em] &=\frac{\var{t}(\var{d+p}+\sqrt{\var{p}})}{\simplify{{d+p}^2}-\var{p}} \\[0.5em] &=\frac{\var{t}(\var{d+p}+\sqrt{\var{p}})}{\simplify{{d+p}^2-{p}}} \\[0.5em] &=\simplify{{t}/{(d+p)^2-p}}(\var{d+p}+\sqrt{\var{p}}) \\[0.5em] &= \simplify[all,!noleadingMinus]{({t*(d+p)}+{t}*sqrt({p}))/({(d+p)^2-p})} \text{.} \end{align}$

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Christian Lawson-Perfect 1 year, 7 months ago

francisco Glezortiz 4 years, 3 months ago

francisco Glezortiz 4 years, 3 months ago

Christian Lawson-Perfect 5 years, 11 months ago

Elliott Fletcher 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Elliott Fletcher 7 years, 10 months ago

Lauren Richards 7 years, 10 months ago

Lauren Richards commented 7 years, 10 months ago

Elliott Fletcher commented 7 years, 10 months ago

Elliott Fletcher 7 years, 10 months ago

Chris Graham commented 7 years, 10 months ago

Chris Graham 7 years, 10 months ago

Elliott Fletcher 7 years, 10 months ago

Elliott Fletcher 7 years, 10 months ago

Error in variable testing condition

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Generated value: integer

Parts

Match choices with answers

Advanced settings

a)

b)

c)

d)

e)

f)

g)

h)

Generated value: `integer`