Katherine's copy of Using Surds, Rationalising the Denominator

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Katherine Tomlinson

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Katherine Tomlinson 7 years, 8 months ago

Created this as a copy of Using Surds, Rationalising the Denominator.

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.

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			Name	Type	Generated Value

p	integer	3
d	integer	6
n	integer	8
m	integer	2
b	integer	2
a	integer	2
v	integer	3
t	integer	3
c	rational	1
s	integer	3

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

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Description

prime number for parts a,b and h

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Generated value: `integer`

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Advice
"Part a)" - Answer text
"Part b)" - prompt
"Part h)" - prompt
"Part h)" → "Gap 0." - Correct answer
"Part h)" → "Gap 1." - Correct answer

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Part c) Gap-fill
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Match choices with answers

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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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Katherine's copy of Using Surds, Rationalising the Denominator

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Katherine Tomlinson 7 years, 8 months ago

Error in variable testing condition

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Generated value: `integer`

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Match choices with answers

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a)

b)

c)

d)

e)

f)

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h)

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Katherine's copy of Using Surds, Rationalising the Denominator

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Katherine Tomlinson 7 years, 8 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Match choices with answers

Advanced settings

a)

b)

c)

d)

e)

f)

g)

h)

Generated value: `integer`