// Numbas version: exam_results_page_options {"name": "Simon's copy of Using Surds, Rationalising the Denominator", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"tags": [], "metadata": {"description": "

Manipulate surds and rationalise the denominator of a fraction when it is a surd.

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parts d and e

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Parts a, d,e and h

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Fraction in answer for part d.

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Parts c and e

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all numbers from 3-10 for parts a, b, e, g

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Short list of primes for part d.

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prime number for parts a,b and h

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shorter list of primes for parts a,c,e,f and g

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parts b and d

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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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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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Can be simplified further

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Cannot be simplified further

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$\\sqrt{\\var{p}}$

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

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$\\sqrt{\\var{a}}$

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Which of the following can be simplified further?

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Recall the first rule of surds

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

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Simplify $\\sqrt{\\simplify{{n}^2*{p}}}$.

$\\sqrt{\\simplify{{n}^2*{p}}} =$ [[0]]$\\sqrt{\\var{p}}$.

You must simplify your answer further.

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You must simplify your answer further.

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You could use either of the following rules:

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

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

Simplify $\\displaystyle\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}}$.

$\\displaystyle\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}} =$ [[0]].

Simplify $\\displaystyle\\frac{\\sqrt{\\simplify{({b}{m})^2*{s}}}}{\\var{m}}$.

$\\displaystyle\\frac{\\sqrt{\\simplify{({b}*{m})^2*{s}}}}{\\var{m}} =$ [[0]]$\\sqrt{\\var{s}}$.

Simplify $\\simplify{{d}sqrt({a}) - {b}sqrt({v}^2*{a})+{n}sqrt({b}^2*{a})}$.

$\\simplify{{d}sqrt({a}) - {b}sqrt({v}^2*{a})+{n}sqrt({b}^2*{a})} =$ [[0]].

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To rationalise the denominator of fractions in the form $\\frac{1}{\\sqrt{a}}$, multiply the top and bottom by $\\sqrt{a}$.

Rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\sqrt{\\var{a}}}$.

$\\displaystyle\\frac{1}{\\sqrt{\\var{a}}} =$ [[0]] [[1]] .

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To rationalise the denominator of fractions in the form $\\displaystyle\\frac{1}{a+\\sqrt{b}}$, multiply the top and bottom by $a-\\sqrt{b}$.

Rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}}$.

$\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}} =$ [[0]] [[1]] .

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To rationalise the denominator of fractions in the form, $\\displaystyle\\frac{1}{a-\\sqrt{b}}$, multiply the top and bottom by ${a+\\sqrt{b}}$.

Rationalise the denominator of the fraction $\\displaystyle\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}}$.

$\\displaystyle\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}} =$ [[0]] [[1]] .

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