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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", "Cannot be simplified further
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", "$\\sqrt{\\simplify{{a}*{n}^2}}$
", "$\\sqrt{\\var{a}}$
"], "prompt": "Which of the following can be simplified further?
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\n$\\sqrt{\\simplify{{n}^2*{p}}} =$ [[0]]$\\sqrt{\\var{p}}$.
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\n$\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}$.
\n\n", "extendBaseMarkingAlgorithm": true, "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "type": "information", "showFeedbackIcon": true, "marks": 0, "unitTests": [], "useCustomName": false, "variableReplacements": []}], "showFeedbackIcon": true, "marks": 0, "unitTests": [], "useCustomName": false, "gaps": [{"customName": "", "customMarkingAlgorithm": "", "failureRate": 1, "checkingType": "absdiff", "showCorrectAnswer": true, "checkVariableNames": false, "vsetRange": [0, 1], "showFeedbackIcon": true, "mustmatchpattern": {"pattern": "$n", "nameToCompare": "", "partialCredit": 0, "message": "You haven't fully simplified."}, "vsetRangePoints": 5, "checkingAccuracy": 0.001, "scripts": {}, "answer": "{n}", "valuegenerators": [], "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "type": "jme", "answerSimplification": "all", "marks": "2", "unitTests": [], "showPreview": true, "useCustomName": false, "variableReplacements": []}], "variableReplacements": []}, {"customName": "", "customMarkingAlgorithm": "", "sortAnswers": false, "scripts": {}, "prompt": "Simplify $\\displaystyle\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}}$.
\n$\\displaystyle\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}} =$ [[0]].
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\n$\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}$.
\n$\\displaystyle\\sqrt{\\frac{a}{b}} = \\displaystyle\\frac{\\sqrt{a}}{\\sqrt{b}}$.
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\n$\\displaystyle\\frac{\\sqrt{\\simplify{({b}*{m})^2*{s}}}}{\\var{m}} =$ [[0]]$\\sqrt{\\var{s}}$.
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\n$\\simplify{{d}sqrt({a}) - {b}sqrt({v}^2*{a})+{n}sqrt({b}^2*{a})} =$ [[0]].
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$\\displaystyle\\frac{1}{\\sqrt{\\var{a}}} =$
To rationalise the denominator of fractions in the form $\\frac{1}{\\sqrt{a}}$, multiply the top and bottom by $\\sqrt{a}$.
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\n$\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}} =$
To rationalise the denominator of fractions in the form $\\displaystyle\\frac{1}{a+\\sqrt{b}}$, multiply the top and bottom by $a-\\sqrt{b}$.
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\n$\\displaystyle\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}} =$
To rationalise the denominator of fractions in the form, $\\displaystyle\\frac{1}{a-\\sqrt{b}}$, multiply the top and bottom by ${a+\\sqrt{b}}$.
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\n\\[\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}.\\]
\nWe 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.
\nHere, $\\var{p}$ is a prime number which means that its only divisors are $\\var{p}$ and $1$.
\nTherefore, $\\sqrt{\\var{p}}$ cannot be simplified any further.
\nSimilarly, $\\var{a}$ is also a prime number, so $\\sqrt{\\var{a}}$ also cannot be simplified any further.
\nOn 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
\n\\[
\\begin{align}
\\sqrt{\\simplify{{a}*{n}^2}} &= \\sqrt{\\simplify{{n}^2}} \\times \\sqrt{\\var{a}}\\\\
&= \\simplify{{n}*sqrt({a})}.
\\end{align}
\\]
Using the same rule of manipulation as in part a), we can simplify $\\sqrt{\\simplify{{n}^2*{p}}}$ as
\n\\[
\\begin{align}
\\sqrt{\\simplify{{n}^2*{p}}} &= \\sqrt{\\simplify{{n}^2}} \\times \\sqrt{\\var{p}}\\\\
&= \\simplify{{n}*sqrt({p})}.
\\end{align}
\\]
Here, we can use both of the rules for manipulating surds:
\n\\[\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b} \\text{.} \\]
\n\\[ \\sqrt{\\frac{a}{b}} = \\frac{\\sqrt{a}}{\\sqrt{b}} \\text{.} \\]
\nWe can simplify $\\displaystyle\\frac{ \\sqrt{\\simplify{{a}*{v}}} }{ \\sqrt{\\var{a}} }$ as follows.
\n\\[
\\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,
\n\\[
\\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}
\\]
We can simplify the fraction as
\n\\[
\\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}
\\]
\\[
\\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}
\\]
We rationalise the denominator of fractions of the form $\\displaystyle\\frac{1}{\\sqrt{a}}$, by multiplying the top and bottom by $\\sqrt{a}$.
\nTherefore, to rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\sqrt{\\var{a}}}$, we multiply top and bottom by $\\sqrt{\\var{a}}$.
\n\\[
\\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}
\\]
We rationalise the denominator of fractions of the form $\\displaystyle\\frac{1}{a+\\sqrt{b}}$ by multiplying the top and bottom by $a-\\sqrt{b}$.
\nTherefore, to rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}}$, we multiply the top and bottom by $\\var{n} - \\sqrt{\\var{a}}$.
\n\\[
\\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}
\\]
We rationalise the denominator of fractions of the form $\\displaystyle\\frac{1}{a-\\sqrt{b}}$ by multiplying the top and bottom by $a+\\sqrt{b}$.
\nTherefore, to rationalise the denominator of the fraction $\\displaystyle\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}}$, we multiply the top and bottom by $\\var{d+p}+\\sqrt{\\var{p}}$.
\n\\[
\\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}
\\]