Manipulate surds and rationalise the denominator of a fraction when it is a surd.
"}, "name": "Using Surds, Rationalising the Denominator", "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.
Surds can be manipulated using the rule
\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}
\\]
all numbers from 3-10 for parts a, b, e, g
"}, "a": {"name": "a", "definition": "random(2..19 #1 except 4 except 6 except 8 except 9 except 10 except 12 except 14 except 15 except 16 except 18 except v) ", "templateType": "anything", "group": "Ungrouped variables", "description": "shorter list of primes for parts a,c,e,f and g
"}, "b": {"name": "b", "definition": "random(2..5 #1 except v except sqrt(m^2*d) except 3)", "templateType": "anything", "group": "Ungrouped variables", "description": "parts d and e
"}, "m": {"name": "m", "definition": "random(2..5 #1 except n^2)", "templateType": "anything", "group": "Ungrouped variables", "description": "parts b and d
"}, "d": {"name": "d", "definition": "random(2..7 except 4)", "templateType": "anything", "group": "Ungrouped variables", "description": "Parts a, d,e and h
"}, "t": {"name": "t", "definition": "random(2,3)", "templateType": "anything", "group": "Ungrouped variables", "description": ""}, "c": {"name": "c", "definition": "(m)/(b)", "templateType": "anything", "group": "Ungrouped variables", "description": "Fraction in answer for part d.
"}, "v": {"name": "v", "definition": "random(2,3,5)", "templateType": "anything", "group": "Ungrouped variables", "description": "Parts c and e
"}, "p": {"name": "p", "definition": "random(1..20 #2 except 1 except 9 except 15 except a)", "templateType": "anything", "group": "Ungrouped variables", "description": "prime number for parts a,b and h
"}, "s": {"name": "s", "definition": "random(2..7 except 4 except 6)", "templateType": "anything", "group": "Ungrouped variables", "description": "Short list of primes for part d.
"}}, "variable_groups": [], "preamble": {"css": "fraction {\n display: inline-block;\n vertical-align: middle;\n}\nfraction > numerator, fraction > denominator {\n float: left;\n width: 100%;\n text-align: center;\n line-height: 2.5em;\n}\nfraction > numerator {\n border-bottom: 1px solid;\n padding-bottom: 5px;\n}\nfraction > denominator {\n padding-top: 5px;\n}\nfraction input {\n line-height: 1em;\n}\n\nfraction .part {\n margin: 0;\n}\n\n.table-responsive, .fractiontable {\n display:inline-block;\n}\n.fractiontable {\n padding: 0; \n border: 0;\n}\n\n.fractiontable .tddenom \n{\n text-align: center;\n}\n\n.fractiontable .tdnum \n{\n border-bottom: 1px solid black; \n text-align: center;\n}\n\n\n.fractiontable tr {\n height: 3em;\n}\n", "js": ""}, "parts": [{"type": "gapfill", "gaps": [{"vsetrangepoints": 5, "answersimplification": "all", "type": "jme", "showCorrectAnswer": true, "checkvariablenames": false, "showpreview": true, "answer": "{n}", "variableReplacements": [], "checkingaccuracy": 0.001, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "scripts": {"mark": {"script": "// Parse the student's answer as a syntax tree\nvar studentTree = Numbas.jme.compile(this.studentAnswer,Numbas.jme.builtinScope);\n\n// Create the pattern to match against \n// we just want two sets of brackets, each containing two terms\n// or one of the brackets might not have a constant term\n// or for repeated roots, you might write (x+a)^2\nvar rule = Numbas.jme.compile('m_any(m_number)');\n\n// Check the student's answer matches the pattern. \nvar m = Numbas.jme.display.matchTree(rule,studentTree,true);\n// If not, take away marks\nif(!m) {\n this.multCredit(0,'You haven\\'t simplified: your answer is not in the form $\\?$.');\n}", "order": "after"}}, "vsetrange": [0, 1], "marks": "2", "expectedvariablenames": [], "checkingtype": "absdiff"}], "showCorrectAnswer": true, "stepsPenalty": "1", "steps": [{"type": "information", "showCorrectAnswer": true, "marks": 0, "variableReplacements": [], "prompt": "Recall the first rule of surds
\n$\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}$.
\n\n", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "scripts": {}}], "marks": 0, "variableReplacements": [], "prompt": "Simplify $\\sqrt{\\simplify{{n}^2*{p}}}$.
\n\n$\\sqrt{\\simplify{{n}^2*{p}}} =$ [[0]]$\\sqrt{\\var{p}}$.
", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "scripts": {}}, {"type": "gapfill", "gaps": [{"vsetrangepoints": 5, "type": "jme", "showCorrectAnswer": true, "checkvariablenames": false, "showpreview": true, "answer": "sqrt({v})", "variableReplacements": [], "showFeedbackIcon": true, "checkingaccuracy": 0.001, "scripts": {}, "vsetrange": [0, 1], "musthave": {"message": "You must simplify your answer further.
", "partialCredit": 0, "strings": ["sqrt", "(", ")"], "showStrings": true}, "marks": "2", "variableReplacementStrategy": "originalfirst", "notallowed": {"message": "You must simplify your answer further.
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\n$\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}$.
\n$\\displaystyle\\sqrt{\\frac{a}{b}} = \\displaystyle\\frac{\\sqrt{a}}{\\sqrt{b}}$.
", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "scripts": {}}], "marks": 0, "variableReplacements": [], "prompt": "Simplify $\\displaystyle\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}}$.
\n$\\displaystyle\\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}} =$ [[0]].
\n", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "scripts": {}}, {"type": "gapfill", "showCorrectAnswer": true, "gaps": [{"vsetrangepoints": 5, "answersimplification": "all", "type": "jme", "showCorrectAnswer": true, "checkvariablenames": false, "showpreview": true, "answer": "{b}", "variableReplacements": [], "checkingaccuracy": 0.001, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "scripts": {}, "vsetrange": [0, 1], "marks": "2", "expectedvariablenames": [], "checkingtype": "absdiff"}], "marks": 0, "variableReplacements": [], "prompt": "Simplify $\\displaystyle\\frac{\\sqrt{\\simplify{({b}{m})^2*{s}}}}{\\var{m}}$.
\n$\\displaystyle\\frac{\\sqrt{\\simplify{({b}*{m})^2*{s}}}}{\\var{m}} =$ [[0]]$\\sqrt{\\var{s}}$.
\n", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "scripts": {}}, {"type": "gapfill", "gaps": [{"vsetrangepoints": 5, "type": "jme", "showCorrectAnswer": true, "checkvariablenames": false, "showpreview": true, "answer": "sqrt({a})", "variableReplacements": [], "showFeedbackIcon": true, "checkingaccuracy": 0.001, "scripts": {}, "vsetrange": [0, 1], "musthave": {"message": "", "partialCredit": 0, "strings": ["sqrt", "(", ")"], "showStrings": false}, "marks": "1", "variableReplacementStrategy": "originalfirst", "notallowed": {"message": "", "partialCredit": 0, "strings": ["/"], "showStrings": false}, "expectedvariablenames": [], "checkingtype": "absdiff"}, {"vsetrangepoints": 5, "type": "jme", "showCorrectAnswer": true, "checkvariablenames": false, "showpreview": true, "answer": "{a}", "variableReplacements": [], "showFeedbackIcon": true, "checkingaccuracy": 0.001, "scripts": {}, "vsetrange": [0, 1], "marks": 1, "variableReplacementStrategy": "originalfirst", "expectedvariablenames": [], "checkingtype": "absdiff"}], "showCorrectAnswer": true, "stepsPenalty": "1", "steps": [{"type": "information", "showCorrectAnswer": true, "marks": 0, "variableReplacements": [], "prompt": "To rationalise the denominator of fractions in the form $\\frac{1}{\\sqrt{a}}$, multiply the top and bottom by $\\sqrt{a}$.
", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "scripts": {}}], "marks": 0, "variableReplacements": [], "prompt": "Rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\sqrt{\\var{a}}}$.
$\\displaystyle\\frac{1}{\\sqrt{\\var{a}}} =$
To rationalise the denominator of fractions in the form $\\displaystyle\\frac{1}{a+\\sqrt{b}}$, multiply the top and bottom by $a-\\sqrt{b}$.
", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "scripts": {}}], "marks": 0, "variableReplacements": [], "prompt": "Rationalise the denominator of the fraction $\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}}$.
\n$\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}} =$