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\n[[0]]
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\n$\\displaystyle\\sqrt{\\var{c}}$ = [[0]]$\\displaystyle\\sqrt{\\var{b}}$
\n$\\displaystyle\\sqrt{\\var{g}}$ = [[1]]$\\displaystyle\\sqrt{\\var{f}}$
\n", "variableReplacements": [], "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "scripts": {}}], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "This question tests the student's understanding of what is and is not a surd, and on their simplification of surds.
"}, "name": "Katy's copy of Surds simplification", "statement": "Surds are square roots that cannot be simplified to a whole number. They have a decimal equivalent but their decimal representations are never-ending. Therefore, it is often easier to leave surds as they are in algebraic calculations.
", "rulesets": {}, "advice": "$\\sqrt{\\var{square}}$ and $\\sqrt[3]{\\var{cube}}$ are not surds, as they can be simplified to whole integers: $\\simplify{{sqrt(square)}}$ and $\\var{root}$ respectively. They are roots, but not surds. All surds are roots but not all roots are surds.
\n$\\sqrt{\\var{h}}$, $\\sqrt{\\var{j}}$ and $\\sqrt{\\var{k}}$ are surds, as they cannot be simplified to a whole integer. There is no number, $b$, such that $b^2=\\var{h}, \\var{j}$ or $\\var{k}$. Therefore, $\\sqrt{\\var{h}}$, $\\sqrt{\\var{j}}$ and $\\sqrt{\\var{k}}$ are both roots and surds.
\n\n\nThe rule that should be used is $\\sqrt{a}\\times\\sqrt{b}=\\sqrt{ab}$.
\nWe need to try to find a square number that divides $ab$ and rewrite this as $\\sqrt{b^2}\\times\\sqrt{a}$.
\ni)
\n$\\sqrt{48}$ = $\\sqrt{16}\\times\\sqrt3$
\n$\\sqrt{16}$ simplifies down to $4$ so the final answer is: $4\\sqrt3$.
\nii)
\n$\\sqrt{56}$ = $\\sqrt{4}\\times\\sqrt{14}$
\n$\\sqrt4$ simplifies down to $2$ so the final answer is: $2\\sqrt{14}$.
\niii)
\n$\\sqrt{32}$ = $\\sqrt{16}\\times\\sqrt{2}$
\n$\\sqrt{16}$ simplifies down to $4$ so the final answer is: $4\\sqrt2$.
\niv)
\n$\\sqrt{44}$ = $\\sqrt{4}\\times\\sqrt{11}$
\n$\\sqrt4$ simplifies down to $2$ so the final answer is: $2\\sqrt{11}$.
\n\nThis question requires you to notice that $\\sqrt{\\var{a}}$ and $\\sqrt{\\var{d}}$ are squared numbers and can be simplified to integers.
\n$\\sqrt{\\var{a}}$ = $\\var{sqrta}$ such that:
\ni) $\\sqrt{\\var{c}}$ = $\\sqrt{\\var{a}}$ x $\\sqrt{\\var{b}}$ = $\\var{sqrta}\\sqrt{\\var{b}}$ and
\nii) $\\sqrt{\\var{g}}$ = $\\sqrt{\\var{d}}$ x $\\sqrt{\\var{f}}$ = $\\var{sqrtd}\\sqrt{\\var{f}}$.
\n\n", "preamble": {"js": "", "css": ""}, "type": "question", "variablesTest": {"condition": "", "maxRuns": 100}, "functions": {}, "variable_groups": [], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Katy Dobson", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/854/"}, {"name": "Lauren Richards", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1589/"}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Katy Dobson", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/854/"}, {"name": "Lauren Richards", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1589/"}]}