// Numbas version: exam_results_page_options {"name": "Solve equations which include a single root (e.g. \\sqrt{x}=blah)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"showfrontpage": false, "preventleave": false, "allowregen": true}, "question_groups": [{"questions": [{"variable_groups": [{"variables": ["intpower", "intrhs", "intsoln"], "name": "a"}, {"variables": ["bpower", "bnice", "bsoln", "bxcoeff", "bb", "bc"], "name": "b"}], "variables": {"bpower": {"group": "b", "definition": "random(2..9)", "templateType": "anything", "name": "bpower", "description": ""}, "bb": {"group": "b", "definition": "random(1..100)", "templateType": "anything", "name": "bb", "description": ""}, "intpower": {"group": "a", "definition": "random(2..9)", "templateType": "anything", "name": "intpower", "description": ""}, "bnice": {"group": "b", "definition": "switch(bpower=3 or bpower=2, random(-10..10 except -1..1), bpower=5 or bpower =4, random(-4..4 except -1..1), bpower=7 or bpower=6, random(-3..3 except -1..1), 2)", "templateType": "anything", "name": "bnice", "description": "

((bc-bb)/bxcoeff)^(1/bpower)

"}, "ddenom": {"group": "Ungrouped variables", "definition": "random(2..15)", "templateType": "anything", "name": "ddenom", "description": ""}, "bsoln": {"group": "b", "definition": "bnice^bpower", "templateType": "anything", "name": "bsoln", "description": ""}, "bc": {"group": "b", "definition": "bnice*bxcoeff+bb", "templateType": "anything", "name": "bc", "description": ""}, "dc": {"group": "Ungrouped variables", "definition": "random(-100..100 except -1..1)", "templateType": "anything", "name": "dc", "description": ""}, "dpower": {"group": "Ungrouped variables", "definition": "random(2..9)", "templateType": "anything", "name": "dpower", "description": ""}, "bxcoeff": {"group": "b", "definition": "random(-3..3 except 0..1)", "templateType": "anything", "name": "bxcoeff", "description": ""}, "db": {"group": "Ungrouped variables", "definition": "random(-100..100 except -1..1)", "templateType": "anything", "name": "db", "description": ""}, "intsoln": {"group": "a", "definition": "intrhs^intpower", "templateType": "anything", "name": "intsoln", "description": ""}, "intrhs": {"group": "a", "definition": "switch(intpower=3 or intpower=4, random(2..12), intpower=5 or intpower=6, random(2..5), intpower=7 or intpower=8, random(2..3), 2)\n", "templateType": "anything", "name": "intrhs", "description": "

intsoln^intpower

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Questions to test if the student knows the inverse of fractional power or root (and how to solve equations that contain them).

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", "ungrouped_variables": ["dpower", "dc", "db", "ddenom"], "advice": "

a) Given $\\sqrt[\\var{intpower}]{x}=\\var{intrhs}$, we raise both sides to the power of $\\var{intpower}$ to get $x$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $\\sqrt[\\var{intpower}]{x}$ $=$ $\\var{intrhs}$ $\\left(\\sqrt[\\var{intpower}]{x}\\right)^{\\var{intpower}}$ $=$ $\\simplify[basic]{({intrhs})^{intpower}}$ $x$ $=$ $\\var{intsoln}$
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b) Given $\\simplify{{bxcoeff}y^(1/{bpower})+{bb}}=\\var{bc}$, we can rearrange the equation to get $y^\\frac{1}{\\var{bpower}}$ by itself and then we can raise both sides to the power of $\\var{bpower}$ to get $y$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $\\simplify{{bxcoeff}y^(1/{bpower})+{bb}}$ $=$ $\\var{bc}$ $\\simplify{{bxcoeff}y^(1/{bpower})}$ $=$ $\\simplify[basic]{{bc}-{bb}}$ $\\simplify{{bxcoeff}y^(1/{bpower})}$ $=$ $\\simplify{{bc-bb}}$ $y^\\frac{1}{\\var{bpower}}$ $=$ $\\simplify[!basic]{{bc-bb}/{bxcoeff}}$ $y^\\frac{1}{\\var{bpower}}$ $=$ $\\simplify{{bc-bb}/{bxcoeff}}$ $\\left(y^\\frac{1}{\\var{bpower}}\\right)^{\\var{bpower}}$ $=$ $\\simplify[basic]{({(bc-bb)/bxcoeff})^{bpower}}$ $y$ $=$ $\\var{bsoln}$
\n

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c) Given $\\displaystyle{\\simplify{(root(z+{db},{dpower}))/{ddenom}}}=\\var{dc}$, we can rearrange the equation to get $\\simplify{(root(z+{db},{dpower}))}$ by itself, then we can raise both sides to the power of $\\var{dpower}$, and finally rearrange to get $z$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $\\displaystyle{\\simplify{(root(z+{db},{dpower}))/{ddenom}}}$ $=$ $\\var{dc}$ $\\displaystyle{\\simplify{(root(z+{db},{dpower}))}}$ $=$ $\\simplify[basic]{{dc}*{ddenom}}$ $\\displaystyle{\\simplify{(root(z+{db},{dpower}))}}$ $=$ $\\var{dc*ddenom}$ $\\left(\\sqrt[\\var{dpower}]{\\simplify{z+{db}}}\\right)^\\var{dpower}$ $=$ $\\simplify[basic]{({dc*ddenom})^{dpower}}$ $\\simplify{z+{db}}$ $=$ $\\simplify[basic]{-{abs(dc*ddenom)}^{dpower}}$  $\\simplify[basic]{({abs(dc*ddenom)})^{dpower}}$  $\\simplify[basic]{({(dc*ddenom)})^{dpower}}$ $z$ $=$ $\\simplify[basic]{-{abs(dc*ddenom)}^{dpower}-{db}}$  $\\simplify[basic]{({abs(dc*ddenom)})^{dpower}-{db}}$  $\\simplify[basic]{({(dc*ddenom)})^{dpower}-{db}}$
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If  $\\sqrt[\\var{intpower}]{x}=\\var{intrhs}$, then $x=$ [[0]].

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If  $\\simplify{{bxcoeff}y^(1/{bpower})+{bb}}=\\var{bc}$, then $y=$ [[0]].

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For this question, if the answer was $\\left(\\frac{35}{11}\\right)^{11}-24$, then you could enter  (35/11)^(11)-24.

\n

If  $\\displaystyle{\\simplify{(root(z+{db},{dpower}))/{ddenom}}}=\\var{dc}$, then $z=$ [[0]].

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