// Numbas version: exam_results_page_options {"name": "Solve equations which include a single even power (e.g. x^even=blah)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [{"name": "a", "variables": ["powers", "intpower", "intrhs", "intsoln"]}, {"name": "b", "variables": ["bpower", "bsoln", "bxcoeff", "bb", "bc", "brhs"]}, {"name": "c", "variables": ["cpower", "cc", "cb", "cxcoeff"]}], "name": "Solve equations which include a single even power (e.g. x^even=blah)", "functions": {}, "tags": [], "rulesets": {}, "ungrouped_variables": ["dc", "db", "ddenom"], "extensions": [], "parts": [{"scripts": {}, "variableReplacementStrategy": "originalfirst", "useCustomName": false, "sortAnswers": false, "marks": 0, "variableReplacements": [], "customName": "", "prompt": "

If  $x^\\var{intpower}=\\var{intrhs}$, then $x=$ [[0]], or [[1]].

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(Please enter the smaller value in the first gap and the larger value on the second gap)

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

\n

(Please enter the smaller value in the first gap and the larger value on the second gap)

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For this question input the exact value by using a fractional power to indicate a root. For example, if the answer was $\\sqrt[3]{\\frac{35}{11}}$, then enter  (35/11)^(1/3).

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If  $\\simplify{{cxcoeff}z^{cpower}+{cb}}=\\var{cc}$, then $z=$ [[0]], or [[1]].

\n

(Please enter the smaller value in the first gap and the larger value on the second gap)

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For this question input the exact value by using a fractional power to indicate a root. For example, if the answer was $\\sqrt[3]{\\frac{35}{11}}$, then enter  (35/11)^(1/3).

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If  $\\displaystyle{\\simplify{((z+{db})^{bpower})/({ddenom})}}=\\var{dc}$, then $z=$ [[0]], or [[1]].

\n

(Please enter the smaller value in the first gap and the larger value on the second gap)

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Suppose you wanted to solve $x^2=4$, that is, you wanted the number that squares to give four. The number $2$ comes to mind, however, $-2$ is also a solution since $(-2)^2=(-2)\\times(-2)=4$.  Recall the product of two negatives is a positive, so the product of any even number of negative numbers is positive. This means when we find a positive solution to an equation like $x^4=10\\,000$ there will also be a negative solution, in particular, the solution to $x^4=10\\,000$ would be $x=\\pm\\sqrt[4]{10\\,000}=\\pm10$.

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a) Since the power in $x^\\var{intpower}=\\var{intrhs}$ is even we will take the plus or minus $\\var{intpower}$nd rd th  root to get two solutions.

\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
 $x^\\var{intpower}$ $=$ $\\var{intrhs}$ $\\sqrt[\\var{intpower}]{x^\\var{intpower}}$ $=$ $\\pm\\sqrt[\\var{intpower}]{\\var{intrhs}}$ $x$ $=$ $\\pm\\var{intsoln}$
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That is, $x$ equals $-\\var{intsoln}$ or $\\var{intsoln}$.

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b) Given $\\simplify{{bxcoeff}y^{bpower}+{bb}}=\\var{bc}$, we can rearrange the equation to get $y^\\var{bpower}$ by itself and then we can take the plus or minus $\\var{bpower}$nd rd th  root 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
 $\\simplify{{bxcoeff}y^{bpower}+{bb}}$ $=$ $\\var{bc}$ $\\simplify{{bxcoeff}y^{bpower}}$ $=$ $\\simplify[basic]{{bc}-{bb}}$ $\\simplify{{bxcoeff}y^{bpower}}$ $=$ $\\simplify{{bc-bb}}$ $y^\\var{bpower}$ $=$ $\\simplify[!basic]{{bc-bb}/{bxcoeff}}$ $y^\\var{bpower}$ $=$ $\\simplify{{bc-bb}/{bxcoeff}}$ $\\sqrt[\\var{bpower}]{y^\\var{bpower}}$ $=$ $\\pm\\sqrt[\\var{bpower}]{\\var{brhs}}$ $y$ $=$ $\\pm\\var{bsoln}$
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That is, $y$ equals $-\\var{bsoln}$, or $\\var{bsoln}$.

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c) Given $\\simplify{{cxcoeff}z^{cpower}+{cb}}=\\var{cc}$, we can rearrange the equation to get $z^\\var{cpower}$ by itself and then we can take the plus or minus $\\var{cpower}$nd rd th  root 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\n\n\n\n\n\n\n\n\n\n
 $\\simplify{{cxcoeff}z^{cpower}+{cb}}$ $=$ $\\var{cc}$ $\\simplify{{cxcoeff}z^{cpower}}$ $=$ $\\simplify[basic]{{cc}-{cb}}$ $\\simplify{{cxcoeff}z^{cpower}}$ $=$ $\\simplify{{cc-cb}}$ $z^\\var{cpower}$ $=$ $\\simplify[!basic]{{cc-cb}/{cxcoeff}}$ $z^\\var{cpower}$ $=$ $\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}}$ $\\sqrt[\\var{cpower}]{z^\\var{cpower}}$ $=$ $\\pm\\sqrt[\\var{cpower}]{\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}}}$ $z$ $=$ $\\pm\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}^{1/{cpower}}}$
\n

That is, $z$ equals $-\\sqrt[\\var{cpower}]{\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}}}$, or $\\sqrt[\\var{cpower}]{\\simplify[fractionNumbers]{{(cc-cb)/cxcoeff}}}$.

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d) Given $\\displaystyle{\\simplify{((z+{db})^{bpower})/({ddenom})}}=\\var{dc}$, we can rearrange the equation to get $\\simplify{(z+{db})^{bpower}}$ by itself, then we can take the plus or minus $\\var{bpower}$nd rd th  root of both sides to get $\\simplify{z+{db}}$ by itself, and then 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\n\n\n\n\n\n\n\n\n\n
 $\\displaystyle{\\simplify{((z+{db})^{bpower})/({ddenom})}}$ $=$ $\\var{dc}$ $\\simplify{(z+{db})^{bpower}}$ $=$ $\\simplify[basic]{{dc}*{ddenom}}$ $\\simplify{(z+{db})^{bpower}}$ $=$ $\\var{dc*ddenom}$ $\\sqrt[\\var{bpower}]{\\simplify{(z+{db})^{bpower}}}$ $=$ $\\pm\\sqrt[\\var{bpower}]{\\var{dc*ddenom}}$ $\\simplify{z+{db}}$ $=$ $\\pm\\sqrt[\\var{bpower}]{\\var{dc*ddenom}}$ $z$ $=$ $\\pm\\simplify{root({dc*ddenom},{bpower})-{db}}$ $z$ $=$ $\\pm\\simplify{{dc*ddenom}^(1/{bpower})-{db}}$
\n

That is, $z$ equals $-\\simplify{root({dc*ddenom},{bpower})-{db}}$, or $\\simplify{root({dc*ddenom},{bpower})-{db}}$.

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Questions to test if the student knows the inverse of an even power (and how to solve equations that contain a single power that is even).

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