// Numbas version: finer_feedback_settings {"name": "Solving a perfect square by factorising", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Solving a perfect square by factorising", "tags": ["binomial", "factorisation", "Factorisation", "factorise", "perfect square", "quadratic", "quadratics", "solving", "square"], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "", "rulesets": {}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"a": {"name": "a", "group": "Ungrouped variables", "definition": "random(2..12)", "description": "", "templateType": "anything", "can_override": false}, "aa": {"name": "aa", "group": "Ungrouped variables", "definition": "a^2", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "shuffle(2..12)[0..3]", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "random(-12..12 except 0)", "description": "", "templateType": "anything", "can_override": false}, "d": {"name": "d", "group": "Ungrouped variables", "definition": "random(-12..-2)", "description": "", "templateType": "anything", "can_override": false}, "bb": {"name": "bb", "group": "Ungrouped variables", "definition": "b^2", "description": "", "templateType": "anything", "can_override": false}, "mid": {"name": "mid", "group": "Ungrouped variables", "definition": "2*a*b", "description": "", "templateType": "anything", "can_override": false}, "gg": {"name": "gg", "group": "Ungrouped variables", "definition": "g^2", "description": "", "templateType": "anything", "can_override": false}, "g": {"name": "g", "group": "Ungrouped variables", "definition": "gcd(a,b)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "127"}, "ungrouped_variables": ["a", "b", "aa", "bb", "mid", "g", "gg", "c", "d"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "
Solve the following quadratic by factorisation:
\n| \n | $\\simplify{{aa}x^2+{mid}x+{bb}}$ | \n$=$ | \n0 | \n
| $\\Longrightarrow$ | \n[[0]] | \n$=$ | \n0 | \n
| $\\Longrightarrow$ | \n$x$ | \n$=$ | \n[[1]] | \n
Note: In the first gap, enter the quadratic in factored form.
\nNote: In the second gap, there should only be one solution.
", "stepsPenalty": 0, "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Recall that $(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$ is called a perfect square.
\nIn fact, $\\simplify{{aa}x^2+{mid}x+{bb}}$ is also a perfect square, since $\\var{aa}x^2=(\\var{a}x)^2$, $\\var{bb}=\\simplify{({b})^2}$, and $\\var{mid}x=2(\\var{a}x)(\\var{b})$.
\nThat is, $\\simplify{{aa}x^2+{mid}x+{bb}}=\\simplify{({a}x+{b})^2}$.
\nNow, since $\\simplify{({a}x+{b})^2}=0$ we can take the (plus or minus) square root of both sides to get $\\simplify{({a}x+{b})}=0$. We can then solve this for $x$ to find $x=\\simplify{-{b}/{a}}$.
\n\nEach bracket has a common factor of $\\var{g}$, so we can move both of them to the front, to write
\n$\\simplify{{aa}x^2+{mid}x+{bb}}=\\simplify{{gg}({a/g}x+{b/g})^2}$.
\nNow, using the null factor law, since $\\simplify{{gg}({a/g}x+{b/g})^2}=0$ we must have $\\simplify{{a/g}x+{b/g}}=0$. We can then solve this for $x$ to find $x=\\simplify{{-b}/{a}}$.
\nEnsure you factorise the expression.
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\n| \n | $\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}$ | \n$=$ | \n0 | \n
| $\\Longrightarrow$ | \n[[0]] | \n$=$ | \n0 | \n
| $\\Longrightarrow$ | \n$x$ | \n$=$ | \n[[1]] | \n
Note: In the first gap, enter the quadratic in factored form.
\nNote: In the second gap, there should only be one solution.
\n", "stepsPenalty": 0, "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Recall that $(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$ is called a perfect square.
\nIn fact, $\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}$ is also a perfect square, since
\nThat is,
\n$\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}=\\simplify{({c[0]}/{c[1]}x+{d}/{c[2]})^2}$.
\n\nNow, since $\\simplify{({c[0]}/{c[1]}x+{d}/{c[2]})^2}=0$ we can take the (plus or minus) square root of both sides to get $\\simplify{{c[0]}/{c[1]}x+{d}/{c[2]}}=0$. We can then solve this for $x$ to find $x=\\simplify{{-d*c[1]}/{c[0]*c[2]}}$.
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