// Numbas version: finer_feedback_settings
{"name": "10.h. Solving ax^2+bx+c=0 for arbitrary a", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "10.h. Solving ax^2+bx+c=0 for arbitrary a", "tags": [], "metadata": {"description": "<p>Factorising where a has multiple factor pairs</p>", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "<p>Solve the following equation for $x$.</p>\n<p>\\[\\simplify{{a}x^2+{b}x+{c}}=0\\]</p>", "advice": "<p>To solve a quadratic equation of the form \\[ ax^2+bx+c=0\\] by factorisation, we want to factorise the equation into the form \\[(mx+p)(nx+q)=0,\\] where $m\\times n=a$, $m\\times q+n\\times p=b$ and $p\\times q=c$.</p>\n<p>For the equation \\[\\simplify{{a}x^2+{b}x+{c}=0},\\]</p>\n<p>the best strategy is to test different factors of $\\var{a}$ and $\\var{c}$. Writing out a table of different factors is a good way to keep track of what guesses you have tried. In this case our polynomial can be factorised to \\[\\simplify{({a1}x+{Fact1})({a2}x+{Fact2})=0}.\\] This equation is satisfied when either \\[\\simplify{{a1}x+{Fact1}=0} \\quad \\text{or} \\quad \\simplify{{a2}x+{Fact2}=0}, \\] which implies the solutions to this quadratic equation are \\[ \\simplify[fractionNumbers]{x={-x1}} \\quad \\text{and} \\quad \\simplify[fractionNumbers]{x={-x2}} .\\]</p>\n<p><a href=\"https://www.sheffield.ac.uk/mash/maths-resources/first-steps#Quadratics\" target=\"_blank\" rel=\"noopener\">Use this link to find resources to help you revise how to solve quadratic equations.</a></p>", "rulesets": {}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"x1": {"name": "x1", "group": "Ungrouped variables", "definition": "Fact1/a1", "description": "<p>polynomial factorises as (a1x+Fact1)(a2x+Fact2), x1=Fact1/a1, x2=Fact2/a2</p>", "templateType": "anything", "can_override": false}, "x2": 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