// Numbas version: exam_results_page_options
{"name": "Quadratic equations", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Quadratic equations", "tags": [], "metadata": {"description": "<p>Tags: algebra, equations (linear), equations (quadratic)</p><p>Last updated Sep 2019</p>", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "<p>Solve the following quadratic equations by factorising and/or using the quadratic formula below. Make sure to write your answers in ascending order.</p>\n<p></p>\n<table style=\"height: 100px; background-color: #c9e0f0;\" border=\"2\">\n<tbody>\n<tr style=\"height: 100px;\">\n<td style=\"height: 100px; width: 490.422px;\">\n<p style=\"text-align: center;\"></p>\n<p style=\"text-align: center;\">if $\\color{red}{a}\\var{partcvariable}^2+\\color{blue}{b}\\var{partcvariable}+\\color{green}{c} = 0$, then:</p>\n<p style=\"text-align: center;\">$ $</p>\n<p style=\"text-align: center;\">$\\var{partcvariable}=\\dfrac{-(\\color{blue}{b})\\pm \\sqrt {(\\color{blue}{b})^2-4(\\color{red}{a})(\\color{green}{c})}}{2(\\color{red}{a})}$</p>\n</td>\n</tr>\n</tbody>\n</table>\n<p></p>\n<p></p>\n<p></p>", "advice": "<p></p>\n<p>a) $\\simplify{{PartCVariable}^2+({c1+c2}){PartCVariable}+{c1*c2}}=0$</p>\n<p>We try to solve the above equation by&nbsp;factorising it, that is, we want to write it as</p>\n<p>$(\\simplify{{PartCVariable}}+\\alpha)(\\simplify{{PartCVariable}}+\\beta)=0$</p>\n<p>for some numbers $\\alpha$ and $\\beta$; we need these two numbers to multiply to make $\\simplify[]{{c1*c2}}$&nbsp;and sum to make $\\simplify{{c1+c2}}$; we see that we can take $\\simplify{{c1}}$ and&nbsp;$\\simplify{{c2}}$, since $\\simplify[!basic]{{c1}*{c2}={c1*c2}}$ and&nbsp;$\\simplify[!basic]{{c1}+{c2}={c1+c2}}$. Then our factorisation is</p>\n<p>\\[\\simplify{{PartCVariable}^2+{c1+c2}{PartCVariable}+{c1*c2}}=\\simplify{({PartCVariable}+{c1})({PartCVariable}+{c2})}\\text{.}\\]</p>\n<p></p>\n<p>b)&nbsp;$\\simplify{{PartEVariable}^2+{e2}{PartEVariable}+{e3}}=0$</p>\n<p>We find we are unable to factorise the above equation into the form&nbsp;$(\\simplify{{PartEVariable}}+\\alpha)(\\simplify{{PartEVariable}}+\\beta)=0$, therefore we use the quadratic formula.&nbsp;Then,</p>\n<p>$\\simplify{{PartEVariable}}=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$,</p>\n<p>where $a=1$, $b=\\var{e2}$ and $c=\\var{e3}$. Therefore</p>\n<p>$\\simplify{{PartEVariable}}=\\frac{-(\\var{e2})\\pm\\sqrt{(\\var{e2})^2-(4\\times&nbsp;1\\times\\var{e3}})}{2\\times 1}=\\frac{\\simplify{-{e2}}\\pm\\sqrt{\\simplify[all,!collectNumbers]{{e2*e2}-{4*e3}}}}{2}=\\frac{\\simplify{-{e2}}\\pm\\sqrt{\\simplify{{e2*e2-4*e3}}}}{2}=\\frac{\\simplify[]{-{e2}}\\pm\\simplify[basicplus]{{edispsurd}}}{2}$, giving the solutions</p>\n<p>$\\simplify[basicplus]{{PartEVariable}={SolutionE1}}=\\var{anse1}$ to 2 d.p.</p>\n<p>and</p>\n<p>$\\simplify[basicplus]{{PartEVariable}={SolutionE2}}=\\var{anse2}$ to 2 d.p.</p>\n<p></p>\n<p></p>", "rulesets": {"basicplus": ["zeroPower", "unitPower", "unitFactor"]}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"PartFVariable": {"name": "PartFVariable", "group": "Unnamed group", "definition": "expression(VariableNames[random(0..length(VariableNames)-1)])", "description": "", "templateType": "anything", "can_override": false}, "a3": 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