Factorise three quadratic equations of the form $x^2+bx+c$.
\nThe first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.
"}, "advice": "Quadratic equations of the form
\n\\[x^2+bx+c=0\\]
\ncan be factorised to create an equation of the form
\n\\[(x+m)(x+n)=0\\text{.}\\]
\nWhen we expand a factorised quadratic expression we obtain
\n\\[(x+m)(x+n)=x^2+(m+n)x+(m \\times n)\\text{.}\\]
\nTo factorise an equation of the form $x^2+bx+c$, we need to find two numbers which add together to make $b$, and multiply together to make $c$.
\n\\[\\simplify{x^2+{v1+v2}x+{v1*v2}=0}\\]
\nWe need to find two values that add together to make $\\var{v1+v2}$ and multiply together to make $\\var{v1*v2}$.
\n\\[\\begin{align}
\\var{v1} \\times \\var{v2}&=\\var{v1*v2}\\\\
\\var{v1}+\\var{v2}&=\\var{v1+v2}\\\\
\\end{align} \\]
So the factorised form of the equation is
\n\\[\\simplify{(x+{v1})(x+{v2})}=0\\text{.}\\]
\n\nWe can begin factorising by finding factors of $\\var{v3*v4}$ that add together to give $\\var{v3+v4}$.
\n\\[\\begin{align}
\\var{v3} \\times \\var{v4}&=\\var{v3*v4}\\\\
\\var{v3}+\\var{v4}&=\\var{v3+v4}\\\\
\\end{align} \\]
So the factorised form of the equation is
\n\\[\\simplify{(x+{v3})(x+{v4})}=0\\text{.}\\]
\nWhen factorising the quadratic expression
\n\\[\\simplify{x^2+{v5*v6}=0}\\]
\nwe need to find two values that add together to make $0$ and multiply together to make $\\var{v5*v6}$.
\n\\begin{align}
\\var{v5} \\times \\var{v6}& = \\var{v5*v6}\\\\
\\simplify[]{ {v5} + {v6}} &= 0 \\\\
\\end{align}
So the factorised form of the equation is
\n\\[\\simplify{(x+{v5})(x+{v6})}=0\\text{.}\\]
", "parts": [{"scripts": {}, "prompt": "$\\simplify{x^2+{v1+v2}x+{v1*v2}=0}$
\n[[0]] $=0$
\n", "variableReplacementStrategy": "originalfirst", "marks": 0, "type": "gapfill", "showFeedbackIcon": true, "variableReplacements": [], "gaps": [{"checkingtype": "absdiff", "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "answer": "(x+{v1})(x+{v2})", "checkvariablenames": false, "vsetrange": [0, 1], "showCorrectAnswer": true, "scripts": {"mark": {"script": "question.is_factorised(this);", "order": "after"}}, "marks": 1, "checkingaccuracy": 0.001, "type": "jme", "expectedvariablenames": [], "showpreview": true, "variableReplacements": [], "showFeedbackIcon": true}], "showCorrectAnswer": true}, {"scripts": {}, "prompt": "
$\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$
\n[[0]] $=0$
\n", "variableReplacementStrategy": "originalfirst", "marks": 0, "type": "gapfill", "showFeedbackIcon": true, "variableReplacements": [], "gaps": [{"checkingtype": "absdiff", "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "answer": "(x+{v3})(x+{v4})", "checkvariablenames": false, "vsetrange": [0, 1], "showCorrectAnswer": true, "scripts": {"mark": {"script": "question.is_factorised(this);", "order": "after"}}, "marks": 1, "checkingaccuracy": 0.001, "type": "jme", "expectedvariablenames": [], "showpreview": true, "variableReplacements": [], "showFeedbackIcon": true}], "showCorrectAnswer": true}, {"scripts": {}, "prompt": "
$\\simplify{x^2+{v5*v6}}=0$
\n[[0]] $=0$
", "variableReplacementStrategy": "originalfirst", "marks": 0, "type": "gapfill", "showFeedbackIcon": true, "variableReplacements": [], "gaps": [{"checkingtype": "absdiff", "variableReplacementStrategy": "originalfirst", "vsetrangepoints": 5, "answer": "(x+{v5})(x+{v6})", "checkvariablenames": false, "vsetrange": [0, 1], "showCorrectAnswer": true, "scripts": {"mark": {"script": "question.is_factorised(this);", "order": "after"}}, "marks": 1, "checkingaccuracy": 0.001, "type": "jme", "expectedvariablenames": [], "showpreview": true, "variableReplacements": [], "showFeedbackIcon": true}], "showCorrectAnswer": true}], "tags": ["Factorisation", "factorisation", "Factorising quadratic equations", "factorising quadratic equations", "taxonomy"], "statement": "Factorise the following quadratic equations.
\n", "functions": {}, "variable_groups": [{"variables": ["v1", "v2", "v3", "v4", "v5", "v6"], "name": "Part A "}], "extensions": [], "ungrouped_variables": [], "rulesets": {}, "variables": {"v4": {"group": "Part A ", "name": "v4", "description": "", "templateType": "anything", "definition": "random(1..10 except -v3)"}, "v5": {"group": "Part A ", "name": "v5", "description": "", "templateType": "anything", "definition": "random(2..10)"}, "v3": {"group": "Part A ", "name": "v3", "description": "", "templateType": "anything", "definition": "random(-8..-1)"}, "v1": {"group": "Part A ", "name": "v1", "description": "", "templateType": "anything", "definition": "random(1..10)"}, "v6": {"group": "Part A ", "name": "v6", "description": "", "templateType": "anything", "definition": "-v5"}, "v2": {"group": "Part A ", "name": "v2", "description": "", "templateType": "anything", "definition": "random(2..6 except v1)"}}, "name": "Katherine's copy of Factorising Quadratic Equations with $x^2$ Coefficients of 1", "preamble": {"js": "question.is_factorised = function(part,penalty) {\n penalty = penalty || 0;\n if(part.credit>0) {\n // Parse the student's answer as a syntax tree\n var studentTree = Numbas.jme.compile(part.studentAnswer,Numbas.jme.builtinScope);\n\n // Create the pattern to match against \n // we just want two sets of brackets, each containing two terms\n // or one of the brackets might not have a constant term\n // or for repeated roots, you might write (x+a)^2\n var rule = Numbas.jme.compile('m_all(m_any(x,x+m_pm(m_number),x^m_number,(x+m_pm(m_number))^m_number))*m_nothing');\n\n // Check the student's answer matches the pattern. \n var m = Numbas.jme.display.matchTree(rule,studentTree,true);\n // If not, take away marks\n if(!m) {\n part.multCredit(penalty,'Your answer is not fully factorised.');\n }\n }\n}", "css": ""}, "type": "question", "contributors": [{"name": "Katherine Tomlinson", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1652/"}]}]}], "contributors": [{"name": "Katherine Tomlinson", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1652/"}]}