// Numbas version: exam_results_page_options {"percentPass": 0, "duration": 0, "name": "Ann's copy of NUMBAS - Quadratics", "timing": {"allowPause": true, "timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "
Questions involving various techniques for rearranging and solving quadratic expressions and equations
"}, "showstudentname": true, "feedback": {"showtotalmark": true, "advicethreshold": 0, "intro": "", "showactualmark": true, "allowrevealanswer": true, "showanswerstate": true, "feedbackmessages": []}, "navigation": {"onleave": {"action": "none", "message": ""}, "allowregen": true, "preventleave": true, "reverse": true, "browse": true, "showfrontpage": true, "showresultspage": "oncompletion"}, "question_groups": [{"pickingStrategy": "all-ordered", "pickQuestions": 1, "name": "Group", "questions": [{"name": "Factorising Quadratic Equations with $x^2$ Coefficients Greater than 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Hannah Aldous", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1594/"}], "advice": "As this question involves a number greater than $1$ before the $x^2$ value it has a factorised form $(ax+b)(cx+d)$.
\nTo find $a$ and $c$, we need to consider the factors of $\\var{a*c}$.
\nWe are already given that one of them is $\\var{a}$, so we know that the other one must be $\\var{c}$.
\nThis means our factorised equation must take the form
\n\\[(\\var{a}x+b)(\\var{c}x+d)=0\\text{.}\\]
\nThis expands to
\n\\[ \\simplify{ {a*c}x^2 + ({a}*d+{c}*b)x + a*b} \\]
\nSo we must find two numbers which add together to make $\\var{a*d+b*c}$, and multiply together to make $\\var{b*d}$.
\nTherefore $b$ and $d$ must satisfy
\n\\begin{align}
b \\times d &=\\var{b*d}\\\\
\\simplify{{a}d+{c}b} &= \\var{a*d+b*c}\\text{.}
\\end{align}
$b = \\var{b}$ and $d = \\var{d}$ satisfy these equations:
\n\\begin{align}
\\var{b} \\times \\var{d} &=\\var{b*d}\\\\
\\simplify[]{ {a}*{d} + {b}*{c} } &= \\var{a*d+b*c}
\\end{align}
So the factorised form of the equation is
\n\\[ \\simplify{({a}x+{b})({c}x+{d}) = 0} \\text{.}\\]
\n$\\simplify{({a}x+{b})({c}x+{d}) = 0}$ when either $\\var{a}x+\\var{b} = 0$ or $\\var{c}x+ \\var{d} = 0$.
\nSo the roots of the equation are $\\var[fractionnumbers]{-b/a}$ and $\\var[fractionnumbers]{-d/c}$.
\n", "statement": "", "variables": {"b": {"templateType": "anything", "name": "b", "description": "$b$ in $(ax+b)(cx+d)$
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", "group": "last q", "definition": "random(-8..8 except 0)"}}, "tags": ["coefficient of x^2 greater than 1", "Factorisation", "factorisation", "factorising", "factorising quadratic equations", "Factorising quadratic equations", "factorising quadratic equations with x^2 coefficients greater than 1", "taxonomy"], "ungrouped_variables": [], "functions": {}, "rulesets": {}, "metadata": {"description": "Factorise a quadratic equation where the coefficient of the $x^2$ term is greater than 1 and then write down the roots of the equation
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\n$\\simplify{{a*c}x^2+{a*d+b*c}x+{b*d}=0}\\text{.}$
\n$(\\var{a}x+\\phantom{.}$[[0]]$) ($[[1]]$x+\\phantom{.}$[[2]]$)\\; = 0$
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\nInput your answer as $x_1$ and $x_2$, where $x_1<x_2$.
\n$x_1=$ [[0]]
\n$x_2=$ [[1]]
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\n$\\simplify {x^2+{sml}x+{big}} = $ [[0]]
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\n\\[ \\simplify {x^2+{sml}x+{big}} = 0\\text{.} \\]
\n$x_1=$ [[0]]
\nor
\n$x_2=$ [[1]]
", "showFeedbackIcon": true, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "sortAnswers": true, "scripts": {}, "customMarkingAlgorithm": "", "gaps": [{"correctAnswerFraction": false, "mustBeReduced": false, "showCorrectAnswer": true, "unitTests": [], "showFeedbackIcon": true, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "minValue": "{-bits[0]-bits[1]}", "maxValue": "{-bits[0]-bits[1]}", "showFractionHint": true, "variableReplacements": [], "customMarkingAlgorithm": "", "adaptiveMarkingPenalty": 0, "customName": "", "type": "numberentry", "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "useCustomName": false, "correctAnswerStyle": "plain", "scripts": {}, "marks": 1}, {"correctAnswerFraction": false, "mustBeReduced": false, "showCorrectAnswer": true, "unitTests": [], "showFeedbackIcon": true, "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "allowFractions": false, "minValue": "{-bits[0]+bits[1]}", "maxValue": "{-bits[0]+bits[1]}", "showFractionHint": true, "variableReplacements": [], "customMarkingAlgorithm": "", "adaptiveMarkingPenalty": 0, "customName": "", "type": "numberentry", "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "useCustomName": false, "correctAnswerStyle": "plain", "scripts": {}, "marks": 1}], "marks": 0, "variableReplacements": [], "useCustomName": false, "adaptiveMarkingPenalty": 0, "customName": ""}], "advice": "Completing the square works by noticing that
\n\\[ (x+a)^2 = x^2 + 2ax + a^2 \\]
\nSo when we see an expression of the form $x^2 + 2ax$, we can rewrite it as $(x+a)^2-a^2$.
\nRewrite $x^2+\\var{sml}x$ as $\\simplify[basic]{ (x+{sml/2})^2 - {sml/2}^2}$.
\n\\begin{align}
\\simplify[basic]{x^2+{sml}x+{big}} &= \\simplify[basic]{(x+{sml/2})^2-{(sml/2)}^2+{big}} \\\\
&= \\simplify[basic]{(x+{sml/2})^2+{-(sml/2)^2+big}} \\text{.}
\\end{align}
We showed above that
\n\\[ \\simplify[basic]{x^2+{sml}x+{big}} = 0 \\]
\nis equivalent to
\n\\[ \\simplify[basic]{(x+{bits[0]})^2-{bits[1]^2}} = 0 \\text{.} \\]
\nWe can then rearrange this equation to solve for $x$.
\n\\begin{align}
\\simplify{(x+{bits[0]})^2-{(bits[1])^2} } &= 0 \\\\
(x+\\var{bits[0]})^2 &= \\var{bits[1]^2} \\\\
x+\\var{bits[0]} &= \\pm \\var{bits[1]} \\\\
x &= -\\var{bits[0]} \\pm \\var{bits[1]} \\\\[2em]
x_1 &= \\var{-bits[0]-bits[1]} \\text{,}\\\\
x_2 &= \\var{-bits[0]+bits[1]} \\text{.}
\\end{align}
We can rewrite quadratic equations given in the form $ax^2+bx+c$ as a square plus another term - this is called \"completing the square\".
\nThis can be useful when it isn't obvious how to fully factorise a quadratic equation.
", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Solve a quadratic equation by completing the square. The roots are not pretty!
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", "licence": "Creative Commons Attribution 4.0 International"}, "ungrouped_variables": ["all", "all2", "multiall", "big", "sml", "multiall2"], "type": "question", "advice": "Completing the square works by noticing that
\n\\[ (x+a)^2 = x^2 + 2ax + a^2 \\]
\nSo when we see an expression of the form $x^2 + 2ax$, we can rewrite it as $(x+a)^2-a^2$.
\nWe have $x^2+ \\var{evens1}x$, so we can replace it with $(x+\\var{evens1/2})^2-\\var{evens1/2}^2 = (x+\\var{evens1/2})^2 - \\var{evens1^2/4}$.
\nCheck that this is equivalent to the original expression by expanding the brackets:
\n\\begin{align}
(x+\\var{evens1/2})^2 - \\var{evens1^2/4} &= \\simplify[basic]{ x^2 + 2*{evens1/2}*x + {evens1/2}^2 - {evens1^2/4} } \\\\
&= x^2 + \\var{evens1}x \\text{.}
\\end{align}
Replace $x^2 + \\var{odds}x$ by $\\simplify[basic]{(x+{odds}/2)^2-({odds}/2)^2}$, to obtain
\n\\begin{align}
x^2 + \\var{odds}x &= \\simplify[basic]{(x+{odds}/2)^2-({odds}/2)^2} \\\\[0.5em]
&= \\simplify[basic]{ (x+{odds}/2)^2 - {odds^2}/4} \\text{.}
\\end{align}
Replace $x^2+\\var{evens2}x$ with $(x+\\var{evens2/2})^2 - \\var{evens2/2}^2$. Remember to keep the $\\var{evens2-evens1}$ term on the end!
\n\\begin{align}
\\simplify[basic]{ x^2 + {evens2}x + {evens2-evens1}} &= \\simplify[basic]{ (x+{evens2/2})^2 - {evens2/2}^2 + {evens2-evens1} } \\\\
&= \\simplify[basic]{ (x+{evens2/2})^2 + {evens2-evens1 - evens2^2/4} }
\\end{align}
First, notice that $\\simplify[basic]{ {all}x^2 + {multiall}x } = \\simplify[basic]{ {all}*( x^2 + {multiall/all} x)}$.
\nThen, we can replace $x^2 + \\var{multiall/all}x$ with $(x+\\var{multiall/all/2})^2 - \\var{multiall/all/2}^2$.
\n\\begin{align}
\\simplify[basic]{ {all}x^2 + {multiall}x + {odds3-evens2}} &= \\simplify[basic]{ {all}*( x^2 + {multiall/all} x) + {odds3-evens2}} & \\text{Extract the common factor of } \\var{all} \\\\
&= \\simplify[basic]{ {all}*( (x+{multiall/all/2})^2 - {multiall/all/2}^2) + {odds3-evens2} } & \\text{Complete the square}\\\\
&= \\simplify[basic]{ {all}*(x+{multiall/all/2})^2 - {all}*{(multiall/all/2)^2} + {odds3-evens2} } & \\text{Expand the constant term}\\\\
&= \\simplify[basic]{ {all}*(x+{multiall/all/2})^2 + {odds3-evens2 - (multiall/2)^2/all}} & \\text{Collect constants}
\\end{align}
We can rewrite quadratic equations given in the form $ax^2+bx+c$ as a square plus another term - this is called \"completing the square\".
\nThis can be useful when it isn't obvious how to fully factorise a quadratic equation.
\nRewrite the following expressions in the form \\[(x+b)^2-c\\] or \\[a(x+b)^2-c\\]
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It doesn't look like you've completed the square.
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It doesn't look like you've completed the square.
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", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "ungrouped_variables": ["a1", "a2", "a3", "a4", "b1", "b2", "b3", "b4", "x1", "p1", "p2", "x2", "a", "m"], "rulesets": {}, "advice": "The quadratic formula is
\n\\[x={\\frac {-b\\pm\\sqrt{b^2-4\\times a\\times c}}{2a}}\\text{.}\\]
\nFrom the equation, we can read off values for $a$, $b$ and $c$:
\n\\[\\begin{align}
a&=1\\text{,}\\\\
b&=\\var{a+m}\\text{,}\\\\
c&=\\var{a*m} \\text{.}
\\end{align}\\]
Substituting these values into the quadratic formula,
\n\\[x = \\frac {-\\var{a+m}\\pm\\sqrt{\\var{a+m}^2-4\\times \\var{a*m}}}{2}\\text{.}\\]
\nNote the $\\pm$ symbol in the formula. This means there are two solutions: one using $+$, the other using $-$.
\nThe two solutions are
\n\\[\\begin{align}
x_1&=\\var{m}\\text{,}\\\\
x_2&=\\var{a}\\text{.}
\\end{align}\\]
Note that the right-hand side of the given equation is not zero. We need to rewrite it in the form $ax^2+bx+c=0$:
\n\\[\\begin{align}
\\simplify{{a1}x^2+{a2}x+{a3}}&=\\var{a4}\\\\
\\simplify{{a1}x^2+{a2}x+{a3-a4}}&=0\\text{.}
\\end{align}\\]
Then we can read off values for $a$, $b$ and $c$:
\n\\[\\begin{align}
a&=\\var{a1}\\\\
b&=\\var{a2}\\\\
c&=\\var{a3-a4} \\text{.}
\\end{align}\\]
We can now substitute these values into the quadratic formula:
\n\\[x = {\\frac {-\\var{a2}\\pm\\sqrt{\\var{a2}^2-4\\times \\var{a1}\\times \\var{a3-a4}}}{2\\times\\var{a1}}}\\text{.}\\]
\nSo the two solutions are
\n\\[\\begin{align}
x_1&=\\var{dpformat(x1,2)}\\\\
x_2&=\\var{dpformat(x2,2)}\\text{.}
\\end{align}\\]
We first rearrange our equation into the form $ax^2+bx+c=0$:
\n\\[\\begin{align}
\\simplify{{b1}x^2+{b2}x+{b3}}&=0=\\var{b4}x\\\\
\\simplify{{b1}x^2+{b2-b4}x+{b3}}&=0\\text{.}
\\end{align}\\]
We can then read off the values for $a, b$ and $c$, which are
\n\\[\\begin{align}
a&=\\var{b1}\\text{,}\\\\
b&=\\var{b2-b4}\\text{,}\\\\
c&=\\var{b3}\\text{.}
\\end{align}\\]
Substituting these values into the quadratic formula,
\n\\[x = {\\frac {-\\var{b2-b4}\\pm\\sqrt{\\var{b2-b4}^2-4\\times \\var{b1}\\times \\var{b3}}}{2\\times\\var{b1}}},\\]
\nwe obtain solutions
\n\\[\\begin{align}
x_1&=\\var{dpformat(p1,2)}\\text{,}\\\\
x_2&=\\var{dpformat(p2,2)}\\text{.}
\\end{align}\\]
When quadratic equations can't be factorised, or if equations are difficult to factorise (perhaps if the coefficients are large), we need to use the quadratic formula to solve the equations.
\nUse the quadratic formula to calculate values for $x$ in these equations. Input the possible values as $x_1$ and $x_2$, where $x_1<x_2$.
", "parts": [{"scripts": {}, "variableReplacements": [], "type": "gapfill", "prompt": "$\\simplify{x^2+{a+m}x+{a*m}=0}$
\n$x_1=$ [[0]]
\n$x_2=$ [[1]]
An equation of the form
\n\\[ax^2+bx+c=0\\text{,}\\]
\n\ncan be solved using the quadratic formula
\n\\[x={\\frac {-b\\pm\\sqrt{b^2-4\\times a\\times c}}{2a}}\\text{.}\\]
\n"}], "variableReplacementStrategy": "originalfirst", "gaps": [{"allowFractions": false, "correctAnswerFraction": false, "precision": "2", "precisionPartialCredit": 0, "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "precisionMessage": "You have not given your answer to the correct precision.", "showFeedbackIcon": true, "precisionType": "dp", "minValue": "-a", "scripts": {}, "variableReplacementStrategy": "originalfirst", "type": "numberentry", "maxValue": "-a", "mustBeReduced": false, "marks": 1, "variableReplacements": [], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "strictPrecision": false, "showPrecisionHint": true}, {"allowFractions": false, "correctAnswerFraction": false, "precision": "2", "precisionPartialCredit": 0, "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "precisionMessage": "You have not given your answer to the correct precision.", "showFeedbackIcon": true, "precisionType": "dp", "minValue": "-m", "scripts": {}, "variableReplacementStrategy": "originalfirst", "type": "numberentry", "maxValue": "-m", "mustBeReduced": false, "marks": 1, "variableReplacements": [], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "strictPrecision": false, "showPrecisionHint": true}], "showFeedbackIcon": true, "marks": 0, "showCorrectAnswer": true}, {"scripts": {}, "variableReplacements": [], "type": "gapfill", "prompt": "\n$\\simplify{{a1}x^2+{a2}x+{a3}={a4}}$
\n$x_1=$ [[0]]
\n$x_2=$ [[1]]
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$\\simplify{{b1}x^2+{b2}x+{b3}={b4}x}$
\n$x_1=$ [[0]]
\n$x_2=$ [[1]]
", "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "gaps": [{"allowFractions": false, "correctAnswerFraction": false, "precision": "2", "precisionPartialCredit": 0, "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "precisionMessage": "You have not given your answer to the correct precision.", "showFeedbackIcon": true, "precisionType": "dp", "minValue": "p1", "scripts": {}, "variableReplacementStrategy": "originalfirst", "type": "numberentry", "maxValue": "p1", "mustBeReduced": false, "marks": 1, "variableReplacements": [], "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "strictPrecision": false, "showPrecisionHint": true}, {"allowFractions": false, "correctAnswerFraction": false, "precision": "2", "precisionPartialCredit": 0, "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "precisionMessage": "You have not given your answer to the correct precision.", "showFeedbackIcon": true, "precisionType": "dp", 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"extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Hannah Aldous", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1594/"}], "variable_groups": [{"variables": ["b", "c", "n2", "b_2", "c_2", "b_3"], "name": "part 2"}], "preamble": {"css": "", "js": ""}, "type": "question", "parts": [{"marks": 0, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "showCorrectAnswer": true, "variableReplacements": [], "scripts": {}, "gaps": [{"checkingtype": "absdiff", "type": "jme", "vsetrangepoints": 5, "vsetrange": [0, 1], "showpreview": true, "answer": "({-b_3/2-sqrt(b_3^2-4*c_2)/2})k", "scripts": {}, "showFeedbackIcon": true, "checkvariablenames": false, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "marks": 1, "checkingaccuracy": 0.001, "variableReplacements": [], "expectedvariablenames": []}, {"checkingtype": "absdiff", "type": "jme", "vsetrangepoints": 5, "vsetrange": [0, 1], "showpreview": true, "answer": "({-b_3/2+sqrt(b_3^2-4*c_2)/2})k", "scripts": {}, "showFeedbackIcon": true, "checkvariablenames": false, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "marks": 1, "checkingaccuracy": 0.001, "variableReplacements": [], "expectedvariablenames": []}], "stepsPenalty": 0, "showFeedbackIcon": true, "prompt": "Solve the equation $\\simplify {x^2+{b_3}*k*x+{c_2}k^2=0}$. Give your answer in terms of $k$. Assuming $k$ is positive, enter the lowest root first.
\n$x_1=$ [[0]]
\n$x_2=$ [[1]]
", "steps": [{"variableReplacementStrategy": "originalfirst", "type": "information", "showCorrectAnswer": true, "marks": 0, "variableReplacements": [], "scripts": {}, "showFeedbackIcon": true, "prompt": "The quadratic formula is
\n\\[{\\frac {-b\\pm\\sqrt{b^2-4\\times a\\times c}}{2a}}\\text{.}\\]
"}]}], "advice": "The quadratic formula is
\n\\[{\\frac {-b\\pm\\sqrt{b^2-4\\times a\\times c}}{2a}}\\text{.}\\]
\nWe can list our values for $a, b$ and $c$.
\n\\[\\begin{align}
a&=1\\\\
b&=\\var{b_3}k\\\\
c&=\\var{c_2}k^2
\\end{align}\\]
Then by substituting them into the quadratic formula, we obtain
\n\\[x=\\frac {-\\var{b_3}k\\pm\\sqrt{\\var{b_3}^2k^2-4\\times \\var{c_2}k^2}}{2}\\]
\nWe can then simplify this equation to
\n\\[\\begin{align}
x&=\\frac {-\\var{b_3}k\\pm k\\sqrt{\\var{b_3}^2-\\var{4c_2}}}{2}\\\\
\\end{align}\\]
\\[\\begin{align}
&=k\\left(\\frac{-\\var{b_3}}{2}\\pm \\frac {\\sqrt{\\var{b_3}^2-\\var{4c_2}}}{2}\\right)\\\\
\\end{align}\\]
\\[\\begin{align}
&=k\\left(-\\frac{\\var{b_3}}{2} \\pm\\frac{\\sqrt{\\var{(b_3^2-4c_2)}}}{2}\\right)\\\\
\\end{align}\\]
This means our possible values for $x$ in terms of $k$ are,
\n\\begin{align}
x_1 &= \\left( \\simplify[all,!noleadingminus,!collectnumbers,!simplifyfractions]{-{b_3}/2 - {sqrt(b_3^2-4c_2)}/2} \\right) k = \\var[fractionnumbers]{-b_3/2 - sqrt(b_3^2-4*c_2)/2}k \\\\
x_2 &= \\left( \\simplify[all,!noleadingminus,!collectnumbers,!simplifyfractions]{-{b_3}/2 + {sqrt(b_3^2-4c_2)}/2} \\right) k = \\var[fractionnumbers]{-b_3/2 + sqrt(b_3^2-4*c_2)/2}k
\\end{align}
The coefficients of the following equation involve the unknown value $k$. We can use the quadratic formula to find expressions for the values of $x$ in terms of $k$.
", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Factorise a quadratic expression of the form $x^2+akx+bk^2$ for $x$, in terms of $k$. $a$ and $b$ are constants.
"}, "variablesTest": {"condition": "", "maxRuns": 100}}, {"name": "Factorising Quadratic Equations with $x^2$ Coefficients of 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Hannah Aldous", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1594/"}], "metadata": {"description": "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.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Factorise the following quadratic equations.
\n", "variables": {"v1": {"name": "v1", "group": "Part A ", "definition": "random(1..10)", "templateType": "anything", "description": ""}, "v2": {"name": "v2", "group": "Part A ", "definition": "random(2..6 except v1)", "templateType": "anything", "description": ""}, "v4": {"name": "v4", "group": "Part A ", "definition": "random(1..10 except -v3)", "templateType": "anything", "description": ""}, "v5": {"name": "v5", "group": "Part A ", "definition": "random(2..10)", "templateType": "anything", "description": ""}, "v3": {"name": "v3", "group": "Part A ", "definition": "random(-8..-1)", "templateType": "anything", "description": ""}, "v6": {"name": "v6", "group": "Part A ", "definition": "-v5", "templateType": "anything", "description": ""}}, "tags": ["Factorisation", "factorisation", "factorising quadratic equations", "Factorising quadratic equations", "taxonomy"], "ungrouped_variables": [], "functions": {}, "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": ""}, "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{.}\\]
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\n[[0]] $=0$
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$\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$
\n[[0]] $=0$
\n", "variableReplacementStrategy": "originalfirst"}, {"scripts": {}, "variableReplacements": [], "customName": "", "useCustomName": false, "unitTests": [], "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"mustmatchpattern": {"pattern": "(`+-x^$n`? + `+- $n)`* * $z", "message": "
$\\simplify{x^2+{v5*v6}}=0$
\n[[0]] $=0$
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