// Numbas version: exam_results_page_options {"navigation": {"preventleave": true, "showresultspage": "oncompletion", "allowregen": true, "onleave": {"message": "", "action": "none"}, "reverse": true, "showfrontpage": true, "browse": true}, "showQuestionGroupNames": false, "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International", "notes": ""}, "allQuestions": true, "percentPass": 0, "name": "Loughborough workshop demo", "questions": [], "pickQuestions": 0, "timing": {"timedwarning": {"message": "", "action": "none"}, "allowPause": true, "timeout": {"message": "", "action": "none"}}, "feedback": {"allowrevealanswer": true, "showtotalmark": true, "showactualmark": true, "advicethreshold": 0, "showanswerstate": true}, "question_groups": [{"questions": [{"name": "4.8 equations with logs", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Jennifer Koenig", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/455/"}], "functions": {}, "ungrouped_variables": ["a", "b", "c", "d"], "tags": ["algebra", "logarithms"], "preamble": {"css": "", "js": ""}, "advice": "

Take an example, $log(x-3)=3$

\n

raise both sides so that they are powers to the base 10

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on the left hand side you are doing $10^{log(x-3)}$ leaves you with just x-3.

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$10^{log(x-3)}=10^3$

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$x-3=10^3$

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$x-3=1000$

\n

$x=1003$

\n

-------------

\n

Another example, $log(x+4)=-2$

\n

$10^{log(x+4)}=10^{-2}$

\n

$x+4=10^{-2}$

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$x+4=0.01$

\n

$x=-3.99$

", "rulesets": {}, "parts": [{"prompt": "

Solve for $x$:

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$log(x-\\var{a}) = \\var{b}$

\n

$x$ = [[0]]

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Solve for $x$:

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$log(x-\\var{c}) = \\var{d}$

\n

$x$ = [[0]]

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solve for x where logs are involved

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Algebra: Simplifying Expressions (2 unknowns)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Michael Proudman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/269/"}], "tags": ["algebra", "algebraic manipulation", "expanding brackets", "simplification", "simplifying an expression"], "variables": {"a": {"name": "a", "definition": "random(-6..6 except 0)", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "d": {"name": "d", "definition": "random(1..9 except [c,b])", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "c": {"name": "c", "definition": "random(-6..6 except [0,a])", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "b": {"name": "b", "definition": "random(1..9 except a)", "templateType": "anything", "description": "", "group": "Ungrouped variables"}}, "parts": [{"extendBaseMarkingAlgorithm": true, "prompt": "\n

Simplify:

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$\\simplify[std]{({a}x+{b}y)({c}x+{d}y)-({a}x+{d}y)({c}x+{b}y)}=\\;$[[0]]

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Do not include brackets in your answer.

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Input $xy$ as $x*y$.

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Do not include brackets in your answer.

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You can simplify the expression further.

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Simplify $(ax+by)(cx+dy)-(ax+dy)(cx+by)$. Answer is a multiple of $xy$.

"}, "variable_groups": [], "statement": "

Simplify the following expression.

", "rulesets": {"std": ["all", "!collectNumbers", "!noLeadingMinus"]}, "functions": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "preamble": {"css": "", "js": ""}, "advice": "\n

Expanding the brackets we have:

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\\[\\begin{eqnarray*}\\simplify[std]{({a}x+{b}y)({c}x+{d}y)-({a}x+{d}y)({c}x+{b}y)}&=&(\\simplify[std]{{a*c}x^2+{b*c+a*d}x*y+{b*d}y^2})-(\\simplify[std]{{a*c}x^2+{b*a+c*d}x*y+{b*d}y^2})\\\\&=&\\simplify[std]{{b*c+a*d}x*y-{b*a+c*d}x*y}\\\\&=&\\var{(a-c)*(d-b)}xy\\end{eqnarray*}\\]

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The formula for {nb} fractions is :

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\\[\\simplify[std]{a / b + {s1} * (c / d) = (ad + {s1} * bc) / bd}\\]

\n

and for this exercise we have $\\simplify{b=x+{b}}$, $\\simplify{d=x+{d}}$.
Hence we have:
\\[\\simplify[std]{{a} / ({a1}*x + {b}) + ({c} / ({a2}*x + {d})) = ({a} * ({a2}*x + {d}) + {c} * ({a1}*x + {b})) / (({a1}*x + {b}) * ({a2}*x + {d})) = ({a*a2 + c*a1} * x + {a * d + c * b}) / (({a1}*x + {b}) * ({a2}*x + {d}))}\\]

\n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "parts": [{"stepsPenalty": 1, "prompt": "

Express \\[\\simplify{{a} / ({a1}x + {b}) + ({c} / ({a2}x + {d}))}\\] as a single fraction.

\n

Input the fraction here: [[0]]

\n

Input your answer in the form $\\displaystyle \\frac{(ax+b)}{((cx+d)(ex+f))}$ with no other brackets than those shown.

\n

Click on Show steps if you need help. You will lose one mark if you do so.

\n

You will also find a video which goes through a similar example.

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Input as a single fraction.

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The formula for {nb} fractions is:

\n

\\[\\simplify[std]{a / b + {s1} * (c / d) = (ad + {s1} * bc) / bd}\\]

\n

and for this exercise we have $\\simplify{b=x+{b}}$, $\\simplify{d=x+{d}}$.

\n

Note that in your answer you do not need to expand the denominator.

\n

The following video goes through an example similar to this one.

\n

", "type": "information", "scripts": {}, "showCorrectAnswer": true, "marks": 0}], "type": "gapfill"}], "statement": "\n

Add the following two fractions together and express as a single fraction over a common denominator.

\n

 

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5/08/2012:

\n

Added tags.

\n

Added description.

\n

Changed to two questions, for the numerator and denomimator, rather than one as difficult to trap student input for this example. Still some ambiguity however.

\n

12/08/2012:

\n

Back to one input of a fraction and trapped input in Forbidden Strings.

\n

Used the except feature of ranges to get non-degenerate examples.

\n

Checked calculation.OK.

\n

Improved display in content areas.

\n

02/02/2013:

\n

Modified variable c so that the coefficient of $x$ in the answer is not 0. 

\n

Checked input again, OK.

", "description": "

Express $\\displaystyle \\frac{a}{x + b} \\pm  \\frac{c}{x + d}$ as an algebraic single fraction over a common denominator. 

\n

Contains a video in Show steps.

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Direct Factorisation

\n

If you can spot a direct factorisation then this is the quickest way to do this question.

\n

For this example we have the factorisation

\n

\\[\\simplify{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d} = ({a} * x + { -c}) * ({b} * x + { -d})}\\]

\n

Hence we find the roots:
\\[\\begin{eqnarray} x&=& \\simplify{{n1-n4}/{2*a*b}}\\\\ x&=& \\simplify{{n1+n4}/{2*a*b}} \\end{eqnarray} \\]

\n

Other Methods.

\n

There are several methods of finding the roots – here are the main methods.

\n

Finding the roots of a quadratic using the standard formula.

\n

We can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

\n

The two roots are

\n

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

\n

1. $\\Delta \\gt 0$. The roots are real and distinct

\n

2. $\\Delta=0$. The roots are real and equal. Their common value is $\\displaystyle -\\frac{b}{2a}$

\n

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

\n

For this question the discriminant of $\\simplify{{a*b}x^2+{-b*c-a*d}x+{c*d}}$ is $\\Delta = \\simplify[std]{{-n1}^2-4*{a*b*c*d}}=\\var{disc}$

\n

{rdis}.

\n

So the {rep} roots are:

\n

\\[\\begin{eqnarray} x = \\frac{\\var{n1} - \\sqrt{\\var{disc}}}{\\var{n3}} &=& \\frac{\\var{n1} - \\var{n4} }{\\var{n3}} &=& \\simplify{{n1 - n4}/ {n3}}\\\\ x = \\frac{\\var{n1} + \\sqrt{\\var{disc}}}{\\var{n3}} &=& \\frac{\\var{n1} + \\var{n4} }{\\var{n3}} &=& \\simplify{{n1 + n4}/ {n3}} \\end{eqnarray}\\]

\n

Completing the square.

\n

First we complete the square for the quadratic expression $\\simplify{{a*b}x^2+{-n1}x+{c*d}}$
\\[\\begin{eqnarray} \\simplify{{a*b}x^2+{-n1}x+{c*d}}&=&\\var{n5}\\left(\\simplify{x^2+({-n1}/{a*b})x+ {c*d}/{a*b}}\\right)\\\\ &=&\\var{n5}\\left(\\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2+ \\simplify{{c*d}/{a*b}-({-n1}/({2*a*b}))^2}\\right)\\\\ &=&\\var{n5}\\left(\\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2 -\\simplify{ {n2^2}/{4*(a*b)^2}}\\right) \\end{eqnarray} \\]
So to solve $\\simplify{{a*b}x^2+{-n1}x+{c*d}}=0$ we have to solve:
\\[\\begin{eqnarray} \\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2&\\phantom{{}}& -\\simplify{ {n2^2}/{4*(a*b)^2}}=0\\Rightarrow\\\\ \\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2&\\phantom{{}}&=\\simplify{ {n2^2}/{4*(a*b)^2}=({abs(n2)}/{2*a*b})^2} \\end{eqnarray}\\]
So we get the two {rep} solutions:
\\[\\begin{eqnarray} \\simplify{x+({-n1}/{2*a*b})}&=&\\simplify{-{abs(n2)}/{2*a*b}} \\Rightarrow &x& = \\simplify{({-abs(n2)+n1}/{2*a*b})}\\\\ \\simplify{x+({-n1}/{2*a*b})}&=&\\simplify{({abs(n2)}/{2*a*b})} \\Rightarrow &x& = \\simplify{({n1+abs(n2)}/{2*a*b})} \\end{eqnarray}\\]

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Solve for $x$:  \\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]
The least root is $x=\\;$ [[0]]. The greatest root is $x=\\;$ [[1]]

\n

You can get more information on solving a quadratic by clicking on Show steps. You will lose 1 mark if you do so.

\n

Enter the least root first. If the roots are equal, enter the root in both input boxes.

\n

Enter the roots as fractions or integers, not as decimals.

", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "steps": [{"prompt": "\n

Finding the roots by factorisation.

\n

Finding a factorisation of a quadratic $q(x)=a(x-r)(x-s)$ where $a$ is the coefficient of $x^2$ gives the roots $x=r$, $x=s$ immendiately.

\n

If you cannot find a factorisation then there are several other methods you can use.

\n

Using the formula for the roots.

\n

You can find the roots by using the formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

\n

The two roots are:

\n

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

\n

1. $\\Delta \\gt 0$. The roots are real and distinct

\n

2. $\\Delta=0$. The roots are real and equal. Their value is $\\displaystyle \\frac{-b}{2a}$

\n

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

\n

 

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Input numbers as fractions or integers not as a decimals.

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Input numbers as fractions or integers not as a decimals.

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Find the roots of the following quadratic equation.

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Solve for $x$: $\\displaystyle ax ^ 2 + bx + c=0$.

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