// Numbas version: exam_results_page_options {"name": "Maths and Stats Week 3 Knowledge Check", "metadata": {"description": "
This is a formative assessment to see how well you have understood the content from the first 3 weeks of the course. It will not count towards your final mark. If you get less than 40%, then you should take immediate steps to improve your knowledge, such as attending Study Clinics.
", "licence": "None specified"}, "duration": 0, "percentPass": "0", "showQuestionGroupNames": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", "", "", "", "", "", ""], "questions": [{"name": "Rounding numbers to a given number of significant figures", "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": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}], "type": "question", "statement": "", "variablesTest": {"condition": "", "maxRuns": "100"}, "variables": {"edig": {"description": "", "name": "edig", "group": "Ungrouped variables", "templateType": "anything", "definition": "repeat(random(1..9), 5)"}, "d1": {"description": "Random integer.
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\ni) $\\var{d1}$ rounded to 1 significant figure is: [[0]]
\nii) $\\var{d1}$ rounded to {sf} significant figures is: [[1]]
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\niii) $\\var{e1}$ rounded to 1 significant figure is: [[0]]
\niv) $\\var{e1}$ rounded to {sf} significant figures is: [[1]]
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"}, "preamble": {"css": "", "js": ""}, "advice": "The first thing to do when we are rounding numbers is to identify the last digit we are keeping.
\nWhen you're asked to round your answer to a number of significant figures, you need to decide whether to keep the last digit same (rounding down) or increase it by 1 (rounding up). If the following digit is less than 5 we round down and we round up when the next digit is 5 or more.
\nTo write it down in steps:
\nIt is important to keep in mind that if the digit we are increasing is 9, it becomes zero and we increase the previous digit instead. If this digit is 9 as well, we move along to the left side until we find a digit less than 9.
\nWhen asked to round to significant figures, it may seem tricky at first to identify the last digit we want to keep. If we round to one significant figure, we look at the first non-zero digit from left to right. If we round to $n$ significant figures, we want to keep all digits up to the $n$th number of places next to the last zero number. Once we identify this digit, the rounding method is the same as previously. It may be easier to understand with an example:
\n$0.02554$ rounded to 2 significant figures is $0.026$ as digits $25$ are the 2 significant figures we keep.
\nSimilarly, $52124$ rounded to 3 significant figures is $52100$ since digits $521$ are the 3 significant figures we keep.
\ni)
\nWe round $\\var{d1}$ to 1 significant figure. The first non-zero digit is $\\var{ddig[5]}$. The following digit is $\\var{ddig[4]}$ so we round updown to get $\\var{dpformat(siground(d1, 1), 0)}$.
\nii)
\nWe round $\\var{d1}$ to {sf} significant figures. The first non-zero digit is $\\var{ddig[5]}$. The second following digit is $\\var{ddig[4]}$, the third following digit is $\\var{ddig[3]}$ and the fourth following digit is $\\var{ddig[2]}$. The digit following the last digit we are keeping is $\\var{ddig[3]}$$\\var{ddig[2]}$$\\var{ddig[1]}$, so we round to get $\\var{sigformat(d1, sf)}$. These are our {sf} significant figures.
\n\n
i)
\nWe round $\\var{e1}$ to 1 significant figure. The first non-zero digit is $\\var{edig[4]}$, followed by $\\var{edig[3]}$. This is lower than 5 so we round downmore than 5 so we round up to get $\\var{sigformat(e1,1)}$.
\nii)
\nWe round $\\var{e1}$ to {sf} significant figures. The first non-zero digit is $\\var{edig[4]}$. The second following digit is $\\var{edig[3]}$, the third following digit is $\\var{edig[2]}$ and the fourth following digit is $\\var{edig[1]}$. The digit following the last digit we are keeping is $\\var{edig[2]}$$\\var{edig[1]}$$\\var{edig[0]}$, so we round to get $\\var{sigformat(e1, sf)}$. These are our {sf} significant figures.
\n"}, {"name": "Converting from standard index form to decimal.", "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": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}], "rulesets": {}, "functions": {}, "ungrouped_variables": ["A", "ran"], "metadata": {"description": "Given some numbers in standard index form, convert to decimal form.
", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "variable_groups": [], "advice": "When given a number in the form $A \\times 10^n$, we can think of $n$ as a number telling us how many places to move the decimal point.
\nWhen $n$ is positive, we move the decimal point to the right side, for example:
\n\\[ 1.5 \\times 10^3 = 1500.0 \\text{ .} \\]
\nWhen $n$ is negative, we move the decimal point to the left side, for example:
\n\\[ 1.5 \\times 10^{-3} = 0.0015 \\text{ .} \\]
\nWhen $n = 0$, we do not move the decimal point:
\n\\[ 1.5 \\times 10^0 = 1.5 \\text{ .}\\]
\n\n
In $\\var{A[0]} \\times 10^\\var{ran}$, $n = \\var{ran}$ and so we move the decimal point {ran} places to the right.
\n\\[\\var{A[0]} ⇒ \\var{precround((A[0] * 10^ran), 0)}\\]
\n\nIn $\\var{A[1]} \\times 10^\\var{-ran + 4}$, $n = \\var{-ran +4}$ and so we move the decimal point {ran -4} places to the left.
\n\\[\\var{A[1]} ⇒ \\var{A[1]*10^(-ran+4)}\\]
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\n", "customName": "", "gaps": [{"answer": "{A[1]*10^(-ran+4)}", "failureRate": 1, "extendBaseMarkingAlgorithm": true, "checkingAccuracy": 0.001, "valuegenerators": [], "variableReplacements": [], "unitTests": [], "customName": "", "checkingType": "absdiff", "vsetRange": [0, 1], "checkVariableNames": false, "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "type": "jme", "mustmatchpattern": {"message": "", "nameToCompare": "", "partialCredit": 0, "pattern": "$n"}, "showPreview": true, "vsetRangePoints": 5, "useCustomName": false, "showFeedbackIcon": true, "showCorrectAnswer": true, "marks": "0.5", "scripts": {}}], "useCustomName": false, "variableReplacements": [], "unitTests": [], "showFeedbackIcon": true, "showCorrectAnswer": true, "marks": 0, "scripts": {}}], "tags": ["conversion", "converting", "standard form", "standard index form", "taxonomy"], "variablesTest": {"maxRuns": 100, "condition": ""}}, {"name": "Working with standard index form ", "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": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}], "type": "question", "statement": "Working with numbers that are very large or very small can be tricky.
\nStandard form allows us to simplify these numbers, using powers of 10.
\n\n
Write the following in standard index form (for example, for $2.01\\times 10^5$ we would write 2.01*10^5
in the gap).
$\\var{A[2]*10^2} = $ [[0]]
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"}, "preamble": {"css": "", "js": ""}, "advice": "Converting from decimal to a standard form, we are looking for $A \\times 10^n$.
\nWe need make the first number ($A$) between 1 and 10, so we put the decimal place after the first non-zero digit.
\n\n
a)
\nIn $\\var{A[2]*10^2}$, the first non-zero digit is $\\var{siground(A[2] - 0.5, 1)}$ so we get $A = \\var{A[2]}$.
\nIf we moved the decimal place in $\\var{A[2]}$ so it matches our original number $\\var{A[2]*10^2}$, we would go 2 places to the right, so $n = 2$.
\n\n
b)
\nIn $\\var{A3dp*10^(-1)}$, the first non-zero digit is $\\var{siground(A3dp - 0.5, 1)}$ so we get $A = \\var{A3dp}$.
\nIf we moved the decimal place in $\\var{A3dp}$ so it matches our original number $\\var{A3dp*10^(-1)}$, we would go 1 place to the left, so $n = -1$.
\n\n
c)
\nIn $\\var{precround(A5dp*10^7,0)}$ the first non-zero digit is $\\var{siground(A5dp - 0.5, 1)}$ so we get $A = \\var{A5dp}$.
\nIf we moved the decimal place in $\\var{A5dp}$ so it matches our original number $\\var{precround(A5dp*10^7,0)}$, we would go 7 places to the right, so $n = 7$.
\n\n
In $\\var{small5}$ the first non-zero digit is {siground({{small5}*10^5} - 0.5, 1)} so we get $A = \\var{small5*10^5}$.
\nIf we moved the decimal place in $\\var{small5*10^5}$ so it matches our original number $\\var{small5}$, we would go 5 places to the left, so $n = -5$.
\n"}, {"name": "Multiplication and division of numbers in standard index form", "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": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}], "tags": ["division", "indices", "multiplication", "standard form", "standard index form", "taxonomy"], "metadata": {"description": "Multiply two numbers in standard form, then divide two numbers in standard form.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Simplify, leaving the result in standard index form.
", "advice": "We use the rules for multiplying and dividing powers.
\nTo multiply powers we add, for example $10^2 \\times 10^6 = 10^{(2+6)} = 10^{8}$.
\nTo divide powers we subtract, for example $10^2 \\div 10^6 = 10^{(2-6)} = 10^{-4}$.
\n\n
\\[ \\begin{align} (\\var{int} \\times 10^\\var{ran - 3} ) \\times (\\var{int - 11} \\times 10^\\var{ran}) &= \\var{int} \\times (\\var{int-11}) \\times 10^{(\\var{ran - 3} + \\var{ran})} \\\\&= (\\var{int*(int-11)}) \\times 10^{\\var{ran - 3 + ran}} \\end{align} \\]
\nWe want to write this number in standard form so we move decimal place one place to the left to get
\n\\[(\\var{int*(int-11)/10}) \\times 10^{(\\var{ran - 3 + ran} + 1)} = (\\var{int*(int-11)/10}) \\times 10^{\\var{ran - 2 + ran}} \\text{.}\\]
\nSimilarly,
\n\\[ \\begin{align} (\\var{int*int} \\times 10^\\var{ran -4} ) \\div (\\var{int} \\times 10^\\var{ran - 2}) &= \\var{int*int} \\div \\var{int} \\times 10^{(\\var{ran - 4} - \\var{ran -2})} \\\\&= \\var{int} \\times 10^{-2} \\text{.}\\end{align} \\]
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", "stepsPenalty": "1", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Molar concentration is the number of moles divided by the volume of solution in litres.
\nIn particular, if a {litres} L solution is made by mixing {number} moles of a solute then the molar concentration is calculated as follows:
\n\n$\\dfrac{\\text{number of moles}}{\\text{litres}}$ | \n$=$ | \n$\\dfrac{\\var{number}}{\\var{litres}}$ mol/L | \n
\n | $=$ | \n$\\var{onedec}$ mol/L (1 dec. pl) | \n
This question determines the students ability to understand the relationship between moles and concentration.
", "advice": "The formula mass of NaOH is 40 g/mol. That is, 40 g of NaOH dissolved into 1000ml (1 litre) of solvent will give a 1 Molar solution (1 M). We need only {vol} mL ({decimal(vol/1000)} L) at {molar} M so 40 x {decimal(vol/1000)} L x {molar} M = {correct} g.
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\n[[0]] moles
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\n[[0]] moles
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\n$\\text{volume of liquid (in litres)} \\times \\text{concentration (in mol/L)} = \\text{number of moles of substance}$.
\nHow many moles of glucose are there in $\\var{a}$L of a $\\var{0.25 * b}$M (mol/L) solution?
\nSolution:
\nPutting our numbers into the formula, we find that there are
\n$\\var{a} \\times \\var{0.25 * b} = \\var{a * 0.25 * b}$ moles
\nof glucose in $\\var{a}$L of a $\\var{0.25 * b}$M (mol/L) solution.
\nHow many moles of glucose are there in $\\var{25 * c}$mL of a $\\var{0.25 * d}$M (mol/L) solution?
\nSolution:
\nOur formula uses the volume of liquid in litres so first we have to convert $\\var{25 * c}$mL to a volume in litres. There are 1000ml in 1L so $\\var{25 * c}$mL is equal to
\n$\\dfrac{\\var{25 * c}}{1000} = \\var{25 * c / 1000}$L.
\nPutting our numbers into the formula, we find that there are
\n$\\begin{align}\\var{25 * c / 1000} \\times \\var{0.25 * d} & = \\var{(25 * c / 1000) * 0.25 * d} \\text{ moles} \\\\ & = \\var{precround(((25 * c / 1000) * 0.25 * d),2 )} \\text{ moles to 2 d.p.}\\end{align}$
\n\nof glucose in $\\var{25 * c}$mL of a $\\var{0.25 * d}$M (mol/L) solution.
", "functions": {}, "statement": "Answer the following questions. Please enter your answers as decimals, not as fractions. Give your answers to 2 decimal places.
\nIf you would like to see how to do this question, click on 'Reveal answers' at the bottom of the page.
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\n$x+2y=\\var{x+2y}$
\n$x^2+y^2=\\var{x^2+y^2}$
", "advice": "Rearranging the first equation gives $x=\\var{x+2y}-2y$
\nSubstituting this expression for $x$ into our second equation gives $(\\var{x+2y}-2y)^2+y^2=\\var{x^2+y^2}$
\nExpanding brackets gives $\\simplify[]{{(x+2y)^2}-{4*(x+2y)}}$ $y+4y^2+y^2=\\var{x^2+y^2}$
\nRearranging gives $\\simplify[]{5y^2-{4*(x+2y)}y-{x^2+y^2-(x+2y)^2}}=0$
\nSolving by factorising or using the quadratic formula gives $y=\\frac{\\var{4a} \\pm \\sqrt{\\var{16a^2+20(b-a^2)}}}{10}$
\nHence $y = \\var{y1}$ or $\\var{y2}$
\nand the corresponding values of $x$ are $\\var{x1}$ and $\\var{x2}$ respectively (found by substituting each value of $y$ into $x+2y=\\var{x+2y}$)
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\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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\n[[0]] $=0$
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\nThe first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Solve the following simultaneous equations.
", "advice": "To solve simultaneous equations, first rewrite the equations in matrix form of $Ab = C$, where the matrix $A$ will contain the coefficients of the equation, $b$ will contain the unknown values and $C$ will contain the constants.
\nWe rearrange $Ab = C$ to get $A^{-1}C = b$.
\nSo, we need to find the inverse of $A$ and multiple it with $C$. This will give us $b$, i.e. the values for $x$ and $y$.
\n\nFor the first question:
\n$A = \\simplify{{maA}}$ $b = \\begin{pmatrix}x\\\\y\\end{pmatrix}$ $C = \\begin{pmatrix}\\var{a1}\\\\ \\var{a2}\\end{pmatrix}$
\nTo find the inverse of $A$, switch the elements on the leading diagonal, change the signs on the non-leading diagonal, and divide all elements by the determinant.
\nThe determinant of $A$ is $|A| = (\\var{maA[0][0]}\\times\\var{maA[1][1]}) - (\\var{maA[0][1]}\\times\\var{maA[1][0]}) = \\var{detA}$
\nThe inverse of $A$ is $A^{-1} = \\dfrac{1}{\\var{detA}}\\times\\begin{pmatrix}
\\var{maAA[0][0]} & \\var{maAA[0][1]} \\\\
\\var{maAA[1][0]} & \\var{maAA[1][1]}
\\end{pmatrix} =
\\begin{pmatrix}
\\frac{\\var{maAA[0][0]}}{\\var{detA}} & \\frac{\\var{maAA[0][1]}}{\\var{detA}} \\\\
\\frac{\\var{maAA[1][0]}}{\\var{detA}} & \\frac{\\var{maAA[1][1]}}{\\var{detA}}
\\end{pmatrix}$
Rearranging for $A^{-1}C = b$:
\n$A^{-1}\\times C =
\\begin{pmatrix}
\\frac{\\var{maAA[0][0]}}{\\var{detA}} & \\frac{\\var{maAA[0][1]}}{\\var{detA}} \\\\
\\frac{\\var{maAA[1][0]}}{\\var{detA}} & \\frac{\\var{maAA[1][1]}}{\\var{detA}}
\\end{pmatrix}
\\times\\begin{pmatrix}\\var{a1}\\\\ \\var{a2}\\end{pmatrix} =
\\begin{pmatrix}
\\frac{\\var{maAA[0][0]}}{\\var{detA}}\\times\\var{a1} + \\frac{\\var{maAA[0][1]}}{\\var{detA}}\\times\\var{a2} \\\\
\\frac{\\var{maAA[1][0]}}{\\var{detA}}\\times\\var{a1} + \\frac{\\var{maAA[1][1]}}{\\var{detA}}\\times\\var{a2}
\\end{pmatrix} =
\\begin{pmatrix}\\var{ax}\\\\ \\var{ay}\\end{pmatrix}$
Therefore $x = \\var{ax}$ and $y = \\var{ay}$.
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\n\n$x =$ [[0]]
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\nrebelmaths
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\n\n$ V=\\frac{\\var{a}S}{S+\\var{b}}$
\n\nto write a fraction you type (numerator)/(denominator)
\nS=[[0]]
\n", "extendBaseMarkingAlgorithm": true}], "rulesets": {}, "tags": [], "advice": "start by multiplying both sides by the denominator
\nfor example if you have $V=\\frac{5S}{S+12}$ then multiply both sides by $(S+12)$
\nthis gives: $V(S+12)=\\frac{5S}{S+12} (S+12) $
\nthe (S+12) term on the right hand side cancels out to give: $V(S+12)=5S$
\nnow expand out the brackets: $VS+12V=5S$
\nthen collect the like terms, you want to get all the terms with S in them onto one side, so subtract VS from both sides:
\n$VS-VS+12V=5S-VS$
\nthis becomes $12V=5S-VS$
\nnow you can factorise the right hand side: $12V=S(5-V)$
\nthen divide both sides by (5-V) to leave S on its own: $\\frac{12V}{5-V}=S$
\n"}, {"name": "Laws of Indices", "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": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}], "tags": ["indices", "laws of indices", "powers", "taxonomy"], "metadata": {"description": "This question aims to test understanding and ability to use the laws of indices.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Using the laws of indices, simplify each expression down to its simplest form. Recall that $a^{0} = 1$ for any number $a$.
", "advice": "Here we are using the rule of indices: $a^m \\times a^n = a^{m+n}$.
\nUsing this rule,
\n\\[
\\begin{align}
a^\\var{x} \\times a^\\var{y}\\ &= a^\\simplify[all, !collectNumbers]{{x}+{y}}\\\\
&= a^\\var{x+y}.
\\end{align}
\\]
We are asked to find $\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}$.
\nNotice there is a constant in front of each of the terms.
\nTo do this, write the product out explicitly, as
\n\\[\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q} = \\var{c} \\times \\var{d} \\times a^\\var{p} \\times a^\\var{q}.\\]
\nWe know that $\\var{c} \\times \\var{d} = \\var{c*d}$, and using the rule of indices: $a^\\var{p} \\times a^\\var{q} = a^\\var{p+q}$.
\nTherefore:
\n\\begin{align}
\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}&= \\var{c*d} \\times a^\\var{p+q} \\\\
&= \\simplify{{c*d}*a^{p+q}}.
\\end{align}
Here we are using: $a^m \\div a^n = a^{m-n}$.
\nWe are asked to simplify the expression, $\\displaystyle\\simplify{{b}*a^{x}/({g}*a^{y})}$.
\nTo do this, we just have to use the previously mentioned rule of indices. We write this out explicity as
\n\\[\\simplify{{b}*a^{x}/({g}*a^{y})} = \\simplify{{b}/{g}} \\times \\simplify{a^{x}/(a^{y})}.\\]
\nUsing rules of indices,
\n\\begin{align} \\frac{a^\\var{x}}{a^\\var{y}} &= a^\\var{x} \\div a^\\var{y}\\\\
&= a^\\simplify[all, !collectNumbers]{{x}-{y}}\\\\
&= a^\\var{x-y}.
\\end{align}
Therefore,
\n\\begin{align}
\\frac{\\var{b}a^\\var{x}}{\\var{g}a^\\var{y}} &= \\simplify{{b}/{g}} \\times \\simplify{a^{{x}-{y}}}\\\\
&= \\simplify{{b}/{g}*a^{x-y}}.
\\end{align}
Alternatively,
\nUsing the rule of indices: $a^{-m} = \\displaystyle\\frac{1}{a^{m}}$, we can rewrite the question as:
\n\\begin{align}
\\frac{\\var{b}a^\\var{x}}{\\var{g}a^\\var{y}} &= \\simplify{{b}/{g}} \\times \\frac{a^\\var{x}}{a^\\var{y}}\\\\
&= \\simplify{{b}/{g}} \\times a^\\var{x} \\times a^{-\\var{y}}.
\\end{align}
And then using the rule: $a^m \\times a^n = a^{m+n}$, this becomes:
\n\\begin{align}
\\simplify{{b}/{g}} \\times a^\\var{x} \\times a^{-\\var{y}} &= \\simplify{{b}/{g}} \\times a^\\simplify[all,!collectNumbers]{{x}+(-{y})}\\\\
&= \\simplify{{b}/{g}*a^{x-y}}.
\\end{align}
The question asks us to simplify $(\\simplify{{c}*a^{p}})^{\\var{q}}$.
\nTo do this we use the rules:
\n\\[(a^{m})^{n} = a^{mn},\\]
\n\\[(ab)^m = a^mb^m.\\]
\nWe can then expand the equation as
\n\\[(\\simplify{{c}*a^{p}})^{\\var{q}}= \\var{c}^{\\var{q}} \\times (a^{\\var{p}})^{\\var{q}}.\\]
\nThen using the rule of indices mentioned previously,
\n\\[
\\begin{align}
(\\simplify{{c}*a^{p}})^{\\var{q}}&= \\simplify{{c}^{q}} \\times a^\\var{p*q}\\\\
&= \\simplify{{c}^{q}*a^{p*q}}.
\\end{align}
\\]
The question asks us to simplify $\\sqrt[\\var{d}]{\\var{x}^\\var{d}a}$.
\nTo do this we use the rules:
\n\\[a^\\frac{1}{m} = \\sqrt[m]{a},\\]
\n\\[(ab)^m = a^mb^m.\\]
\nWe can expand the expression as follows:
\n\\[
\\begin{align}
\\sqrt[\\var{d}]{a} &= (\\simplify{a})^\\frac{1}{\\var{d}}\\\\
&= a^\\frac{1}{\\var{d}}.
\\end{align}
\\]
The question requires us to simplify $\\sqrt[\\var{c}]{a^\\var{q}}$.
\nHere, we use the rule of indices: $a^\\frac{n}{m} = \\sqrt[m]{a^n}$, allowing us to expand the expression as follows:
\n\\[
\\begin{align}
\\sqrt[\\var{c}]{\\simplify{a^{q}}} &= \\simplify[fractionnumbers,all]{(a^{q})^{{1}/{{c}}}}\\\\
&= \\simplify[fractionnumbers,all]{a^{{q}/{c}}}.
\\end{align}
\\]
Used in part c
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\nFinding a factorisation of a quadratic $q(x)=a(x-r)(x-s)$ where $a$ is the coefficient of $x^2$ gives the roots $x=r$, $x=s$ immediately.
\nIf you cannot find a factorisation then there are several other methods you can use.
\nUsing the formula for the roots.
\nYou can find the roots by using the formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$
\nThe 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 $\\Delta=b^2-4ac$
1. $\\Delta \\gt 0$. The roots are real and distinct
\n2. $\\Delta=0$. The roots are real and equal. Their value is $\\displaystyle \\frac{-b}{2a}$
\n3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.
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Solve for $x$:
\n\\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]
\n$x=$ [[0]] or [[1]].
\nYou can get more information on solving a quadratic by clicking on Show steps. You will lose 1 mark if you do so.
\nEnter the roots as fractions or integers, not as decimals.
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", "tags": ["Algebra", "algebra", "checked2015", "Factorisation", "factorisation", "find roots of a quadratic equation", "Quadratic formula", "quadratic formula", "quadratics", "roots of a quadratic equation", "solving a quadratic equation", "Solving equations", "solving equations", "steps", "Steps"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Solve for $x$: $\\displaystyle ax ^ 2 + bx + c=0$.
\nEntering the correct roots in any order is marked as correct. However, entering one correct and the other incorrect gives feedback stating that both are incorrect.
"}, "advice": "\n\tDirect Factorisation
\n\tIf you can spot a direct factorisation then this is the quickest way to do this question.
\n\tFor this example we have the factorisation
\n\t\\[\\simplify{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d} = ({a} * x + { -c}) * ({b} * x + { -d})}\\]
\n\tHence we find the roots:
\\[\\begin{eqnarray} x&=& \\simplify{{n1-n4}/{2*a*b}}\\\\ x&=& \\simplify{{n1+n4}/{2*a*b}} \\end{eqnarray} \\]
Other Methods.
\n\tThere are several methods of finding the roots – here are the main methods.
\n\tFinding the roots of a quadratic using the standard formula.
\n\tWe can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$
\n\tThe two roots are
\n\t\\[ 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 $\\Delta=b^2-4ac$
1. $\\Delta \\gt 0$. The roots are real and distinct
\n\t2. $\\Delta=0$. The roots are real and equal. Their common value is $\\displaystyle -\\frac{b}{2a}$
\n\t3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.
\n\tFor 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\t{rdis}.
\n\tSo the {rep} roots are:
\n\t\\[\\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\tCompleting the square.
\n\tFirst 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}\\]