// Numbas version: finer_feedback_settings {"name": "Practice Test FY001", "metadata": {"description": "
NUMBAS practice exam for FY001 - Your score won't be counted for your mark.
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\n$\\simplify{{f1}x^2+{f2}*x+{f3}}=0$
\nThese are solved in the exact same way as the previous set of quadratics, but the 'roots' (solutions) are the same.
\nThis type of equation is described as having 'repeated roots'.
\nThe graphed functions would only touch the $x$-axis once either at their peak or trough.
\nLooking at the above equation, you may be able to instantly see that the brackets must both include the quantity $\\var{c}$.
\nIf not, don't worry, just go through the same process of trial and error as taught previously until you reach your solution of
\n$\\simplify{({a}x+{c})({b}x+{d})}$.
\nThen, noticing this can be further simplified, write the final simplification as:
\n$\\simplify{({a}x+{c})^2}$.
\n\n\n
\n
Note: If the $x^2$ term has a coefficient and simplifies to, for example, $(2x+4)(x+4)$, this is not defined as having repeated roots, since the solutions are:
\n$2x+4=0$ so $x=\\frac{-4}{2}=-2$
\nand
\n$x+4=0$ so $x=-4$
\nso are in fact different.
\n\n\n\nZero constants equations, using Part d:
\n$\\simplify{{f14}x^2+{f24}*x+{f34}}=0$
\nSome equations will not have a constant at the end, meaning one of the constants is zero.
\nIn this case, if there is still an $x$ term, you know one of the constants must still be non-zero.
\nYou must therefore, take out a factor of a number $\\times x$.
\nHere, the factor could be $\\simplify{{a4}x}$ or $\\simplify{{b4}x}$.
\nIt doesn't matter which you choose, but in this example we will use $\\simplify{{a4}x}$.
\nIn order to pull out this factor, the rest of the equation must be divided by $\\simplify{{a4}x}$:
\n$\\frac{\\simplify{{f14}x^2+{f24}*x+{f34}}}{\\simplify{{a4}*x}}=\\simplify{{f14}x/{a4}+{f24}/{a4}}$.
\nTherefore, the solution is:
\n$\\simplify{{a4}x*({f14}x/{a4}+{f24}/{a4})}$.
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\nQuadratics are defined as having 'repeated roots' if the both roots in the solution are the same. This can be written as $(x+a)(x+a)$, but can be simplified to $(x+a)^2$.
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\nRemember to use brackets appropriately and to use a * for multiplication for letters.
\n-----------------------------------
", "advice": "See Lecture 4.1 and Workshop 4.2 for background and examples.
\nFor background on fractions see 2.3. For background on algebraic manipulations see 2.6 and 3.3.
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\n$x^a \\times x^b = x^{a+b}$
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", "type": "jme", "customMarkingAlgorithm": "", "customName": "", "failureRate": 1, "useCustomName": false, "checkVariableNames": false, "answerSimplification": "all", "vsetRangePoints": 5, "checkingAccuracy": 0.001}, {"checkingType": "absdiff", "steps": [{"unitTests": [], "showCorrectAnswer": true, "adaptiveMarkingPenalty": 0, "extendBaseMarkingAlgorithm": true, "prompt": "Use the following law:
\n$x^a \\div x^b = x^{a-b}$
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\n$\\frac{ax^c}{bx^d}= \\frac{a}{b}x^{(c-d)}$
", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "scripts": {}, "variableReplacements": [], "customName": "", "useCustomName": false, "showFeedbackIcon": true, "marks": 0, "type": "information"}], "valuegenerators": [{"name": "h", "value": ""}], "stepsPenalty": 0, "scripts": {}, "variableReplacements": [], "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "showPreview": true, "marks": 1, "variableReplacementStrategy": "originalfirst", "unitTests": [], "showCorrectAnswer": true, "vsetRange": [0, 1], "answer": "{a}/{b}*h^({c}-{d})", "adaptiveMarkingPenalty": 0, "prompt": "This question differs from part f due to the brackets. Using principles of BODMAS, the brackets need to be expanded first.
\n$(4d)^\\var{g}$ expands to $4^\\var{g}d^\\var{g}$ and $(2d)^\\var{h}$ expands to $2^\\var{h}d^\\var{h}$
\nNow you are left with a simple division question as follows:
\n$\\frac{4^{\\var{g}}d^{\\var{g}}}{2^{\\var{h}}d^{\\var{h}}}$
\n\nUse the principle:
\n$\\frac{ax^c}{bx^d}= \\frac{a}{b}x^{(c-d)}$ to answer the question.
\n", "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "scripts": {}, "variableReplacements": [], "customName": "", "useCustomName": false, "showFeedbackIcon": true, "marks": 0, "type": "information"}], "valuegenerators": [{"name": "d", "value": ""}], "notallowed": {"strings": ["/"], "showStrings": false, "partialCredit": 0, "message": ""}, "stepsPenalty": 0, "scripts": {}, "variableReplacements": [], "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "showPreview": true, "marks": 1, "variableReplacementStrategy": "originalfirst", "unitTests": [], "showCorrectAnswer": true, "vsetRange": [0, 1], "answer": "(4^{g})/(2^{h})*d^{g-h}", "adaptiveMarkingPenalty": 0, "prompt": "Simplify each of the following expressions, giving your answer in its simplest form.
\nClick 'Show steps' for guidance on which index law is applicable.
", "ungrouped_variables": ["a", "b", "c", "d", "f", "g", "h", "j"], "advice": "Recall the laws of indices to help solve the problems:
\n$x^a \\times x^b = x^{a+b}$
\n$x^a \\div x^b = x^{a-b}$
\n$x^{-a} = \\frac{1}{x^a}$
\n$(x^a)^b = x^{ab}$
\n$(\\frac{x}{y})^a = \\frac{x^a}{y^a}$
\n$x^\\frac{a}{b} = (\\sqrt[b]{x})^{a}$
\n$x^0 = 1$
\n\nWorked Solutions:
\nPart a) $x^{(\\var{a}+\\var{b})}=\\simplify{x^{({a}+{b})}}$
\nPart b) $p^{(\\var{c}+\\var{d})}=\\simplify{p^{({c}+{d})}}$
\nPart c) $\\var{a}^\\var{f}\\times{k^{(\\var{b}\\times\\var{f})}}=\\simplify{{a}^{f}*k^{({b}*{f})}}$
\nPart d) $y^{((\\var{a}+\\var{b})/(\\var{a}\\times\\var{b}))}=y^{\\frac{\\simplify{{a}+{b}}}{\\simplify{{a}*{b}}}}$
\nPart e) $c^{(\\var{a}-\\var{b})}=c^\\simplify{({a}-{b})}$
\nPart f) $\\frac{\\var{a}}{\\var{b}}h^{\\var{c}-\\var{d}}=\\frac{\\var{a}}{\\var{b}}{\\simplify{h^{{c}-{d}}}}$
\nPart g) $\\frac{4^\\var{g}}{2^\\var{h}}\\times{d^{\\var{g}-\\var{h}}}=\\simplify{(4^{g})/(2^{h})*d^{g-h}}$
\nPart h) $\\frac{6^\\var{g}}{9^\\var{h}}\\times{p^{\\var{h}\\var{j}-\\var{g}\\var{f}}}=\\simplify{(6^{g})/(9^{h})*p^{h*j-g*f}}$
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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}\\]
YOU HAVE RUN OUT OF TIME
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