Input numbers as fractions or integers not as a decimals.
", "strings": ["."], "partialCredit": 0}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "answersimplification": "std", "type": "jme", "showCorrectAnswer": true, "marks": 1, "vsetrangepoints": 5}, {"answer": "{n1+n4}/{2*a*b}", "vsetrange": [0, 1], "checkingaccuracy": 0.0001, "checkvariablenames": false, "expectedvariablenames": [], "notallowed": {"showStrings": false, "message": "Input numbers as fractions or integers not as a decimals.
", "strings": ["."], "partialCredit": 0}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "answersimplification": "std", "type": "jme", "showCorrectAnswer": true, "marks": 1, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "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]]
You can get more information on solving a quadratic by clicking on Show steps. You will lose 1 mark if you do so.
\nEnter the least root first. If the roots are equal, enter the root in both input boxes.
\nEnter the roots as fractions or integers, not as decimals.
", "steps": [{"type": "information", "prompt": "Finding the roots by factorisation.
\nFinding a factorisation of a quadratic $q(x)=a(x-r)(x-s)$ where $a$ is the coefficient of $x^2$ gives the roots $x=r$, $x=s$ immendiately.
\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.
\n", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "stepsPenalty": 1}], "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "
Find the roots of the following quadratic equation.
", "tags": ["algebra", "checked2015", "factorisation", "find roots of a quadratic equation", "mas1601", "MAS1601", "quadratic formula", "quadratics", "roots of a quadratic equation", "solving a quadratic equation", "solving a quadratic equation ", "solving equations", "Solving equations", "Steps", "steps"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "5/08/2012:
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", "licence": "Creative Commons Attribution 4.0 International", "description": "Solve for $x$: $\\displaystyle ax ^ 2 + bx + c=0$.
"}, "advice": "Direct Factorisation
\nIf you can spot a direct factorisation then this is the quickest way to do this question.
\nFor 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})}\\]
\nHence we find the roots:
\\[\\begin{eqnarray} x&=& \\simplify{{n1-n4}/{2*a*b}}\\\\ x&=& \\simplify{{n1+n4}/{2*a*b}} \\end{eqnarray} \\]
Other Methods.
\nThere are several methods of finding the roots – here are the main methods.
\nFinding the roots of a quadratic using the standard formula.
\nWe can use the following 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 common 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.
\nFor 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}.
\nSo 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}\\]
\nCompleting the square.
\nFirst 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}\\]