// Numbas version: finer_feedback_settings {"name": "Francis 's copy of Solving a quadratic equation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "c", "b", "d", "f", "s3", "s2", "s1", "n4", "n2", "disc", "rdis", "n1", "c1", "n3", "rep", "n5", "d1"], "name": "Francis 's copy of Solving a quadratic equation", "tags": ["algebra", "factorisation", "Factorisation", "find roots of a quadratic equation", "quadratic formula", "quadratics", "roots of a quadratic equation", "solving a quadratic equation", "solving equations", "Solving equations"], "preamble": {"css": "", "js": ""}, "advice": "

There are different ways to solve a quadratic equation.

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If you can see how to factorise the quadratic expression then this should be the quickest method.

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For this example we have

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

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So either \\[\\simplify{ ({a} * x + { -c}) =0} \\  \\text{or} \\ \\ \\simplify{ ({b} * x + { -d})}=0\\]

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So the roots are:
\\[x= \\simplify{{n1-n4}/{2*a*b}} \\ \\ \\text{and} \\ \\  x= \\simplify{{n1+n4}/{2*a*b}}  \\]

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If you cannot see a factorisation then you can use the quadratic formula:

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\\[ x = \\frac{-b \\pm \\sqrt{b^2-4ac}}{2a}\\]

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Remember that there are three possible types of solution, depending upon the value of the discriminant $\\Delta=b^2-4ac$.

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

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

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3. $\\Delta \\lt 0$. There are no real roots. (The roots are complex)

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

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So the {rep} roots are:

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\\[ x = \\frac{\\var{n1} - \\sqrt{\\var{disc}}}{\\var{n3}} = \\frac{\\var{n1} - \\var{n4} }{\\var{n3}} = \\simplify{{n1 - n4}/ {n3}} \\ \\ \\text{and} \\ \\  x = \\frac{\\var{n1} + \\sqrt{\\var{disc}}}{\\var{n3}} = \\frac{\\var{n1} + \\var{n4} }{\\var{n3}} = \\simplify{{n1 + n4}/ {n3}} \\]

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You could also use the method of completing the square.

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First we write:
\\[\\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 we need 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}\\]
and 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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See also these Mathcentre leaflets on solving quadratic equations.

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 \\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]

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

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Enter the root with the lower value first. If the equation has two equal roots then enter this value in both input boxes.

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The root with the lower value is $x=\\;$ [[0]]. The root with the higher value is $x=\\;$ [[1]]

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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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