There are different ways to solve a quadratic equation.
\nIf you can see how to factorise the quadratic expression then this should be the quickest method.
\nFor this example we have
\n\\[\\simplify{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d} = ({a} * x + { -c}) * ({b} * x + { -d})}\\]
\nSo either \\[\\simplify{ ({a} * x + { -c}) =0} \\ \\text{or} \\ \\ \\simplify{ ({b} * x + { -d})}=0\\]
\nSo the roots are:
\\[x= \\simplify{{n1-n4}/{2*a*b}} \\ \\ \\text{and} \\ \\ x= \\simplify{{n1+n4}/{2*a*b}} \\]
If you cannot see a factorisation then you can use the quadratic formula:
\n\\[ x = \\frac{-b \\pm \\sqrt{b^2-4ac}}{2a}\\]
\nRemember that there are three possible types of solution, depending upon the value of the discriminant $\\Delta=b^2-4ac$.
\n1. $\\Delta \\gt 0$. 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 roots are complex)
\nFor this question the discriminant is $\\Delta = \\simplify[std]{{-n1}^2-4*{a*b*c*d}}=\\var{disc}$. {rdis}
\nSo the {rep} roots are:
\n\\[ 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}} \\]
\nYou could also use the method of completing the square.
\nFirst 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}\\]
See also these Mathcentre leaflets on solving quadratic equations.
\n", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]
\nEnter the values of the roots as fractions or integers, not as decimals.
\nEnter the root with the lower value first. If the equation has two equal roots then enter this value in both input boxes.
\nThe root with the lower value is $x=\\;$ [[0]]. The root with the higher value is $x=\\;$ [[1]]
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