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Factorise the equation

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$\\simplify{{a*c}x^2+{a*d+b*c}x+{b*d}=0}\\text{.}$

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$(\\var{a}x+\\phantom{.}$[[0]]$) ($[[1]]$x+\\phantom{.}$[[2]]$)\\; = 0$

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Write down the roots of the equation above.

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Input your answer as $x_1$ and $x_2$, where $x_1<x_2$.

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$x_1=$ [[0]]

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$x_2=$ [[1]]

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$d$ in $(ax+b)(cx+d)$

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$c$ in $(ax+b)(cx+d)$

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$b$ in $(ax+b)(cx+d)$

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$a$ in $(ax+b)(cx+d)$

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The roots of the equation

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Factorise a quadratic equation where the coefficient of the $x^2$ term is greater than 1 and then write down the roots of the equation

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#### a)

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As this question involves a number greater than $1$ before the $x^2$ value it has a factorised form $(ax+b)(cx+d)$.

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To find $a$ and $c$, we need to consider the factors of $\\var{a*c}$.

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We are already given that one of them is $\\var{a}$, so we know that the other one must be $\\var{c}$.

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This means our factorised equation must take the form

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\$(\\var{a}x+b)(\\var{c}x+d)=0\\text{.}\$

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This expands to

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\$\\simplify{ {a*c}x^2 + ({a}*d+{c}*b)x + a*b} \$

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So we must find two numbers which add together to make $\\var{a*d+b*c}$, and multiply together to make $\\var{b*d}$.

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Therefore $b$ and $d$ must satisfy

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\\begin{align}
b \\times d &=\\var{b*d}\\\\
\\simplify{{a}d+{c}b} &= \\var{a*d+b*c}\\text{.}
\\end{align}

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$b = \\var{b}$ and $d = \\var{d}$ satisfy these equations:

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\\begin{align}
\\var{b} \\times \\var{d} &=\\var{b*d}\\\\
\\simplify[]{ {a}*{d} + {b}*{c} } &= \\var{a*d+b*c}
\\end{align}

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So the factorised form of the equation is

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\$\\simplify{({a}x+{b})({c}x+{d}) = 0} \\text{.}\$

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#### b)

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$\\simplify{({a}x+{b})({c}x+{d}) = 0}$ when either $\\var{a}x+\\var{b} = 0$ or $\\var{c}x+ \\var{d} = 0$.

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So the roots of the equation are $\\var[fractionnumbers]{-b/a}$ and $\\var[fractionnumbers]{-d/c}$.

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