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quadratics with some set up rearranging needed

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Solve the following equation for $x$.

\n

\\[\\simplify{({g}x+{a})/{b}}=\\simplify{{c}/({f}x+{d})}\\]

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First multiply through by $\\var{b}$ and $\\simplify{{f}x+{d}}$ and re-arrange to get 

\n

\\[\\simplify{{x2Coff}x^2+{xCoff}x+{Const}=0}.\\]

\n

To solve a quadratic equation of the form \\[ ax^2+bx+c=0\\] by factorisation, we want to factorise the equation into the form \\[(mx+p)(nx+q)=0,\\] where $mn=a$, $np+mq=b$, and $p \\times q = c$. 

\n

The equation \\[\\simplify{{x2Coff}x^2+{xCoff}x+{Const}=0}\\]

\n

can be factorised to \\[\\simplify{({a1}x+{Fact1})({a2}x+{Fact2})=0}.\\] This equation is satisfied when either \\[\\simplify{{a1}x+{Fact1}=0} \\quad \\text{or} \\quad \\simplify{{a2}x+{Fact2}=0}, \\] which implies the solutions to this quadratic equation are \\[ \\simplify[fractionNumbers]{x={-x1}} \\quad \\text{and} \\quad \\simplify[fractionNumbers]{x={-x2}} .\\]

\n

Use this link to find resources to help you revise how to solve quadratic equations.

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Coefficient of x when written in form ax^2+bx+c

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Coefficient of x^2 when written in form ax^2+bx+c

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polynomial factorises as (a1x+Fact1)(a2x+Fact2)

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x=[[0]]

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x=[[1]]

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