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Students need to solve a quadratic equation and recognise that only the positive root has physical significance. Roots are randomised with one always negative and one positive. Equation can be factorised fairly easily or the quadratic formula can be used to find the solution. Advice gives solution by factorisation.

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We are told that the energy used by the process is given by \\[E=\\simplify{x^2+{b+a}x+{a*b+energy}}\\text{ Joules}\\] and that \\(E=\\var{energy}\\) joules. Hence we need to solve

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\\[\\var{energy}=\\simplify{x^2+{b+a}x+{a*b+energy}}\\] or 

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\\[\\simplify{x^2+{b+a}x+{a*b}}=0\\]

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This equation can be factorised as \\[(x+\\var{a})(x-\\var{-b})=0\\] which has solutions \\[x=\\var{-a}\\quad\\text{or}\\quad x=\\var{-b}.\\]

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Since the process will produce a positive number of components we can ignore the negative root and conclude that \\(\\var{-b}\\) components were produced.

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Note that the quadratic formula could also be used to find the roots of this equation.

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Negative root of quadratic, random between 10 and 90 in steps of 5

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energy used by process, random 100 to 1000 in steps of 50

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In order to produce a quantity \\(x\\) of a certain electronic component an amount of energy, \\(E\\) is required where 

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\\[E=\\simplify{x^2+{b+a}x+{a*b+energy}}\\text{ Joules}\\]

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If the energy used during a production run of this component was \\(\\var{energy}\\) Joules, how many of the components were produced?

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[[0]] components were produced. Show full working on your handwritten notes. 

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