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Question asks student to find zeros of a quadratic equation. In this version students are expected to write up their working and submit it seperately to the Numbas question. Students are expcted to recognise that only the positive solution has physical significance.
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\nUsing the quadratic formula we have \\begin{align} x&=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}\\\\&=\\frac{-\\var{b}\\pm\\sqrt{\\simplify[!collectNumbers]{{b}^2-4*{a}*{-c}}}}{\\var{2*a}}\\\\&=\\frac{-\\var{b}+\\sqrt{\\simplify{{b}^2-4*{a}*{-c}}}}{\\var{2*a}}\\text{ or }\\frac{-\\var{b}-\\sqrt{\\simplify{{b}^2-4*{a}*{-c}}}}{\\var{2*a}}\\\\&=\\var{ans}\\text{ or }\\var{ans2}\\end{align}
\nSince we are told that \\(x\\ge 0\\) we can ignore the negative answer and conclude that the bending moment will be \\(0\\) at a distance approximately \\(\\var{precround({ans},2)}\\) m from the end of the beam.
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\nFind the value of \\(x\\) for which \\(M\\) is \\(0\\). Write your final answer in the box below and show full working on your handwritten notes.
\n\\(x=\\)[[0]]m (round answer to 2 decimal places)
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