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\title{Math 587 Assignment 2}
\date{March 3, 2009}
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\author{John Doe}
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\begin{center}\LARGE\bfseries{}
Math 587 Assignment 2
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\textsl{John Doe}
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March 3, 2009
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\begin{enumerate}

\item  At what point does the line \(2 x - y \, = \, 3\) intersect the line \(x + 2y \, = \, -1\)? \  \par{\emph{Solution.} The point of intersection may be obtained by solving the two equations simultaneously. \   For this multiply the first equation by \(2\) and add that to the second equation obtaining the equation \[ 5 x \ = \  5 \ . \] Thus, \(x \, = \, 1\), and, using either of the two original equations, one finds \(y \, = \, -1\). \   The required point is \((1, -1)\). \  
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\item  Find all solutions of the quadratic equation \(x^{2} - x - 12 \, = \, 0\). \  \par{\emph{Solution.} The well known formula for solution of the quadratic equation \(ax^{2} + b x + c \, = \, 0\) is \[ x \ = \  \frac{-b \pm{} \sqrt{b^{2} - 4ac}}{2a} \ . \] In this case one finds \[ x \ = \  \frac{1 \pm{} \sqrt{1 -4(1)(-12)}}{2}      \ = \  \frac{1 \pm{} 7}{2}      \ = \  4 \ \mbox{or} \ -3 \ .\] 
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