document.write( "Question 1158688: The width of a rectangle is 16 ft less than the length. The area is 10 ft^2. Find the length and the width. \n" ); document.write( "
Algebra.Com's Answer #781653 by Shin123(626)\"\" \"About 
You can put this solution on YOUR website!
Let the length and width of the rectangle be l and w, respectively. We have \"system%28w=l-16%2Clw=10%29\". Substituting l-16 for w in the second equation, we get \"l%28l-16%29=10\". Expanding and rearranging, we get \"l%5E2-16l-10=0\". \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "
Solved by pluggable solver: Completing the Square for Quadratics
To complete the square for the quadratic \"x%5E2-16%2Ax-10=0\", we must first find a square which when expanded, has x2 and -16x in it.
\n" ); document.write( "\"%28x-8%29%5E2\" is the square we are looking for. Expanding \"%28x-8%29%5E2\" gets \"x%5E2-16%2Ax%2B64\". So we have \"x%5E2-16%2Ax%2B-10=%28x-8%29%5E2-74\". So completing the square gives \"x%5E2-16%2Ax%2B-10=highlight%28%28x-8%29%5E2-74%29\".\n" ); document.write( " Adding 74 from both sides, we get \"%28x-8%29%5E2=74\". Taking the square root of both sides gives \"system%28x-8=sqrt%2874%29=8.60232526704263%2Cx-8=-sqrt%2874%29=-8.60232526704263%29\". \"system%28x=16.6023252670426%2Cx=-0.602325267042627%29\" So the solutions are x=16.6023252670426 and x=-0.602325267042627.
\n" ); document.write( " We pick the positive solution, \"x=8%2Bsqrt%2874%29\". So the length is \"8%2Bsqrt%2874%29\". The width is \"%288%2Bsqrt%2874%29%29-16=sqrt%2874%29-8\".
\n" ); document.write( "
\n" );