document.write( "Question 966893: the length of the rectangle is 5 inches more than the width. the area is 33 square inches. find the length and width. round to the nearest tenth if necessary. \n" ); document.write( "
Algebra.Com's Answer #590915 by amarjeeth123(569)\"\" \"About 
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Let the width be x.
\n" ); document.write( "Then the length is (x+5).
\n" ); document.write( "Area=length*width
\n" ); document.write( "x(x+5)=33
\n" ); document.write( "x^2+5x-33=0
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation \"ax%5E2%2Bbx%2Bc=0\" (in our case \"1x%5E2%2B5x%2B-33+=+0\") has the following solutons:
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\n" ); document.write( " \"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
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\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%285%29%5E2-4%2A1%2A-33=157\".
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\n" ); document.write( " Discriminant d=157 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28-5%2B-sqrt%28+157+%29%29%2F2%5Ca\".
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\n" ); document.write( " \"x%5B1%5D+=+%28-%285%29%2Bsqrt%28+157+%29%29%2F2%5C1+=+3.76498204307083\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%285%29-sqrt%28+157+%29%29%2F2%5C1+=+-8.76498204307083\"
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\n" ); document.write( " Quadratic expression \"1x%5E2%2B5x%2B-33\" can be factored:
\n" ); document.write( " \"1x%5E2%2B5x%2B-33+=+1%28x-3.76498204307083%29%2A%28x--8.76498204307083%29\"
\n" ); document.write( " Again, the answer is: 3.76498204307083, -8.76498204307083.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B5%2Ax%2B-33+%29\"

\n" ); document.write( "\n" ); document.write( "x=3.764=3.8
\n" ); document.write( "The width is 3.8 units.
\n" ); document.write( "The length is 8.8 units.
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