document.write( "Question 993402: The length of a rectangle is one less than twice the rectangle's width. The rectangle's area equals 703 in^2. Find the rectangle's width (in inches) \n" ); document.write( "
Algebra.Com's Answer #612681 by macston(5194)\"\" \"About 
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\n" ); document.write( "W=width; L=length=2W-1
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\n" ); document.write( "A+LW
\n" ); document.write( "703in^2=(2W-1)(W)
\n" ); document.write( "703=2W^2-W
\n" ); document.write( "0=2W^2-W-703
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation \"aW%5E2%2BbW%2Bc=0\" (in our case \"2W%5E2%2B-1W%2B-703+=+0\") has the following solutons:
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\n" ); document.write( " \"W%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=%28-1%29%5E2-4%2A2%2A-703=5625\".
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\n" ); document.write( " Discriminant d=5625 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28--1%2B-sqrt%28+5625+%29%29%2F2%5Ca\".
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\n" ); document.write( " \"W%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+5625+%29%29%2F2%5C2+=+19\"
\n" ); document.write( " \"W%5B2%5D+=+%28-%28-1%29-sqrt%28+5625+%29%29%2F2%5C2+=+-18.5\"
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\n" ); document.write( " Quadratic expression \"2W%5E2%2B-1W%2B-703\" can be factored:
\n" ); document.write( " \"2W%5E2%2B-1W%2B-703+=+2%28W-19%29%2A%28W--18.5%29\"
\n" ); document.write( " Again, the answer is: 19, -18.5.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-1%2Ax%2B-703+%29\"

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\n" ); document.write( "ANSWER: The width is 19 inches.
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