document.write( "Question 954418: A photograph has a length that is 2 inches longer than its width, x. So its area is given by the expression x(x+2) square inches. If the area of the photograph is 48 square inches, what is the width of the photograph \n" ); document.write( "
Algebra.Com's Answer #582882 by macston(5194)\"\" \"About 
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\"x%28x%2B2%29=48\"
\n" ); document.write( "\"x%5E2%2B2x=48\"
\n" ); document.write( "(((x^2+2x-48=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%2B2x%2B-48+=+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=%282%29%5E2-4%2A1%2A-48=196\".
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\n" ); document.write( " Discriminant d=196 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28-2%2B-sqrt%28+196+%29%29%2F2%5Ca\".
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\n" ); document.write( " \"x%5B1%5D+=+%28-%282%29%2Bsqrt%28+196+%29%29%2F2%5C1+=+6\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%282%29-sqrt%28+196+%29%29%2F2%5C1+=+-8\"
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\n" ); document.write( " Quadratic expression \"1x%5E2%2B2x%2B-48\" can be factored:
\n" ); document.write( " \"1x%5E2%2B2x%2B-48+=+1%28x-6%29%2A%28x--8%29\"
\n" ); document.write( " Again, the answer is: 6, -8.\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%2B2%2Ax%2B-48+%29\"

\n" ); document.write( "\n" ); document.write( "ANSWER: The width is 6 inches.
\n" ); document.write( "CHECK:
\n" ); document.write( "Length=Width+2=6 in + 2 in=8 in
\n" ); document.write( "Area=L*W
\n" ); document.write( "48 sq in=8 in*6 in
\n" ); document.write( "48 sq in=48 sq in
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