document.write( "Question 527520: What is the area of a rectangle that is 5 inches longer than it is wide and if you double the length and subtract 2 inches from the width the area is increased by 52 inches squared?\r
\n" ); document.write( "\n" ); document.write( "I tried...
\n" ); document.write( "(x-2)(2x+10)=x^2+5x+52
\n" ); document.write( "but I don't thin Im setting it up right...\r
\n" ); document.write( "\n" ); document.write( "Can you help me set it up?
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Algebra.Com's Answer #348930 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
You are properly setting it up.
\n" ); document.write( "The original rectangle has a width that you called \"x\"
\n" ); document.write( "and we should add a requirement that it should be a real number, and that
\n" ); document.write( "\"x%3E0\"
\n" ); document.write( "The length of the original rectangle is \"x%2B5\"
\n" ); document.write( "Its area is \"%28x%28x%2B5%29=x%5E2%2B5x\"
\n" ); document.write( "The new rectangle has a length of \"2%28x%2B5%29=2x%2B10\"
\n" ); document.write( "a width of \"x-2\"
\n" ); document.write( "(so we are also going to ask for \"x%3E2\")
\n" ); document.write( "The area of the new rectangle is
\n" ); document.write( "\"%28x-2%29%282x%2B10%29=2x%5E2%2B6x-20\"
\n" ); document.write( "Your equation transforms into
\n" ); document.write( "\"2x%5E2%2B6x-20=x%5E2%2B5x%2B52\" and into
\n" ); document.write( "(((x^2+x-72=0}}}
\n" ); document.write( "which can be solved by factoring, or by completing the square, or by using the quadratic formula.
\n" ); document.write( "One of the solutions for x is negative. The other solution is the width of the original rectangle, which you would need to use to find the area of the original rectangle.
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