document.write( "Question 1188894: A rectangle is 5 cm longer than its width. If its width is x cm and its area is
\n" ); document.write( "14 cm2 \n" ); document.write( "

Algebra.Com's Answer #820087 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "If this problem is for a basic algebra class, then you will want to do something like this:

\n" ); document.write( "let x = width
\n" ); document.write( "then x+5 = length

\n" ); document.write( "The area (length times width) is 14:

\n" ); document.write( " $\"x%28x%2B5%29=14\"$
\n" ); document.write( " $\"x%5E2%2B5x=14\"$

\n" ); document.write( "It's a quadratic equation; get everything on one side and solve by factoring:

\n" ); document.write( " $\"x%5E2%2B5x-14=0\"$
\n" ); document.write( " $\"%28x%2B7%29%28x-2%29=0\"$
\n" ); document.write( " $\"x=-7\"$ or $\"x=2\"$

\n" ); document.write( "x = -7 makes no sense in the problem, so reject it; the solution is

\n" ); document.write( "width = x = 2
\n" ); document.write( "length = x+5 = 7

\n" ); document.write( "While that is a good exercise in algebra, note that the formal algebra doesn't help you solve the problem, because in factoring $\"x%5E2%2B5x-14\"$ you need to find two numbers whose difference is 5 and whose sum is 14 -- but that's what the original problem required you to do.

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