document.write( "Question 286089: The width of a rectangle is two less than its length. If the area is 48 cm squared, what are the measures of the rectangle's length and width? \n" ); document.write( "
Algebra.Com's Answer #207463 by mgmoeab(37)\"\" \"About 
You can put this solution on YOUR website!
Given:\r
\n" ); document.write( "\n" ); document.write( "A= 48cm²\r
\n" ); document.write( "\n" ); document.write( "W = L - 2\r
\n" ); document.write( "\n" ); document.write( "You know the area of a rectangle is A = W x L\r
\n" ); document.write( "\n" ); document.write( "Now, you can say:\r
\n" ); document.write( "\n" ); document.write( "48cm² = W x L\r
\n" ); document.write( "\n" ); document.write( "Substituting:\r
\n" ); document.write( "\n" ); document.write( "48cm² = (L - 2) L\r
\n" ); document.write( "\n" ); document.write( "You end up with a quadratic equation:\r
\n" ); document.write( "\n" ); document.write( "L² - 2L - 48 = 0\r
\n" ); document.write( "\n" ); document.write( "You can write this as:\r
\n" ); document.write( "\n" ); document.write( "L² - 8L + 6L - 48 = 0
\n" ); document.write( "(L² - 8L) + (6L - 48) = 0
\n" ); document.write( "L(L - 8) + 6 (L - 8) = 0
\n" ); document.write( "(L - 8)(L + 6) = 0
\n" ); document.write( "For this product to be true, one of them must be zero, so:\r
\n" ); document.write( "\n" ); document.write( "L - 8 = 0 ; L = 8
\n" ); document.write( "L + 6 = 0 ; L = -6 ** WE NEGLECT THE NEGATIVE ROOT.\r
\n" ); document.write( "\n" ); document.write( "Now that we have the length of the rectangle, we substitute this in W = L - 2, and:\r
\n" ); document.write( "\n" ); document.write( "W = 8 - 6 = 2\r
\n" ); document.write( "\n" ); document.write( "w = 8 - 2 = 6
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