document.write( "Question 87337: A painting canvas has a length 9 inches longer than its width. If the area is 90in to the second power, what are the lengths of the legs of the field? \n" ); document.write( "

Algebra.Com's Answer #63256 by tutorcecilia(2152) $\"\"$ $\"About$

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Use the formula for the area of a rectangle
\n" ); document.write( "Area = (length) ( width)
\n" ); document.write( "Area = 90
\n" ); document.write( "Length = Width + 9
\n" ); document.write( "Width = W
\n" ); document.write( ".
\n" ); document.write( "Plug-in the values and solve for the w-term]
\n" ); document.write( "90=(w+9)(w)
\n" ); document.write( "90 = w^2+9w [set the equation equal to zero]
\n" ); document.write( "90-90 = w^2+9w-90
\n" ); document.write( "0=w^2+9w-90 [solve for the w-term]
\n" ); document.write( "0=(w+15)(w-6)[factor; set each factor equal to zero and solve for w]
\n" ); document.write( "w+15=0
\n" ); document.write( "w=-15 [disregard a negative measurement]
\n" ); document.write( "or
\n" ); document.write( "w-6=0
\n" ); document.write( "w=6 [use this measurement because it is positive]
\n" ); document.write( ".
\n" ); document.write( "Length = w+90=6+90=96
\n" ); document.write( "Check by plugging all values back into the original equation:
\n" ); document.write( "A=lw
\n" ); document.write( "90=(w+9)(w)
\n" ); document.write( "90=(6+9)(6)
\n" ); document.write( "90=(15)(6)
\n" ); document.write( "90=90 [checks out]
\n" ); document.write( " \n" ); document.write( "

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