document.write( "Question 129109: The area of a rectangle is 154 sq. Ft. If its width is 3 ft. less than its length what are its dimensions? I need to use a quadratic formula to solve it.\r
\n" ); document.write( "\n" ); document.write( "ty
\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #94390 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Start with the formula for the area of a rectangle:
\n" ); document.write( "\"A+=+L%2AW\"
\n" ); document.write( "You are told that the width, W is 3 ft. less than the length, L, so this can be written as:
\n" ); document.write( "\"W+=+L-3\" so now you can replace W with (L-3) and write:
\n" ); document.write( "\"A+=+L%2A%28L-3%29\" and you are also told that the area is 154 sq.ft., so...
\n" ); document.write( "\"L%2A%28L-3%29+=+154\" Simplify.
\n" ); document.write( "\"L%5E2-3L+=+154\" Subtract 154 from both sides.
\n" ); document.write( "\"L%5E2-3L-154+=+0\" There's your quadratic equation and it can be solve by factoring.
\n" ); document.write( "\"%28L-14%29%28L%2B11%29+=+0\" Applying the zero products principle, we get:
\n" ); document.write( "\"L-14+=+0\" or \"L%2B11+=+0\" so that...
\n" ); document.write( "\"L+=+14\" or \"L+=+-11\" Use only the positive value as lengths are positive quantities.
\n" ); document.write( "\"L+=+14\" and...
\n" ); document.write( "\"W+=+L-3\"
\n" ); document.write( "\"W+=+14-3\"
\n" ); document.write( "\"W+=+11\"
\n" ); document.write( "The dimensions of the rectangle are:
\n" ); document.write( "Length = 14 ft. and width = 11 ft.
\n" ); document.write( "Check:
\n" ); document.write( "\"A+=+L%2AW\"
\n" ); document.write( "\"A+=+14%2A11\"
\n" ); document.write( "\"A+=+154\"sq.ft.
\n" ); document.write( "
\n" );