document.write( "Question 221224: The length of a rectangle is 6 cm more than the width. The area is 11 cm2. Find the length and width. \n" ); document.write( "

Algebra.Com's Answer #165987 by sofiyac(983) $\"\"$ $\"About$

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The length of a rectangle is 6 cm more than the width. The area is 11 cm2. Find the length and width.
\n" ); document.write( "lets call width w, then length will be w+6 since area is 11 then
\n" ); document.write( "w(w+6)=11
\n" ); document.write( "w^2+6w-11=0
\n" ); document.write( " $\"w+=+%28-6+%2B-+sqrt%28+6%5E2-4%2A1%2A%28-11%29+%29%29%2F%282%2A1%29+\"$
\n" ); document.write( " $\"w+=+%28-6+%2B-+sqrt%28+36%2B44+%29%29%2F2+\"$
\n" ); document.write( " $\"w+=+%28-6+%2B-+sqrt%28+80+%29%29%2F2+\"$
\n" ); document.write( " $\"w+=+%28-3+%2B-+4%2Asqrt%28+5+%29%29+\"$ since our answer can't be negative because width can't be negative our answer would be
\n" ); document.write( " $\"w+=+%28-3+%2B+4%2Asqrt%28+5+%29%29+\"$ (about 5.9) and since length was 6 more than width
\n" ); document.write( "then length is
\n" ); document.write( " $\"l+=+%28-3+%2B+4%2Asqrt%28+5+%29%29%2B6+=+%283+%2B+4%2Asqrt%28+5+%29%29+\"$ (about 11.9) \n" ); document.write( "

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