SOLUTION: plleae help me for this equation A certain rectangle's length is 3 feet longer than its width. If the area of the rectangle is 54 square feet, find its dimensions

Algebra ->  Rectangles -> SOLUTION: plleae help me for this equation A certain rectangle's length is 3 feet longer than its width. If the area of the rectangle is 54 square feet, find its dimensions      Log On


   

Question 798780: plleae help me for this equation A certain rectangle's length is 3 feet longer than its width. If the area of the
rectangle is 54 square feet, find its dimensions
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +x+ = the width
+x+%2B+3+ = the length
+x%2A%28+x%2B+3+%29+=+54+
+x%5E2+%2B+3x+=+54+
You can solve this by completing the square
+x%5E2+%2B+3x+%2B+%283%2F2+%29%5E2+=+54+%2B+%283%2F2%29%5E2+
+x%5E2+%2B+3x+%2B+9%2F4++=++216%2F4+%2B+9%2F4
+x%5E2+%2B+3x+%2B+9%2F4+=+225%2F4+
+%28+x+%2B+3%2F2+%29%5E2+=+%28+15%2F2+%29%5E2+
+x+%2B+3%2F2+=+15%2F2+
+x+=+15%2F2+-+3%2F2+
+x+=+12%2F2+
+x+=+6+
and
+x+%2B+3+=+9+
The dimensions are 6' x 9'
check:
+x+=+6+
+x+-+6+=+0+
also, there is a negative square root in the answer, too
+x+%2B+3%2F2+=+-15%2F2+
+x+=+-18%2F2+
+x+=+-9+
+x+%2B+9+=+0+
So, the factors are:
+%28+x+-+6+%29%2A%28+x+%2B+9+%29+=+0+
+x%5E2+-+6x+%2B+9x+-+54+=+0+
+x%5E2+%2B+3x+=+54+
+x%2A%28+x+%2B+3+%29+=+54+
This is the original equation