Question

If the length of a rectangle is 3 feet less then twice its width and the area is 54square feet what are its dimensions

Answer 1

The length (L) of the rectangle can be written as a function of the width (W)$L = 2W - 3$:
Now since we know Area = Width*Length, we can write the area as a function of the width:
$A = L*W = (2W-3)*W$
Distributing the W inside the parentheses we have:
$A = 2W^2 - 3W$
We know the area is 54 ft^2, so we can rewrite it as:
$2W^2 - 3W - 54 = 0$
Now solve for W by factoring (or by applying the quadratic formula):
$2W^2 - 12W + 9W - 54 = 0$
Factor out a common 2W from the first two terms and a 9 from the last two terms:
$2W(W-6) + 9(W-6) = 0$
Regroup the terms to get our fully factored equation:
$(2W + 9)(W-6) = 0$
This gives us the roots W = 6 and W = -9/2, but width can't be negative so we have width = 6 ft. Then remember that the length L = 2W - 3, so our length is:
$L = 2W - 3 = 2(6) - 3 = 12 - 3 = 9$
So now we know that our rectangle is 9 feet long and 6 feet wide.