Question

the length of a rectangle is 11yd more than twice the width,and the area of the rectangle is 63yd^2 what are the dimensions

Answer 1

Let the width be w.
Then the length = 2w + 11.
The area is w * l = w(2w + 11).
Now we can write an equation for the area as follows:
$2w^{2}+11w=63$
This equation can be rearranged to make the following quadratic:
$2w^{2}+11w-63=0$
The factors of the quadratic are:
(2w - 7)(w + 9) = 0
The positive solution is w = 3.5.
Plugging in this value for w into the equation for the length we get:
Length = 18 yards, width = 3.5 yards.