Question

The area of a rectangle is
44 yd2
, and the length of the rectangle is
3 yd
less than twice the width. Find the dimensions of the rectangle.

Answer 1

@campbell_st

Answer 2

let the width = w

then the length is 2w - 3

so then you can find the area

\[A = w(2w - 3) ~~~or~~~A = 2w^2 - 3w\]

you know the area so

\[44 = 2w^2 - 3w \]

rewriting the equation

\[2w^2 - 3w - 44 = 0\]

use the general quadratic formula to solve for w,

only consider the positive value for w.

hope it helps...

p.s. the with is a mixed number....

Answer 3

thank you!

Answer 4

so the width is 11/2 ? @campbell_st

Answer 5

and the length is 8?

Answer 6

Here is how I would do it:
A = L*W
L = 2W - 3
A = (2W-3)*W
44 = 2W^2 - 3W
2W^2 - 3W - 44 = 0
((3 +- sqrt((-3)^2 - 4*2*-44))/4



(3 +- sqrt( 9 + 352))/4
(3 +- 19)/4
W= -16/4 and 22/4
However, W must be positive as we are talking about a measurement.
So W = 22/4 = 5.5
Plug that back into our original 44 = L * 5.5
44/5.5 = 8
So our dimensions are 8 x 5.5.

Answer 7

oh ok got it! thank you!!! =)

Answer 8

You're Welcome :)