Since L = 2·W+1 and area = L x W = 21 yd², we can write this as:

(2·W+1) x W = 21

2·W² + W = 21

2·W² + W - 21 = 0

We use the use the quadratic formula to solve (ignoring the negative root)

and get

W = 3 yds

L = 2·W+1 = 7 yds

Crystal B.

asked • 08/06/14the length of a rectangle is 1 yd less than twice the width, and the area of the rectangle is 21 yards squared. Find the dimensions of the rectangle.

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Since L = 2·W+1 and area = L x W = 21 yd², we can write this as:

(2·W+1) x W = 21

2·W² + W = 21

2·W² + W - 21 = 0

We use the use the quadratic formula to solve (ignoring the negative root)

and get

W = 3 yds

L = 2·W+1 = 7 yds

Kelly K. answered • 08/06/14

Tutor

New to Wyzant
Mathematics and SAT Tutoring

L=length

W=Width

A= Area

So from the information given, we know that:

L*W=A

L=2W-1

So, we know A is 21, and if we plug in the equation for L into the first equation we get:

(2W-1)*W=21

Let’s then rearrange this equation to equal 0 by multiplying the W by 2W-1 and moving the 21 to the other side to get:

2W^2 –W -21 = 0

This can be rewritten as:

(2W-7)(W+3)=0

If you solve each part of the equation above so that each equals zero, you will get W=-3 and 3.5

The width cannot be negative, so it must be 3.5. plug that into the above equation for L, and you get:

L=2(3.5) – 1 where L =6

Check our work with L*W = A, where 6*3.5=21

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Stephen K.

08/06/14