The length of s rectangle is 3m less than twice the width, and the area of the rectangle is 44m^2 . Find the dimensions of the rectangle

Question

i need help on this problem can anybody help me. Needs to be done by tonight please help. Help help

Answer 1

Length of Rectangle is 3 cm less than twice the width

Area is 44 m² ; let W = width;  2W - 3 = Length

 

A_rect = L times W

 

44 = W ( 2W - 3 )

 

44 = 2W² - 3W

 

0 = 2W² - 3W - 44

 

Factoring

 

0 = ( 2W - 11 )( W +4)

 

2W - 11 =0   ;            W + 4 = 0

 

        W = 11/2                 W = -4  (width can not be a negative number)

 

Therefor: width is 11/2  and length is 8

Answer 2

The length of s rectangle is 3m less than twice the width:

L = 2W - 3

 

The area of the rectangle is 44m²

A = 44

 

The area of a rectangle is L(W) so

 

LW = 44

Replace L with 2W-3

(2W-3)W = 44

multiply this out

2W² - 3W = 44

subtract 44 from both sides to create a quadratic

2W² - 3W - 44 = 0

 

This can be factored by grouping

2(-44) = -88

Factors of -88 that add to -3 are (8)(-11)

Replace -3W with 8W -11W 

2W² + 8W - 11W - 44 = 0

 

Group and factor

 

(2W² + 8W) + (-11W - 44) = 0

2W(W + 4) -11(W + 4) = 0

 

(2W-11)(W+4) = 0

 

Either 2W-11 = 0       OR  W + 4 = 0 

          2W = 11                W = -4

           W = 11/2

  

Discard the negative since width will be positive

The width is 11/2 meters or 5 1/2 meters as you prefer

 

L = 2W - 3

L = 2(11/2) - 3

L = 11-3

L = 8m

 

Your rectangle has dimensions

length = 8m

width = 5 1/2 m

 

5.5 by 8 meters