Question 2 The length of a rectangle is 5 yd more than twice the width, and the area of the rectangle is 63 yd2 . Find the dimensions of the rectangle.
Mathematics
dmb123
Question
Question 2 The length of a rectangle is 5 yd more than twice the width, and the area of the rectangle is 63 yd2 . Find the dimensions of the rectangle.
2 Answer

1. User Answers PHDFlopper
L = 2W + 5
A = LW = 63
L = 63/W
W=63/L
L = 2W + 5
63/W = 2W + 5
63 = 2WW + 5W
2WW + 5W  63 = 0
2WW + 5W  63 = 0
is already in quadratic equation form: ax^2 + bx + c = 0
use the quadratic formula to solve for two solutions:
x = ( b + sqrt( bb  4ac ) ) / 2a
the two solutions are:
x = 9/2
x = 7
you cant have a negative solution for length so use the positive one
solve for length:
L=5+2(9/2)
L=5+18/2
L=5+9
L=14
W = 9/2
L = 14 
2. User Answers topeadeniran2
The length and the width of the rectangle is 14 yards and 4.5 yards respectively.
Let the width of the rectangle be represented by w
Therefore, the length of the rectangle will be represented by 2w + 5.
Therefore, the area of the rectangle will be: = Length × Width
w(2w + 5) = 63
2w² + 5w = 63
2w² + 5w  63 = 0
2w² + 14w  9w  63 = 0
2w(w + 7)  9(w + 7) = 0
Therefore, 2w  9 = 0
2w = 9
w = 9/2
w = 4.5
Length = 2w + 5 = 2(4.5) + 5 = 14
The length and the width of the rectangle is 14 yards and 4.5 yards respectively.
Read related link on:
https://brainly.com/question/781158