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.

Answer 1

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

Answer 2

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.

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