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 dimen

Question

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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 dimen

sions of the rectangle.

Mathematics

1 answer:

Sedaia [141] · Answer 1 · 2022-04-27T04:48:52+03:00

<span>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</span>