sarahc33
Answered

Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

The length of a rectangle is 5 ft more than twice the width, and the area of the rectangle is 63 ft^2. Find the dimensions of the rectangle.​

Sagot :

reinhard10158

Step-by-step explanation:

l = length

w = width

l = 2w + 5

l × w = 63

using the identity of the first equation in the second :

(2w + 5) × w = 63

2w² + 5w = 63

2w² + 5w - 63 = 0

the general solution to such a squared equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = w

a = 2

b = 5

c = -63

w = (-5 ± sqrt(5² - 4×2×-63))/(2×2) =

= (-5 ± sqrt(25 + 504))/4 = (-5 ± sqrt(529))/4 =

= (-5 ± 23)/4

w1 = (-5 + 23)/4 = 18/4 = 4.5 ft

we = (-5 - 23)/4 = -28/4 = -7

a negative Stilton does not make any sense for a side length, so, the only valid solution is w = 4.5 ft.

l = 2w + 5 = 2×4.5 + 5 = 9 + 5 = 14 ft

so,

the length = 14 ft

the width = 4.5 ft

Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.