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The dimensions of a square space are altered so that one dimension is increased by 5 feet, while the other is decreased by 3 feet. The area of the resulting rectangle is 84ft^2. What was the perimeter of the original square

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Explanation:

s = side length of the original square.

the area of a rectangle is

length × width

length = s + 5

width = s - 3

(s + 5)(s - 3) = 84

s² - 3s + 5s - 15 = 84

s² + 2s - 15 = 84

s² + 2s - 99 = 0

for a quadratic equation

ax² + bx + c = 0

the general solution is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = s

a = 1

b = 2

c - -99

s = (-2 ± sqrt(2² - 4×1×-99))/(2×1) =

= (-2 ± sqrt(4 + 396))/2 =

= (-2 ± sqrt(400))/2 = (-2 ± 20)/2 =

= -1 ± 10

s1 = -1 + 10 = 9 ft

s2 = -1 - 10 = -11 ft

a negative solution for an actual length does not make any sense. so, s = 9 ft is our solution.

the perimeter of the original square was then

4 × 9 = 36 ft

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