SOLUTION: the length and width of a rectangle that measures 8 inches by 6 inches are both increased by the same amount. The area of the larger rectangle is twice the area of the original rec
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Question 307452: the length and width of a rectangle that measures 8 inches by 6 inches are both increased by the same amount. The area of the larger rectangle is twice the area of the original rectangle.How much was added to each dimension of the original rectangle? round to the nearest hundredth of an inch. Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! the length and width of a rectangle that measures 8 inches by 6 inches are both increased by the same amount. The area of the larger rectangle is twice the area of the original rectangle.How much was added to each dimension of the original rectangle? round to the nearest hundredth of an inch.
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Let x = amount added
then
length is 8+2x
width is 6+2x
.
(8+2x)(6+2x) = 2(8*6)
48+16x+12x+4x^2 = 2(48)
48+28x+4x^2 = 96
4x^2+28x+48= 96
4x^2+28x-48= 0
x^2+7x-12= 0
Solve by applying the quadratic formula. Doing so, yields:
x = {1.42, -8.42}
Tossing out the negative solution leaves:
x = 1.42 inches
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Details of quadratic:
