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

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.
.
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
.
Details of quadratic:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=97 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 1.42442890089805, -8.42442890089805. Here's your graph:

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