the area of a rectangle is 120*2. if the length is 8 centimeters more than twice the width, what are the dimensions of the rectangle?

Answer 1

Let "L" be the length of the rectangle and "W" be the width of the rectangle. The problem is describing a scenario in which the area of the rectangle is 240 cm^2 and the length of the rectangle is 8 cm more than twice the width. This gives us two equations:

L*W=240 cm^2

L=2*W+8 cm.

We can now use the method of substitution to solve this system of equations. L is already described in terms of W by the second equation, so we can substitute L in the first equation with the right hand side of the second equation.

(2*W+8 cm)*W=240 cm^2

Next, we simplify the new equation:

2*W^2+W*8 cm=240 cm^2 (distributing W)

W^2+W*4 cm=120 cm^2

(dividing both sides of the equation by 2)

W^2+W*4 cm-120 cm^2=0 (subtracting 120 from both sides).

The zeros of this polynomial are

(-4+sqrt(496))/2 and (-4-sqrt(496))/2.

Since negative width does not exist, the first solution (-4+sqrt(496))/2 (the only positive one) is the width of the rectangle. Lastly, we plug in the width of the rectangle into the first equation of the original system and find the length.

W*L=240 cm^2

L*(-4+sqrt(496))/2=240 cm^2 (substitution)

L=480/(-4+sqrt(496)) cm.

Thus, the length is 480/(-4+sqrt(496)) cm and width is (-4+sqrt(496))/2 cm.

Answer 2

w(2w + 8) = 120

2w2 + 8w = 120

w2 + 4w -60 = 0

(w + 10)(w - 6) = 0

w + 10 = 0

w = -10

Well, we don't measure in negative numbers, so we disregard this solution.

w - 6 = 0

w = 6 cm

l = 2(6) + 8

l = 20 cm