Question

The length of a rectangle is 4 centimeters less than twice its width. Find the dimensions if the area of the rectangle is 96 square centimeters.

Answer 1

8 and 12

Answer 2

I could've guessed that, but without showing work I'm afraid my teacher will give me no credit. I've tried setting it up algebraically but can't seem to get the correct answer.

L= 2w-4cm
w= w

(2w-4)(w)=96

Don't know what I have wrong so if anyone could tell me it'd be appreciated.

Answer 3

"length is 4 cm less than twice its width"
L = 2W - 4 cm

"area is 96 square centimeters"
LW = 96 cm²
(2W-4)W = 96
2W² - 4W = 96
2W² - 4W - 96 = 0

Use the quadratic formula to solve
aW²+bW+c=0
where
a=2
b=-4
c=-96

W = [-b ± √(b²-4ac)] / (2a)
= [-(-4) ± √((-4)² - 4(2)(-96))] / 2(2)
= [4 ± √(784)] / 4
= [4 ± √(2⁴)√(7²)] / 4
= [4 ± 4×7] / 4
= [4 ± 28] / 4
= 1 ± 7

W₊ = 1 + 7 = 8
W₋ = 1 - 7 = -6 invalid

W = 8 cm
L = 2W-4 = 12 cm

Answer 4

Thanks!