the area of a rectangle is 30 square inches. determine both dimensions if the length is 4 inches less than twice the width.

Question

write the equation in (x) to solve problem

find the solution

Answer 1

Hello Kevin,

For a rectangle, Area = Length * Width: A = L * W

 

Let W = the width.

Then L = 2W - 4,

 

which gives us:

30 = (2W - 4)(W)

--> 2W² - 4W - 30 = 0, by distributing the W, and subtracting 30 from both sides,

--> W² - 2W - 15 = 0, dividing everything by 2 (Why?  Because we can. :-) )

--> (W - 5)(W + 3) = 0, by factoring

--> W - 5 = 0 or W + 3 = 0, by setting each factor equal to 0,

--> W = 5 or W = -3.

However, W = -3 wouldn't make sense in the context of the problem, since we wouldn't want a negative width,

--> W = 5 only,

--> L = 2(5) - 4 = 10 - 4 = 6.

 

So the dimensions are 6 by 5.

 

Hope the above helps, and thank you for posting your question.

~Michael E.

Answer 2

Let w = the width

Let 2w - 4 = the length

A = lw

30 = w(2w-4)

30 = 2w² - 4w       Divide both sides by 2.

w² - 2w - 15 = 0

(w - 5)(w + 3) = 0

w - 5 = 0     or    w + 3 = 0

w = 5 inches    w = -3 (illogical answer)

2w - 4 = 10 - 4

length = 6 inches