Word Problem-area of a rectangle

[HeidiS]

The problem reads: The length of a rectangle is 4 feet more than 2 times the width. The area of the rectangle is 48. What is the length and width of the rectangle?

I know this should be simple, but I have been racking my brain all day and can't figure it out algebraically. The worst part is I know the answer, I just can't solve it algebraically. Here is what I have done:

Area = Length * width
Length = 2w + 4
Width = w

so,

w(2w + 4) = 48

2w^2 + 4w = 48

2w^2 + 4w - 48 = 0

2(w^2 + 2w - 24) = 0

(w + 6)(w - 4) = 0

w + 6 = 0 and w - 4 = 0

w = -6 and w = 4

I know that -6 can't be the width of the triangle, and I know that 4 is the width of the triangle, making the length (2w + 4) equal to 12. But what about the -6? Or am I totally off base with the way I solved this??? :?

 

[stapel]

But what about the -6?

What about it? :wink:

The extra and non-sensical result is called an "extraneous" solution, and is ignored. It just goes to prove that math is not the same as "real life". :lol:

Eliz.

P.S. Thank you for showing your work and reasoning so nicely!

 

[HeidiS]

Thank you! I hate it when I second quess myself.