please help difficult word problem (for me, at least)

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
quaz7829
Posts: 1
Joined: Tue Sep 25, 2012 12:55 am
Contact:

please help difficult word problem (for me, at least)

Postby quaz7829 » Tue Sep 25, 2012 12:59 am

Can anyone help me? It is a word problem I am stuck on.

A square poster has sides measuring 2 feet less than the sides of a square sign. If the difference between there areas is 12 sq feet, find the lengths of the sides of the poster and sign.

so far all i have figured out is (x-2)*(x-2)=x^2 if some one could help i would appreciate that. :wave:

User avatar
maggiemagnet
Posts: 310
Joined: Mon Dec 08, 2008 12:32 am
Contact:

Re: please help difficult word problem (for me, at least)

Postby maggiemagnet » Tue Sep 25, 2012 8:23 pm

quaz7829 wrote:A square poster has sides measuring 2 feet less than the sides of a square sign. If the difference between there areas is 12 sq feet, find the lengths of the sides of the poster and sign.

so far all i have figured out is (x-2)*(x-2)=x^2 if some one could help i would appreciate that. :wave:

I think you are using "x" to be "the length of the sides of the sign", so "x - 2" is "the length of the sides of the poster". The left side of your equation is "the area of the poster" and the right side is "the area of the sign". But are these areas the same?
:clap:


Return to “Intermediate Algebra”