SOLUTION: Hello I am not sure how to go about this. If the sides of a square are decreased by 3cm, the area is decreased by 81cm^2. What were the dimensions of the original square. I am thin

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hello I am not sure how to go about this. If the sides of a square are decreased by 3cm, the area is decreased by 81cm^2. What were the dimensions of the original square. I am thin      Log On


   

Question 92180: Hello I am not sure how to go about this. If the sides of a square are decreased by 3cm, the area is decreased by 81cm^2. What were the dimensions of the original square. I am thinking (x-3)^2=a-81cm^3, but where do I go from here?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, your initial idea is right on!
%28x-3%29%5E2+=+A-81 Where A = original area and x = the original length of the side of the square.
The next step is to recall that A+=+x%5E2, that is; the original area (A) is equal to the original sides squared %28x%5E2%29, so substitute the A with x%5E2 in the first equation then solve for x.
%28x-3%29%5E2+=+x%5E2-81 Expand the left side.
x%5E2-6x%2B9+=+x%5E2-81 Subtract x%5E2 from both sides.
-6x%2B9+=+-81 Subtract 9 from both sides.
-6x+=+-90 Finally, divide both sides by -6.
x+=+15cm.
Check:
%2815-3%292+=+15%5E2-81
12%5E2+=+225-81
144+=+144