SOLUTION: the width of a rectangle is 3 inches less than its length. the area of the rectangle is 340 square inches. what are the length and width of the rectangle? thank you!!!!

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: the width of a rectangle is 3 inches less than its length. the area of the rectangle is 340 square inches. what are the length and width of the rectangle? thank you!!!!      Log On


   



Question 7208: the width of a rectangle is 3 inches less than its length. the area of the rectangle is 340 square inches. what are the length and width of the rectangle?
thank you!!!!

Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
"The width of the rectangle is 3 inches less than it's length" translates to w = l - 3. The length is longer, so it needs to be knocked off by 3 to equal the shorter w.

Since l*w = area, our equation will be +l%28l+-+3%29+=+340+ which is expanded to +l%5E2+-+3l+=+340+. We need to transfer the 340 to the left side so that the right side will be equal to zero. We now have +l%5E2+-+3l+-+340+=+0+

We now have to factor that equation so that it will look like a (x ± ?)(x ± ?) = 0. This is a bit tough. What times what = 340 whose sum or difference gives you the -3? It turns out to be -20 and 17 after trial and error. So, our equation now should be +%28l-20%29%28l%2B17%29+=+0+.

As you can see, the solutions are 20 and -17, since either value, when plugged in, will make the equation true. BUT, he have to throw out the -17 because it's negative, and you can't have a negative measure.

So the length turned out to be 20 inches. The width, then, if 3 inches shorter, would be 17 inches.