SOLUTION: The area of a rectangle in square feet is x^2+13x+36. How much longer is the length than the width of the rectangle?

Algebra ->  Rectangles -> SOLUTION: The area of a rectangle in square feet is x^2+13x+36. How much longer is the length than the width of the rectangle?      Log On


   

Question 1126136: The area of a rectangle in square feet is x^2+13x+36. How much longer is the
length than the width of the rectangle?
Found 2 solutions by josmiceli, ikleyn:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+x%5E2+%2B+13x+%2B+36+=+%28+x+%2B+4+%29%2A%28+x+%2B+9+%29+ ( by looking at it )
+x+%2B+9+ is the longer side, so the difference is
+x+%2B+9+-+%28+x+%2B+4+%29+=+x+%2B+9+-+x+-+4+
+x+%2B+9+-+x+-+4+=+5+
the length is 5 ft longer than the width

Answer by ikleyn(52880) About Me  (Show Source):
You can put this solution on YOUR website!
.

From the given condition, how it is presented in the post,  IT  IS  NOT  POSSIBLE  to determine on
how much longer is the length than the width of the rectangle.

The fact that the given quadratic polynomial can be factored into the product of the linear binomial,  DOES  NOT  IMPLY
that the factors represent the length and the width.


These measurements might be totally different numbers.


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I'd say that in this problem the question is not relevant to the given data, and, inversely,
the given data is not relevant to the question.