The length of a rectangular floor is 4 feet longer than its width w. The area of the floor is 525 ft^2. (a)Write a quadratic equation in terms of w that represents the situation.
(b)What are the dimensions of the floor?
Show your work.

Thanks!

Step-by-step explanation:

Let w represent the width of the floor. Then the length of the floor is (w+4) and its area is ...

  A = LW

  525 = (w+4)(w)

  w^2 +4w -525 = 0

  (w -21)(w +25) = 0 . . . . factor the above

Solutions are ...

  w = 21, w = -25

We are interested in the positive solution: w = 21.

The floor is 21 feet wide and 25 feet long.

Alternate solution

Sometimes, when the factors aren't obvious, it works well to write an equation for the average of the dimensions. Here, we can represent that with x, and use (x-2) for the width, and (x+2) for the length. Then we have ...

  525 = (x-2)(x+2) = x^2 -4

  529 = x^2

  √529 = 23 = x

Then w=23 -2 = 21, and the length is w+4 = 25.

x = 21

x= -25

Step-by-step explanation:x^2+4-525=0

               x^21x+25x-525=0

           x{_x^2__ - __3.7__) +5^2 (_5^2.x___ -  _3.5^2.7)

                  x                x                     5^2                   5^2

          x(x^2-1-(3.7))+5^2(x-(3.7))=0

             x(x-27)+5^2(x-21)=0

          (x-21)((x(_x-21_)  + _5^2(x-21)_)=0

                       x-21               x-21

(x-21)(x+5^2)=0

(x-21)(x+25)=0

x- 21 =0

x+25= 0

x-21+21=21

(x+25)-25=-25

x=21

x+25-25=-25

x= 21

x= -25

throw the computer

step-by-step explanation:

2/3

step-by-step explanation:

the slope of a line is equal to the rise/run. in this case the line rises 2 units for every +3 units it moves