I'm going to assume that the room is a rectangle.
The area of a rectangle is A = lw, where l=length of the rectangle and w=width of the rectangle.
You're given that the length, l = (x+5)ft and the width, w = (x+4)ft. You're also told that the area, A = 600 sq. ft. Plug these values into the equation for the area of a rectangle and FOIL to multiply the two factors:
[tex]A = lw\\
600 = (x+5)(x+4)\\
600 = x^{2} + 9x + 20
[/tex]
Now subtract 600 from both sides to get a quadratic equation that's equal to zero. That way you can factor the quadratic to find the roots/solutions of your equation. One of the solutions is the value of x that you would use to find the dimensions of the room:
[tex]600 = x^{2} + 9x + 20\\
x^{2} + 9x - 580 = 0\\
(x + 29)(x - 20) = 0\\
x + 29 = 0, \:\: x - 20 = 0\\
x = -29, x = 20[/tex]
Now you know that x could be -29 or 20. For dimensions, the value of x must give you a positive value for length and width. That means x can only be 20. Plugging x=20 into your equations for the length and width, you get:
Length = x + 5 = 20 + 5 = 25 ft.
Width = x + 4 = 20 + 4 = 24 ft.
The dimensions of your room are 25ft (length) by 24ft (width).
