Question

The area of a room is 600 square feet. The length is (x + 5) feet and the width is (x + 4) feet. Find the dimensions of the room.

Answer 1

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:
$A = lw\\ 600 = (x+5)(x+4)\\ 600 = x^{2} + 9x + 20$

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:
$600 = x^{2} + 9x + 20\\ x^{2} + 9x - 580 = 0\\ (x + 29)(x - 20) = 0\\ x + 29 = 0, \:\: x - 20 = 0\\ x = -29, x = 20$

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).

Answer 2

Area = length*width

(x+5)(x+4) = 600

Expand left side by distributing
----> x(x+4) + 5(x+4) = x^2 +4x + 5x + 20 = x^2 + 9x +20

x^2 + 9x +20 = 600

x^2 + 9x -580 = 0

Factor:
(x+a)(x+b) = 0
a*b = -580   and a+b = 9
a = 29, b = -20

(x+29)(x-20) = 0
Solve by setting each factor equal to zero.
x+29 = 0 ----> x = -29
x-20 = 0  ----> x = 20

Note: x cannot be negative because dimensions of room must be positive.

Dimensions are 25 and 24 feet.