MATH SOLVE

4 months ago

Q:
# 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.

Accepted Solution

A:

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).

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).