Question

The length of a rectangle is 4ft less than twice its width. The perimeter of the rectangle is 256 ft. Find the length and width of the rectangle.

Answer 1

48ft^2

Now solve this equation by factoring it out into a polynomial:
48=2W^2-4W --> 24=W^2-2W --> W^2-2W-24=0 --> (W+4)(W-6)=0. Solving this equation gives us W values of 6 and -4, but since the width of a rectangle cannot be negative, the width must be 6.

Since L=2W-4, then L=2(6)-4 --> L=12-4 = 8. Therefore, the dimensions are 6X8, or 6 feet by 8 feet.

To check our work: The length equals four feet less than twice the width: 8=2(6)-4 --> 8=12-4 --> 8=8. This checks out.

Also, the area is 48 ft^2, and (6)(8) = 48, so this also checks out.

Answer 2

For problems like this, change the wording into simple algebraic formulas. We know that the area of a rectangle is its width times its length, or A=LW. The problem tells you that the area is 48, so LW=48. Also, if the length = twice the width minus 4, then L=2W-4.

From here, substitute your new L value for the L in the first equation:
48=LW --> 48=(2W-4)W

Now solve this equation by factoring it out into a polynomial:
48=2W^2-4W --> 24=W^2-2W --> W^2-2W-24=0 --> (W+4)(W-6)=0. Solving this equation gives us W values of 6 and -4, but since the width of a rectangle cannot be negative, the width must be 6.

Since L=2W-4, then L=2(6)-4 --> L=12-4 = 8. Therefore, the dimensions are 6X8, or 6 feet by 8 feet.

To check our work: The length equals four feet less than twice the width: 8=2(6)-4 --> 8=12-4 --> 8=8. This checks out.

Also, the area is 48 ft^2, and (6)(8) = 48, so this also checks out.

Hopefully this helps. If you need any more help or if I went over something too quickly, just let me know.