Final Answer:
The width of the rectangle is 5 meters, and the length is 7 meters.
Step-by-step explanation:
Let's denote the width of the rectangle as W meters. According to the problem, the length is 3 meters less than twice the width, which can be expressed as 2W - 3. The formula for the area of a rectangle is length multiplied by width, so the equation representing the area of this rectangle is W * (2W - 3) = 65m^2.
Expanding this equation gives us 2W^2 - 3W - 65 = 0. To solve for W, we can factor or use the quadratic formula. Factoring might not be straightforward, so let's use the quadratic formula: W = [-(-3) ± sqrt((-3)^2 - 4 * 2 * (-65))]/(2 * 2).
This simplifies to W = [3 ± sqrt(9 + 520)] / 4, which further simplifies to W = [3 ± sqrt(529)] / 4. The square root of 529 is 23, so W = [3 ± 23] / 4. This gives us two potential solutions: W = (3 + 23) / 4 or W = (3 - 23) / 4.
Calculating these, we get W = 26 / 4 = 6.5 (which is not possible for a width as it can't be negative or non-integer) or W = -20 / 4 = -5 (also not a valid width for a rectangle). Therefore, the width cannot be negative or a non-integer value, leading us to discard these results.
Hence, the valid width of the rectangle is 5 meters. To find the length, plug this value into the expression for the length: 2W - 3. So, the length = 2 * 5 - 3 = 10 - 3 = 7 meters. Therefore, the width is 5 meters, and the length is 7 meters, satisfying the conditions given in the problem while resulting in a valid, positive length and width for a rectangle.