Equations and Inequalities Solving a word problem using a quadratic equation with rational roots The area of a rectangle is $21 \mathrm{yd}^{2}$, and the length of the rectangle is $1 \mathrm{yd}$ less than double the width. Find the dimensions of the rectangle. Length : yd Width : $\mathrm{yd}$

Step 1 ：Let's denote the width of the rectangle as \(x\) (in yards). According to the problem, the length of the rectangle is \(2x - 1\) (in yards).

Step 2 ：The area of a rectangle is calculated by multiplying its length by its width. So, we can set up the following equation: \(x * (2x - 1) = 21\)

Step 3 ：This simplifies to: \(2x^2 - x - 21 = 0\)

Step 4 ：This is a quadratic equation, and we can solve it by factoring: \((2x + 7)(x - 3) = 0\)

Step 5 ：Setting each factor equal to zero gives the solutions \(x = -7/2\) and \(x = 3\).

Step 6 ：However, the width of a rectangle cannot be negative, so we discard the solution \(x = -7/2\).

Step 7 ：Therefore, the width of the rectangle is \(3\) yards, and the length is \(2*3 - 1 = 5\) yards.

Step 8 ：So, the dimensions of the rectangle are: Length: \(5\) yards Width: \(3\) yards

Step 9 ：We can check our solution by substituting these values back into the original equation: \(3 * (2*3 - 1) = 21\) \(3 * 5 = 15\) \(15 = 15\)

Step 10 ：This is true, so our solution is correct. The dimensions of the rectangle are indeed \(5\) yards by \(3\) yards.

Step 11 ：Final Answer: \(\boxed{5}\) yards by \(\boxed{3}\) yards