The area of a rectangle is 77 yd2, and the length of the rectangle is 3 yd more than twice the width find the dimensions of the rectangle

Let the width be x.

Length is 3 more than twice the width = 2x + 3

Area = x * (2x + 3)

x * (2x + 3) = 77

2x² + 3x = 77

2x² + 3x - 77 = 0 This is a quadratic equation. Solve by factorization.

Multiply first and last coefficients: 2*-77 = - 154

We look for two numbers that multiply to give - 154, and add to give the middle coefficient + 3

Those two numbers are 14 and - 11.

Check: 14*-11 = - 154 14 + - 11 = 14 - 11 = + 3

We replace the middle term of + 3x in the quadratic expression with 14x - 11x

2x² + 3x - 77 = 0

2x² + 14x - 11x - 77 = 0

2x (x + 7) - 11 (x + 7) = 0

(2x - 11) (x + 7) = 0

2x - 11 = 0 or x + 7 = 0

2x = 0 + 11 x = 0 - 7

2x = 11 x = - 7

x = 11/2 = 5.5

The solutions are x = - 7 or 5.5.

Since we are solving for length of rectangle, x can't be negative. So we do away with x = - 7.

x = 5.5 is the only valid solution.

Recall width = x = 5.5,

Length = 2x + 3 = 2*5.5 + 3 = 11 + 3 = 14

Therefore length = 14 yards, and width = 5.5 yards.