Question

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

Answer 1

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.