Solve with the Quadratic Formula:

The length of a rectangle is 1 less than 2 times the width. The area of the rectangle is 45 cm^2. Find the length and width.

Let the width be x.

Length = 2x - 1

The area of rectangle = 45 cm²

x * (2x - 1) = 45

2x² - x = 45

2x² - x - 45 = 0 This is a quadratic equation

comparing to ax² + bx + c = 0, a = 2, b = - 1, c = - 45

x = (-b + √ (b² - 4ac)) / 2a or (-b - √ (b² - 4ac)) / 2a

x = ( - - 1 + √ ((-1) ² - 4*2*-45)) / 2*2 or ( - - 1 - √ ((-1) ² - 4*2*-45)) / 2*2

x = (1 + √ (1 + 360)) / 2*2 or (1 - √ (1 + 360) / 2*2

x = (1 + √361) / 4 or (1 - √361) / 4

x = (1 + 19) / 4 or (1 - 19) / 4

x = 20/4 or - 18/4

x = 5 or - 4.5

x can't be negative since we are solving for side.

x = 5 as the only valid solution.

Recall, the width = x = 5.

Length = 2x - 1 = 2*5 - 1 = 10 - 1 = 9

Hence the length = 9cm, and width = 5cm