Solving a word problem using a quadratic quadratic with rational root

Question

The area of a rectangle is 27 yd² and the length of the rectangle is 3 yd less than double the width find the dimensions of the rectangle

Answer 1

We have a rectangle that has the length, L, and the width, W. Since we have two unknowns, we need to equations. The first equation will be the equation for the area of a rectangle, A:

L • W = A

L • W = 27 (eqn. 1)

The second equation is to relate the length and width according to the parameters set by the problem (the length of the rectangle is 3 yd less than double the width):

L = 2W - 3 (eqn. 2)

We can then substitute the equality of L from eqn. 2 into eqn. 1 to get 1 variable, and then solve for W using a quadratic equation:

L • W = 27 L = 2W - 3 (2W - 3) • W = 27

multiply W throughout the parenthesis 2W² - 3W = 27

set equal to zero 2W² - 3W - 27 = 0

factor (2W - 9)(W + 3) = 0

solve for each factored term 2W - 9 = 0 W = 4.5 W + 3 = 0 W = -3 (discard, the width cannot be negative).

Above, we see that the width, W, is equivalent to 4.5 (not -3 because it is not possible for the width to be a negative number as it is a measurement). now we can solve for the length, L, by substituting W into eqn. 1 or eqn. 2 (both will give the same answer).

L • W = 27 L • 4.5 = 27 L = 6

L = 2W - 3 L = (2 • 4.5) - 3 L = 6

Therefore, the dimensions of the rectangle are L = 6 yards, W 4.5 yards

Answer 2

Since "the length of the rectangle is 3 yards less than double the width", let "w" be the width. Then double the width is "2w" and 3 yards less than that is "2w - 3" so the length is "2w - 3".

Area = length times width so:

27 = (2w - 3)(w)

27 = 2w² - 3w

2w² - 3w - 27 = 0

(2w - 9 )(w + 3) = 0

2w - 9 = 0 or w = 9/2

w + 3 = 0 or w = -3

Since the width cannot be a negative value, we can throw out the answer w = -3

So w = 9/2 = 4.5

L = 2w - 3 = 2(9/2) - 3 = 9 - 3 = 6

So the dimensions are 4.5 yd x 6 yd