20 !
the length of a rectangle is 3ft less than twice the width, and the area of the rectangle is 44ft^2. what is the length and width of the rectangle?

Work Shown:

w = width

2*w = twice the width

2*w-3 = 3 less than twice the width

L = length = 2w-3

area = L*W = (2w-3)*w = 2w^2-3w

set the area equal to 44, then get everything to one side

2w^2-3w = 44

2w^2-3w-44 = 0

Now use the factor by grouping method

2w^2 - 3w - 44 = 0

2w^2 - 11w + 8w - 44 = 0

(2w^2 - 11w) + (8w - 44) = 0

w(2w - 11) + 4(2w - 11) = 0

(w + 4)(2w - 11) = 0

w+4 = 0 or 2w-11 = 0

w = -4 or w = 11/2 = 5.5

side note: you can use the quadratic formula to solve 2w^2-3w-44 = 0 for w to get the same two solutions as above

A negative width makes no sense, so we'll ignore the first solution. The only practical solution is w = 5.5

Use w = 5.5 to find the length L

L = 2w - 3

L = 2*5.5 - 3

L = 11 - 3

L = 8

So the length is 8 feet and the width is 5.5 feet

As a check: A = L*W = 8*5.5 = 44 which helps confirm the answer.

the dimensions are:

step-by-step explanation:

we want to find the dimensions of a rectangular box with volume of

we factor to obtain;

we rewrite the rightmost factor to obtain;

recall that:

we apply difference of two squares formula to obtain;

hence the dimensions are