Question

The length of a rectangle is 2 feet less than 4 times the width. The area is 256ft2. Find the dimensions of the rectangle.

Answer 1

`W = 8.255`

`L = 31.02`

Explanation:

Let L = Length and W = Width.

So:

`L = 4W - 2`

`Area = 256`

Required

Find L and W

Area is calculated as:

`Area = L * W`

Substitute 4W - 2 for L and 256 for Area

`Area = (4W - 2) * W`

`256 = (4W - 2) * W`

Open Bracket

`256 = 4W^2 - 2W`

Divide through by 2

`128 = 2W^2 - W`

Equate to 0

`2W^2 - W - 128 = 0`

An equation
`aw^2 + bw + c = 0` has the roots

`W = (-b\±√(b^2 - 4ac))/(2a)`

Where

`a = 2`
`b = -1`
`c = -128`

So:

`W = (-(-1)\±√((-1)^2 - 4*2*-128))/(2*2)`

`W = (1\±√(1 +1024))/(4)`

`W = (1\±√(1025))/(4)`

`W = (1\± 32.02)/(4)`

`W = (1+ 32.02)/(4)` or
`W = (1 - 32.02)/(4)`

`W = (33.02)/(4)` or
`W = (-31.02)/(4)`

`W = 8.255` or
`W = -7.755`

But the dimension can not be negative.

So:

`W = 8.255`

Recall:

`L = 4W - 2`

`L = 4 * 8.255 - 2`

`L = 31.02`