Question

The length of a rectangle is 3ft more than twice its width, and the area of the rectangle is 77ft^2, how do you find the dimensions of the rectangle?

Answer 1

Width = `11/2" ft = 5 foot 6 inches"`

Length = `14" feet"`

Explanation:

Breaking the question down into its component parts:

Let length be `L`
Let width be `w`
Let area be `A`

Length is 3 ft more than: `L=" "?+3`
twice`" "L=2?+3`
its width`" "L=2w+3`

Area `=A = 77="width "xx" Length"`

`A=77= wxx(2w+3)`

`2w^2+3w=77`

`2w^2+3w-77=0` This is a quadratic equation
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Standard form `y=ax^2+bx+c`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`a= 2"; "b=3"; "c=-77`

`x=(-(3)+-sqrt((-3)^2-4(2)(-77)))/(2(2))`

`x=(-(3)+-25)/4=-7 or 11/2`

As we can not have a negative area in this context the answer for `x` is `11/2`

But `color(blue)(x=w" so the width is "11/2)`

`color(blue)(L=2w+3 = 11+3=14)`