Question

Question #0089c

Answer 1

width: `6` inches

Explanation:

Let `L` be the length and `W` be the width (both in inches).

We are told that the length is `8` inches longer than `2` times the width:
`color(white)("XXX")L=2W+8`

We are also told that the area is `120 "in"^2`
and since
`color(white)("XXX")"Area" = L xx W`

we have
`color(white)("XXX")(2W+8) xx W = 120`

`color(white)("XXX")2W^2+8W-120=0`

`color(white)("XXX")W^2+4W-60=0`

`color(white)("XXX")(W+10)*(W-6)=0`

`color(white)("XXX")W=-10" (not possible)"color(white)("XXX")or W=6`

(We weren't asked for this, but...)
If `W=6` and `L=2W+8`
then
`color(white)("XXX")L=20`

Answer 2

Width = `6` inches.

Explanation:

Although there are two dimensions involved, we can use one variable to define the length and the width.

Choose the smaller value to be `x`

Let the width be `x`.

The length is `2x+8`
(`8` inches longer than twice the width)

The area is `120` Write an equation: `A = lxx w`

`x(2x+8) = 120`

`2x^2 +8x -120 =0" "larr div 2`

`x^2 +4 -60 =0" "larr` find factors

`(x+10)(x-6)=0`

`x = -10 or x = 6" "larr` reject `-10` as a length.

Width = `6` inches.

Check: `w = 6" "` then `l = 20`

`A = 6 xx 20 = 120 " in"^2`