Question

Jane bought a new rectangular placemat with an area of 126 square inches. The length of the placemat is four times the quantity of nine less than half its width.

Complete the equation that models the area of the placemat, in terms of the width of the placemat, w.
= w2 - w
The width of the placemat is inche

Answer 1

`126in^2=2w^2-36w` and w= 21 inches.

Explanation:

The placemat has an area of
`126in^2`. Now, let's call w to the width, the exercise says that the length of the placemat is four times the quantity of nine less than half its width, that is

length = 4(w/2-9).

Now, the area is length*width, so

`A=w*4((w)/(2)-9)= (4w^2)/(2)-36w = 2w^2-36w.`

and we know that
`A=126in^2`, so

`126in^2=2w^2-36w.`

Now, let's find the solution.

`2w^2-36w-126=0` is a quadratic formula of the form
`ax^2+bx+c` with a=2, b= -36 and c = -126. The solutions will be

`w=(-b\pm√(b^2-4ac))/(2a)`

`w=(-(-36)\pm√(36^2-4(2)(-126)))/(2a)`

`w=(36\pm√(1296+1008))/(2*2)`

`w=(36\pm√(2304))/(4)`

`w=(36\pm48)/(4)`

`w=(36+48)/(4)` or
`w=(36-48)/(4)`

`w=(84)/(4)` or
`w=(-12)/(4)`

as we are searching a distnace, we are going to use the positive solution, that is

`w=(84)/(4)= 21.`

Then, the width is 21 inches.

Answer 2

• 126 = 2w² -36w
• 21 inches

Explanation:

The length is described as 4(-9 +w/2), which simplifies to 2w-36. The area is the product of length and width, so is ...

Area = width×length

126 = w(2w -36) = 2w² -36w

__

This equation can be divided by 2, then the square completed to give ...

w² -18w = 63

w² -18w +81 = 144 . . . . add 81 to complete the square

(w -9)² = 12² . . . . . . . . . show as squares

w -9 = 12 . . . . . . . . . . . . take the positive square root

w = 12 +9 = 21

The width of the placemat is 21 inches.