Question

A rectangular fence with a positive area has length represented by the expression 3x2+5x-8 and width by 2x2+6x

1. Write expressions in terms of x for the perimeter.
2. Write expressions in terms of x for the area. Give your answers in standard polynomial form and show your work.
3. If the perimeter of the rectangular fence is 16 units, prove, using the quadratic formula, that the solutions to the equation are x=1 and x=-3.2.
4. Using the first solution as x, show work and give the dimensions of the rectangular fence.
5. Using the second solution as x, show work and give the dimensions of the rectangular fence.
6. Explain which dimensions of the fence make sense in the context of the problem.
7. Using the dimensions from #6, what is the area of the rectangular fence?

Answer 1

1.
`Perimeter = 10x^2+ 22x - 16`

2.
`Area = 6x^4 +28x^3 + 14x^2 - 48x`

3.
`x = 1` or
`x = -3.2`

4.
`Length = 0` and
`Width = 8`

5.
`Length = 6.72` and
`Width = 1.28`

6. Second Solution

7.
`Area = 8.6016`

Explanation:

Given

`Length = 3x^2 + 5x - 8`

`Width = 2x^2 + 6x`

Solving (1): The perimeter

Perimeter is calculated as thus:

`Perimeter = 2 * (Length + Width)`

Substitute values for Length and Width

`Perimeter = 2 * (3x^2 + 5x - 8 + 2x^2 + 6x)`

Collect Like Terms

`Perimeter = 2 * (3x^2+ 2x^2 + 6x + 5x - 8 )`

`Perimeter = 2 * (5x^2+ 11x - 8 )`

Open Bracket

`Perimeter = 10x^2+ 22x - 16`

Solving (2): The Area:

Area is calculated as thus:

`Area = Length * Width`

Substitute values for Length and Width

`Area = (3x^2 + 5x - 8) * (2x^2 + 6x)`

Expand Bracket

`Area = 2x^2(3x^2 + 5x - 8) + 6x(3x^2 + 5x - 8)`

Open Bracket

`Area = 6x^4 + 10x^3 - 16x^2 + 18x^3 + 30x^2 - 48x`

Collect Like Terms

`Area = 6x^4 + 10x^3 + 18x^3 - 16x^2 + 30x^2 - 48x`

`Area = 6x^4 +28x^3 + 14x^2 - 48x`

Solving (3): If perimeter is 16, show that x = 1 and x = -3.2

In (a)
`Perimeter = 10x^2+ 22x - 16`

Substitute 16 for Perimeter

`16 = 10x^2+ 22x - 16`

Equate to 0

`10x^2+ 22x - 16 -16=0`

`10x^2+ 22x - 32=0`

Expand

`10x^2+ 32x -10x- 32=0`

Factorize

`x(10x + 32) - 1(10x + 32) = 0`

`(x - 1)(10x + 32) = 0`

Split

`x - 1 =0` or
`10x + 32 = 0`

`x = 1` or
`10x = -32`

`x = 1` or
`x = -32/10`

`x = 1` or
`x = -3.2`

Solving (4): Using x = 1; Solve the dimension of the rectangle

We have that:

`Length = 3x^2 + 5x - 8`

`Width = 2x^2 + 6x`

Substitute 1 for x in the given parameters

`Length = 3(1)^2 + 5(1) - 8`

`Length = 3*1 + 5*1 - 8`

`Length = 3 + 5 - 8`

`Length = 0`

`Width = 2(1)^2 + 6(1)`

`Width = 2*1 + 6*1`

`Width = 2 + 6`

`Width = 8`

Solving (5): Using x = -3.2; Solve the dimension of the rectangle

We have that:

`Length = 3x^2 + 5x - 8`

`Width = 2x^2 + 6x`

Substitute -3.2 for x in the given parameters

`Length = 3(-3.2)^2 + 5(-3.2) - 8`

`Length = 3*10.24 + 5(-3.2) - 8`

`Length = 30.72 -16 - 8`

`Length = 6.72`

`Width = 2(-3.2)^2 + 6(-3.2)`

`Width = 2*10.24 + 6(-3.2)`

`Width = 20.48 -19.2`

`Width = 1.28`

`Length = 6.72` and
`Width = 1.28`

Solving (6):

In the first solution of the dimensions in number 4, we have that

`Length = 0` and
`Width = 8`

This dimension can not be considered because the length of a rectangle can not be 0

In the second solution of the dimensions in number 5, we have that

`Length = 6.72` and
`Width = 1.28`

This dimension makes more sense because both dimensions are greater than 0

Solving (7): Using (6), determine the Area

In (6), we conclude that

`Length = 6.72` and
`Width = 1.28`

So, Area is calculated as thus:

`Area = Length * Width`

`Area = 6.72 * 1.28`

`Area = 8.6016`