Question

X -5x+4 =015.x+6=0A rectangular painting is 3 feet shorter in length than it is tall (height).12. Write a polynomial to represent the area of the painting.13. Write a polynomial to represent the perimeter of the painting.14. The painting has the unique quality of having an area that has a value that is equal to the value of theperimeter. Find the height of the painting.15. What is the extraneous solution to the polynomial created when the area is set equal to the perimeter?

Answer 1

The rectangular painting is 3 feet shorter in length than in height.

Let "x" represent the painting height, then its length can be expressed as "x-3"

12.

The area of the rectangular painting can be calculated by multiplying its length by its height.

`A=l\cdot h`

For the painting

h=x

l=x-3

-Replace the formula with the expressions for both measurements:

`A=(x-3)x`

-Distribute the multiplication on the parentheses term:

`\begin{gathered} A=x\cdot x-3\cdot x \\ A=x^2-3x \end{gathered}`

The polynomial that represents the area of the painting is:

`A=x^2-3x`

13.

The perimeter of a rectangle is calculated by adding all of its sides, or two times its length and two times its height. You can calculate the perimeter of the painting as follows:

`P=2l+2h`

We know that

h=x

l=x-3

Then the perimeter can be expressed as follows:

`P=2(x-3)+2(x)`

-Distribute the multiplication on the parentheses term:

`\begin{gathered} P=2\cdot x-2\cdot3+2x \\ P=2x-6+2x \end{gathered}`

-Order the like terms together and simplify them to reach the polynomial:

`\begin{gathered} P=2x+2x-6 \\ P=4x-6 \end{gathered}`

14.

The painting has the unique quality of having an area that is equal to the value of the perimeter, then we can say that:

`A=P`

-Replace this expression with the polynomials that represent both measures:

`x^2-3x=4x-6`

To solve the expression for x, first, you have to zero the equation, which means that you have to pass all therm to the left side of the equation. Do so by applying the opposite operation to both sides of the equal sign.

`\begin{gathered} x^2-3x-4x=4x-4x-6 \\ x^2-7x=-6 \\ x^2-7x+6=-6+6 \\ x^2-7x+6=0 \end{gathered}`

We have determined the following quadratic equation:

`x^2-7x+6=0`

Using the quadratic formula we can calculate the possible values for x. The formula is:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Where

a is the coefficient that multiplies the quadratic term

b is the coefficient that multiplies the x term

c is the constant of the quadratic equation

For our equation the coefficients have the following values:

a=1

b=-7

c=6

Replace these values in the formula and simplify:

`\begin{gathered} x=\frac{-(-7)\pm\sqrt[]{(-7)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{7\pm\sqrt[]{49-24}}{2} \\ x=\frac{7\pm\sqrt[]{25}}{2} \\ x=(7\pm5)/(2) \end{gathered}`

Next is to calculate the addition and subtraction separately:

-Addition

`\begin{gathered} x=(7+5)/(2) \\ x=(12)/(2) \\ x=6 \end{gathered}`

-Subtraction

`\begin{gathered} x=(7-5)/(2) \\ x=(2)/(2) \\ x=1 \end{gathered}`

The possible values of x, i.e., the possible heights of the painting are:

x= 6ft

x=1 ft

15.

To determine the extraneous solution created when the area was set equal to the perimeter you have to calculate the corresponding length for both possible values of the height:

For the height x= 1ft, the corresponding length of the painting would be -2ft. This value, although mathematically correct, is not a possible measurement for the painting's length since these types of measures cannot be negative.