Question

the area of a floor is given as y2+ 4y-5 square feet. if the width of room is (y-1), what is the length?

Answer 1

We are given that the area of a rectangle is given by the following function:

`A=y^2+4y-5`

The area of a rectangle is the product of its width by its length:

`A=wl`

We are given that the width is:

`w=y-1`

Replacing in the formula for the area we get:

`y^2+4y-5=(y-1)l`

Since the area is a quadratic equation this means that for the product of the given width by the length to yield a quadratic equation the length must be of the form:

`l=y-b`

Replacing in the formula for the area:

`y^2+4y-5=(y-1)(y-b)`

Now we need to determine the value of "b" to do that we will first solve the product on the right side.

`y^2+4y-5=y^2-by-y+b`

Now we subtract "y squared" from both sides:

`4y-5=-by-y+b`

Now we associate the terms that are multiplied by "y" on the right side:

`4y-5=(-by-y)+b`

Now we take common factor on the associated terms;

`4y-5=y(-b-1)+b`

Now each coefficient for the expression on the left side and the right side must be the same, therefore we have:

`-b-1=4\text{ and -5=b}`

We get that b = -5. Therefore, the length of the floor must be equal to:

`\begin{gathered} l=y-(-5) \\ l=y+5 \end{gathered}`