Answer:
Answer:
y^{2}-5y=750y
2
−5y=750
750-y(y-5)=0750−y(y−5)=0
(y + 25)(y -30) = 0(y+25)(y−30)=0
Explanation:
Givens
The area of a rectangular room is 750 square feet.
The width of the room is 5 feet less than the length of the room.
Let's call ww the width and ll the length. According to the problem they are related as follows
w=l-5w=l−5 , because the width is 5 feet less than the lenght.
We know that the area of the room is defined as
A=w\times lA=w×l
Where A=750 ft^{2}A=750ft
2
Replacing the given area and the expression, we have
\begin{gathered}750=(l-5)l\\750=l^{2}-5l \\l^{2}-5l-750=0\end{gathered}
750=(l−5)l
750=l
2
−5l
l
2
−5l−750=0
We need to find two number which product is 750 and which difference is 5, those numbers are 30 and 25.
l^{2}-5l-750=(l-30)(l+25)l
2
−5l−750=(l−30)(l+25)
Using the zero property, we have
\begin{gathered}l=30\\l=-25\end{gathered}
l=30
l=−25
Where only the positive number makes sense to the problem because a negative length doesn't make any sense.
Therefore, the length of the room is 30 feet.
Also, the right answers are the second choice where y=ly=l , the third choice and the last choice