Answer with explanation:
We have to find ,those equation which is used to solve for, y ,the length of room.
Measurement of length is not given,but surely it will be a positive Integer , not equal to 0.
We will check from option 1,which of these equation gives positive Integral value.
1. y×(y+5)=750
→y² + 5 y - 750=0
→y² + 30 y - 25 y - 750=0
→y× (y +30) -25 × (y +30)=0
→ (y - 25)(y + 30)=0
→ y -25 =0 ∧ y + 30=0
y= 25 ∧ y = -30
It gives a Positive value.So, this equation can be used to solve for y, the length of the room.
2. y² - 5 y =750
→y² - 5 y - 750=0
→y² - 30 y + 25 y - 750=0
→y× (y -30) +25 × (y -30)=0
→ (y + 25)(y - 30)=0
→ y +25 =0 ∧ y - 30=0
y= -25 ∧ y = 30
It gives a Positive value.So, this equation can be used to solve for y, the length of the room.
3. 750 - y×(y-5)=0
⇒750 - y² + 5 y=0
⇒-1× (y² - 5 y - 750)=0
→y² - 5 y - 750=0
→y² - 30 y + 25 y - 750=0
→y× (y -30) +25 × (y -30)=0
→ (y + 25)(y - 30)=0
→ y +25 =0 ∧ y - 30=0
y= -25 ∧ y = 30
It gives a Positive value.So, this equation can be used to solve for y, the length of the room.
4. y × (y -5) + 750 =0
y² - 5 y + 750=0
you will not get real root,so this equation can't be used to solve for y, the length of the room.
5. (y + 25)(y-30)=0
→y+ 25 =0 ∧→ y -30=0
y= -25 ∧→ y =30
It gives a Positive value.So, this equation can be used to solve for y, the length of the room.
→→Equation 1,2,3,and 5 that is apart from equation 4, all of equation can be used to solve for y, the length of the room.
