Five consecutive odd integers are such that the square root of the middle term is 16 less than the smallest term. What is the second smallest of these five integers?

Select answer A 47
Select answer B 23
Select answer C 62
Select answer D 21
Select answer E 79

Question

Five consecutive odd integers are such that the square root of the middle term is 16 less than the smallest term. What is the second smallest of these five integers?

Select answer A 	47
Select answer B 	23
Select answer C 	62
Select answer D 	21
Select answer E 	79

agent0smith · Answer

Call the second smallest x (since we want to find it) then the odd integers are

x-2, x, x+2, x+4, x+6 (think of it like this... 3, 5, 7, 9 etc are all 2 apart, so keep adding 2)

square root of the middle term (x+2) is 16 less than the smallest term (x-2):

$$\Large \sqrt{x+2} = (x-2) - 16$$

Now try solving for x (you can simplify the right first)