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?

47
23
62
21
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?

 	47
	23
	62
	21
	79

callisto · Answer

Let x be the middle term of the 5 terms, the other terms are x-4, x-2, x+2 and x+4
the smallest term is x-4
From ''square root of the middle term is 16 less than the smallest term.''
we can get 
$$\sqrt{x} = (x-4)-16$$
Square both sides and solve for x
$$\sqrt{x}^2 =( x-20)^2$$$$x =( x^2-40x +400)$$$$0=( x^2-41x +400)$$x =25 or x=16 (rejected)
The second smallest term = middle term - 2 = 25-2 
Can you work it out?

anonymous · Answer

Yes its 23, thank you!

callisto · Answer

Welcome, do you know how to do this type of question next time?