the sum of the squares of four consecutive positive integers is 734. what is the smallest of the integers?
To find the smallest of the four consecutive positive integers, we can set up an equation using their values.
Let's assume the first integer is n.
So, the four consecutive positive integers will be n, n+1, n+2, and n+3.
The sum of the squares of these four numbers can be represented as:
n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 = 734
Expanding the squares, we get:
n^2 + (n^2 + 2n + 1) + (n^2 + 4n + 4) + (n^2 + 6n + 9) = 734
Combine the like terms:
4n^2 + 12n + 14 = 734
Subtracting 734 from both sides:
4n^2 + 12n - 720 = 0
Dividing the equation by 4:
n^2 + 3n - 180 = 0
To factorize this quadratic equation, we need to find two numbers that multiply to give -180 and add up to 3.
After some calculations, we can determine that the factors are 15 and -12.
So, the equation can be factored as:
(n + 15)(n - 12) = 0
Setting each factor equal to zero, we have:
n + 15 = 0 --> n = -15 (not a valid solution for consecutive positive integers)
n - 12 = 0 --> n = 12
Therefore, the smallest of the four consecutive positive integers is 12.