there are three consecutive positive integers such that the sum of the squares of the smallest two is 221.
write and equation to find the three consecutive positive integers let x= the smallest integer
let the 3 consecutive positive integers be
x , x+1, and x+2
x^2 + (x+1)^2 = 221
x^2 + x^2 + 2x + 1 = 221
2x^2 + 2x - 220 = 0
x^2 + x - 110 = 0
(x+11)(x-10) = 0
x = -11 or x = 10
but x must be positive
so the 3 numbers are 10,11, and 12
check: is 10^2 + 11^2 = 221 ?
YES
Let's first use the given information to form an equation.
We are told that "the sum of the squares of the smallest two is 221." Let's break down this statement.
Let x be the smallest of the three consecutive positive integers.
The next two consecutive positive integers would be (x + 1) and (x + 2).
The sum of the squares of the smallest two would be x^2 + (x + 1)^2.
According to the given information, this sum is 221.
So our equation becomes:
x^2 + (x + 1)^2 = 221
By simplifying the equation, we can then solve for x.