Posted by **eddy** on Tuesday, October 30, 2012 at 1:15am.

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

- math -
**Reiny**, Tuesday, October 30, 2012 at 9:34am
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

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