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)(x10) = 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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