# math

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

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asked by eddy
1. 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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posted by Reiny

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