Asked by eddy
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
write and equation to find the three consecutive positive integers let x= the smallest integer
Answers
Answered by
Reiny
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
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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