Find three consecutive integers whose sum is 0?

Question

I am trying to help my daughter with her math and can't figure this one out, please show steps on how to do it. Thanks!

Answer 1

-1, 0, 1 are the only three consecutive integers that will work. (Recall that integers include the negative integers as well.)

-1 + 0 + 1 = 0

Answer 2

-1+0+1=0

Answer 3

-1 +1+=0 +2+-2=0 +3+-3=0

Answer 4

-1 0 and 1 = 0

Answer 5

it rather is beneficial to try this by ability of fairly sturdy judgment. 194/3 = sixty 4.666 So the middle sq. sort is probable sixty 4, sq. root of it rather is 8. So the numbers would good be 7 8 9 attempt it: 7^2 = 40 9 8^2 = sixty 4 9^2 = 80 one 40 9 sixty 4 + 80 one = 194 confident! for useful the numbers would additionally be unfavourable: -9 -8 -7 (-9)^2 = 80 one (-8)^2 = sixty 4 (-7)^2 = 40 9 80 one + sixty 4 + 40 9 = 194 besides the fact that, the question asks you to install writing an equation, so which you would be able to desire to apply a sledgehammer to crack a nut and use algebra: call the 1st integer n, so the subsequent is (n + a million) and the third is (n + 2) So n^2 + (n + a million)^2 + (n + 2)^2 = 194 Multiply out the brackets: n^2 + n^2 + 2n + a million + n^2 + 4n + 4 = 194 carry jointly words: 3n^2 + 6n + 5 = 194 subtract 5 each and each factor 3n^2 + 6n = 189 divide by ability of three each and each factor n^2 + 2n = sixty 3 finished the sq. n^2 + 2n + a million = sixty 4 (n + a million)^2 = sixty 4 n + a million = +or- 8 n = 7 or n = -9 So the three integers are 7 8 9 sanity learn: 40 9 + sixty 4 + 80 one = 194 or -9 -8 -7 sanity learn: 80 one + sixty 4 + 40 9 = 194

Answer 6

LET 1ST INTEGER TO BE X,THEN 2ND WILL BE X+1,THIRD WILL BE X+2.THEN ADD [X]+[X+1]+[X+2]=0.
WHICH GIVES 3X+3=0.
THEN 3X=-3.
X=-1.
PUTTING X=1 IN ABOVE ASSUMPTION WE GET 3 NO'S AS "-1,[-1+1=0],[-1+2=1]

Answer 7

My guess would be: -1, 0, 1

Answer 8

-1, 0, 1, it is rather simple, 0 is an interger

Answer 9

x + x+1 + x+2 = 0
3x = -3
x = -1
x + 1 = 0
x + 2 = 1.