SOLUTION: Three more than the square root of the sum of 3 consecutive odd numbers is 12. Find the three numbers.

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Question 959335: Three more than the square root of the sum of 3 consecutive odd numbers is 12. Find the three numbers.
Found 2 solutions by macston, checkley77:
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
n,n+2,and n+4 are consecutive odd integers
sqrt%28n%2B%28n%2B2%29%2B%28n%2B4%29%29%2B3=12 Subtract 3 from each side.
sqrt%283n%2B6%29=9 Square each side.
3n%2B6=81 Subtract 6 from each side.
3n=75 Divide each side by 3.
n=25 ANSWER 1: The first odd number is 25
n%2B2=25%2B2=27 ANSWER 2: The second number is 27.
N%2B4=25%2B4=29 ANSWER 3: The third number is 29.
CHECK:
sqrt%2825%2B27%2B29%29%2B3=12
sqrt%2881%29%2B3=12
9%2B3=12
12=12

Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
SQRT(X+X+2+X+4)+3=12
SQRT(3X+6)=12-3
SQRT(3X+6)=9 SQ BOTH SIDES
3X+6=81
3X=81-6
3X=75
X=75/3
X=25 ANS. FOR THE FIRST ODD NUMBER
25+2=27 FOR THE SECOND NUMBER.
25+4=29 FOR THE THIRD NUMBER.
PROOF:
SQRT(25+27+29)+3=12
SQRT81+3=12
9+3=12
12=12