SOLUTION: The smallest of three consecutive positive numbers is m. Three times the square of the largest is greater than the sum of the squares of the two other numbers by 67. Find m.
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Question 977116: The smallest of three consecutive positive numbers is m. Three times the square of the largest is greater than the sum of the squares of the two other numbers by 67. Find m. Answer by manishkrs9.4@gmail.com(1) (Show Source):
You can put this solution on YOUR website! smallest = m
next number would be = m+1 (because consecutive positive numbers)
largest number = m+2
3*(m+2)^2 = m^2 + (m+1)^2 + 67 (given in ques)
3*(m^2+4+4m) = m^2+ m^2+1+2m +67
3m^2 + 12m + 12 = 2m^2 + 2m + 68
m^2 + 10m - 56 = 0
Simplify m^2+10m -56 = 0
m^2 - 4m + 14m - 56 = 0
m(m-4) +14(m-4) = 0
(m-4)(m+14)=0
so the factors are 4 and -14
m cannot be -14 as it is positive integer
so m = 4
and numbers are 4,5,6
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