Find the lower of the two successive natural numbers if the square of the sum of those numbers exceeds the sum of their squares by 112.
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Let the two natural numbers be
n and n+1
Square of the sum of those numbers = (n + n + 1)^2 = (2n + 1)^2
Sum of their squares = n^2 + (n + 1)^2
Therefore from the given data we can incur that...
(2n + 1)^2 = n^2 + (n + 1)^2 + 112
4n^2 + 4n + 1 = n^2 + n^2 + 2n + 1 + 112
Rearrange the expression into a quadratic equation
2n^2 + 2n - 112 = 0
2n^2 + 16n - 14n - 112 = 0
2n (n + 8) - 14 (n + 8) = 0
(2n - 14)(n + 8) = 0
From this we can say n is 7 or -8, but since n is a natural number, it cannot be 8.
Hence, n is 7, n + 1 =8.
The numbers are 7 and 8
n and n+1
Square of the sum of those numbers = (n + n + 1)^2 = (2n + 1)^2
Sum of their squares = n^2 + (n + 1)^2
Therefore from the given data we can incur that...
(2n + 1)^2 = n^2 + (n + 1)^2 + 112
4n^2 + 4n + 1 = n^2 + n^2 + 2n + 1 + 112
Rearrange the expression into a quadratic equation
2n^2 + 2n - 112 = 0
2n^2 + 16n - 14n - 112 = 0
2n (n + 8) - 14 (n + 8) = 0
(2n - 14)(n + 8) = 0
From this we can say n is 7 or -8, but since n is a natural number, it cannot be 8.
Hence, n is 7, n + 1 =8.
The numbers are 7 and 8
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