The square of the sum of two consecutive natural numbers is greater than the sum of the squares of these two numbers by 112. find the two numbers.
Accepted Solution
A:
We solve the equation, ( a + a + 1 )^2 = 112 + a^2 + ( a + 1 )^2; Then, ( 2a + 1 )^2 = 112 + a^2 + a^2 + 2a +1; 4a^2 + 4a + 1 = 113 + 2a^2 + 2a; Finally, 2a^2 + 2a - 112 = 0; a^2 + a - 56 = 0; We use Quadratic Formula for this Quadratic Equation; The solutions are a1 = 7 and a2 = -8; But a is a natural number; so, a = 7; The natural consecutive numbers are 7 and 8.