Question

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.

Answer 1

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.