Question

the sum of the squares of two consecutives natural number is 41.find the numbers

Answer 1

Let the first natural number be \(x\) , so, the next (consecutive) natural number is \(x+1\) .

So , according to the question,

\(\sf{x^2 + (x+1)^2 = 41 }\)

Answer 2

Expand \(\sf{(x+1)^2}\) using the algebraic identity : \(\sf{(a+b)^2 = a^2 + 2ab + b^2}\) .