We are given with information that , we have to find two consecutive positive integers , sum of whose squares is 365 .
[tex]{\pmb{\bf Assumption \: :-}}[/tex] Let's assume that the first number is x , and as the other no. is it's consecutive so , the second no. is (x+1)
Now , According to the question ;
[tex]{: \implies \quad \sf x^{2}+x^{2}+1^{2}+ 2\cdot x \cdot 1 = 365\quad \{\because (a+b)^{2}=a^{2}+b^{2}+2ab\}}[/tex]
Can be further written as ;
[tex]{: \implies \quad \sf 2{x^{2}}+2x -364=0}[/tex]
[tex]{: \implies \quad \sf 2(x^{2}+x-182)=0}[/tex]
As , [tex]\sf 2 \neq 0[/tex] . So ;
[tex]{: \implies \quad \sf x^{2}+x-182=0}[/tex]
Using splitting the middle term ;
[tex]{: \implies \quad \sf x^{2} +14x-13x-182=0}[/tex]
[tex]{: \implies \quad \sf x(x+14)-13(x+14)=0}[/tex]
[tex]{: \implies \quad \sf (x+14)(x-13)=0}[/tex]
Case I :-
When (x+14) = 0 , then x = - 14 , which is rejected as it's not +ve
Case II :-
When (x-13) = 0 , then x = 13 , which is +ve
Now ,
- First no. = x = 13
- Second no. = (x+1) = (13+1) = 14
Hence , The required numbers are 13 and 14 respectively .
