Question

Find two consecutive odd natural numbers, the sum of whose squares is $202$.

Answer 1

Let the two odd consecutive numbers ${}^{\prime }{x}^{\prime }\text{and}(x+2)$

According to given question, we have

$x^2+{(x+2)}^2=202$

$\Rightarrow x^2+x^2+4+4x=202$

$\Rightarrow 2x^2+4x=198$

$\Rightarrow x^2+2x=99$

$\Rightarrow x^2+2x-99=0$

$\Rightarrow x^2+(11-9)x-99=0$

$\Rightarrow x^2+11x-9x-99=0$

$\Rightarrow x(x+11)-9(x+11)=0$

$\Rightarrow (x+11)(x-9)=0$

$\Rightarrow (x+11)=0or,(x-9)=0$

$\therefore x=9$ [$\because x\ne -11$,as it is a not a natural number]

And, $x+2=9+2=11$

Hence, two consecutive odd natural numbers are $9$ and $11.$