Question

Find two consecutive odd positive integers, sum of whose square is 290.

Answer 1

Let \(x\) and \(x+2\) are two consecutive odd positive integers.
According to question,
$$
\begin{aligned}
&(x)^{2}+(x+2)^{2}=290 \\
&\Rightarrow x^{2}+x^{2}+4 x+4=290 \\
&\Rightarrow 2 x^{2}+4 x+4=290 \\
&\Rightarrow 2 x^{2}+4 x-286=0 \\
&\Rightarrow x^{2}+2 x-143=0 \\
&\Rightarrow x^{2}+13 x-11 x-143=0 \\
&\Rightarrow x(x+13)-11(x+13)=0 \\
&\Rightarrow(x+13)(x-11)=0 \\
&\Rightarrow x-11=0 \\
&\Rightarrow x=11
\end{aligned}
$$
Consecutive odd numbers \(x=11\)
$$
x+2=11+2=13
$$
Hence, two consecutive odd positive integers are 11,13