Question

Find two consecutive odd positive integers, the sum of whose squares is $290$ by using the quadratic formula.

Answer 1

Find the required consecutive integers

Let the two consecutive odd positive integers be $\left(x\right)$ and $(x+2)$, respectively.

Given, the sum of the square of these two odd positive integers as $290$.

So, ${\left(x\right)}^2+{(x+2)}^2=290$

$$\begin{gathered} \Rightarrow x^2+x^2+4+4x=290 \\ \Rightarrow 2x^2+4x+4-290=0 \\ \Rightarrow 2x^2+4x-286=0 \\ \Rightarrow x^2+2x-143=0 \\ \Rightarrow x^2+13x-11x-143=0 \\ \Rightarrow x(x+13)-11(x+13)=0 \\ \Rightarrow (x-11)(x+13)=0 \\ \Rightarrow x=11,-13 \end{gathered}$$

As it is given that the numbers are positive, so $x$ cannot be $-13$.

So, $x=11$ and $(x+2)=(11+2)=13$

Hence, the two consecutive positive odd integers are $11$ and $13$.