Question

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

Answer 1

Hint: In this problem, first we will formulate the problem and after that we will solve it using the quadratic formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to get the two consecutive odd positive integers.

Complete step-by-step answer:

We know that if $n$ is an odd positive integer, then $n+2$ is the next consecutive odd positive integer.

For example, let $n=3$ then $n+2=5$ is the next consecutive odd positive integer.


In this question let $n$ be the first odd positive integer and $n+2$ be the next consecutive positive odd integer.

Then according to the question, it is given that the sum of the squares of the two consecutive odd positive integers is 290. This can be formulated as follows:

$\Rightarrow {{n}^{2}}+{{(n+2)}^{2}}=290.......(i)$

On simplifying equation (i) we get it as,

$\begin{align}

  & \Rightarrow {{n}^{2}}+({{n}^{2}}+4n+4)=290 \\

 & \Rightarrow 2{{n}^{2}}+4n+4=290 \\

 & \Rightarrow 2{{n}^{2}}+4n+(4-290)=0 \\

 & \Rightarrow 2{{n}^{2}}+4n-286=0.......(ii) \\

\end{align}$

Equation (ii) is of the form $a{{x}^{2}}+bx+c=0$ where $a\ne 0$,$a,b,c\in R$ is called a quadratic equation with real coefficients. Here, $x$ can be solved using the quadratic formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

$\therefore $ in this problem equation (ii) can be solved using the quadratic formula to get the value of $n$.

$\Rightarrow n=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}........(iii)$

According to equation (iii) we will have two solutions for $n$.

First solution, using $n=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$,

$\begin{align}

  & \Rightarrow n=\dfrac{-4+\sqrt{{{4}^{2}}-(4\times 2\times -286)}}{2\times 2} \\

 & \Rightarrow n=\dfrac{-4+\sqrt{{{4}^{2}}-(-2288)}}{4} \\

\end{align}$

$\begin{align}

  & \Rightarrow n=\dfrac{-4+\sqrt{16+2288}}{4} \\

 & \Rightarrow n=\dfrac{-4+\sqrt{2304}}{4} \\

 & \Rightarrow n=\dfrac{-4+48}{4} \\

 & \Rightarrow n=\dfrac{44}{4} \\

 & \Rightarrow n=11........(iv) \\

\end{align}$

Second solution, using $\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$,

$\begin{align}

  & \Rightarrow n=\dfrac{-4-\sqrt{{{4}^{2}}-(4\times 2\times -286)}}{2\times 2} \\

 & \Rightarrow n=\dfrac{-4-\sqrt{{{4}^{2}}-(-2288)}}{4} \\

 & \Rightarrow n=\dfrac{-4+\sqrt{16+2288}}{4} \\

 & \Rightarrow n=\dfrac{-4-\sqrt{2304}}{4} \\

 & \Rightarrow n=\dfrac{-4-48}{4} \\

 & \Rightarrow n=\dfrac{-52}{4} \\

 & \Rightarrow n=-13........(v) \\

\end{align}$

Here, $n=11$ and $n=-13$ are the roots or solutions of the equation (ii).

From (iv) and (v) we observe that, according to (v) $n=-13$ cannot be accepted as a solution because it is not a positive odd integer.

$\therefore n=11$ is the solution.

Hence, the two consecutive odd integers are as follows:

$\begin{align}

  & \Rightarrow n=11 \\

 & \Rightarrow n+2=13 \\

\end{align}$

To check the above answer let us take the squares and add,

$\begin{align}

  & \Rightarrow {{11}^{2}}+{{13}^{2}}=121+169 \\

 & \Rightarrow 290 \\

\end{align}$

Hence, the two consecutive odd positive integers whose sum of squares is 290 are 11 and 13.

Note: In this problem the main idea is about knowing that two consecutive numbers are represented as n and n+2 and also about the usage of quadratic formulas to solve the quadratic equation.