Question

The sum of squares of two consecutive odd numbers is 202. Find the numbers.

Answer 1

Hint:-Let us assume that one of the two numbers is ‘x’. As one of the two numbers is ‘x’, then the second number will be ‘x + 2’.

Complete step-by-step answer:
We shall add the squares of these two numbers and equate it to 202 as the sum of the squares of the two consecutive odd numbers is 202.
We must know the below given factorization identity as we will be using this identity in out solution. We use this identity for solving the square of ‘x + 2’. We will consider ‘x’ as ‘a’ and ‘2’ as ‘b’ in this.
\[{{\left( a\ \ +\ \ b \right)}^{2}}\ \ =\ \ {{a}^{2}}\ \ +\ \ 2ab\ \ +\ \ {{b}^{2}}\]

Let us now solve this question.
We shall firstly form an equation.
\[\begin{align}
  & \Rightarrow {{\left( x \right)}^{2}}\ \ +\ \ {{\left( x\ \ +\ \ 2 \right)}^{2}}\ \ =\ \ 202 \\
 & \Rightarrow {{x}^{2}}\ \ +\ \ \left( {{x}^{2}}\ \ +\ \ 2\ \times \ x\ \times \ 2\ \ +\ \ {{2}^{2}} \right)\ \ =\ \ 202 \\
 & \Rightarrow {{x}^{2}}\ \ +\ \ {{x}^{2}}\ \ +\ \ 4x\ \ +\ \ 4\ \ =\ \ 202 \\
 & \Rightarrow 2{{x}^{2}}\ \ +\ \ 4x\ \ +\ \ 4\ \ =\ \ 202 \\
 & \Rightarrow 2{{x}^{2}}\ \ +\ \ 4x\ \ =\ \ 202\ \ -\ \ 4 \\
 & \Rightarrow 2{{x}^{2}}\ \ +\ \ 4x\ \ =\ \ 198 \\
 & \Rightarrow 2{{x}^{2}}\ \ +\ \ 4x\ \ -\ \ 198\ \ =\ \ 0 \\
 & \Rightarrow 2\left( {{x}^{2}}\ \ +\ \ 2x\ \ -\ \ 99 \right)\ \ =\ \ 0 \\
 & \Rightarrow 2\left( {{x}^{2}}\ \ +\ \ 11x\ \ -\ \ 9x\ \ -\ \ 99 \right)\ \ =\ \ 0 \\
\end{align}\]

\[\begin{align}
  & \Rightarrow 2\ \left[ \ x\ \left( x\ +\ 11 \right)\ -\ 9\ \left( x\ +\ 11 \right)\ \right]\ \ =\ \ 0 \\
 & \Rightarrow \left( x\ \ +\ \ 11 \right)\ \left( x\ \ -\ \ 9 \right)\ \ =\ \ 0 \\
 & x\ \ =\ \ -11\ \ or\ \ 9 \\
\end{align}\]
 Let us now verify our answer.
For verification, we will put the value of ‘x’ in the equation \[{{\left( x \right)}^{2}}\ \ +\ \ {{\left( x\ \ +\ \ 2 \right)}^{2}}\ \ =\ \ 202\] to check whether the answer that we have obtained after these calculations is correct or not.
Firstly, we shall take ‘x’ as -11.
Case I: \[x\ \ =\ \ -11\]
\[\begin{align}
  & \Rightarrow {{\left( x \right)}^{2}}\ \ +\ \ {{\left( x\ \ +\ \ 2 \right)}^{2}}\ \ =\ \ 202 \\
 & \Rightarrow {{\left( -11 \right)}^{2}}\ \ +\ \ {{\left( -11\ \ +\ \ 2 \right)}^{2}}\ \ =\ \ 202 \\
 & \Rightarrow 121\ \ +\ \ \left( -{{11}^{2}}\ \ +\ \ 2\ \times \ -11\ \times \ 2\ \ +\ \ {{2}^{2}} \right)\ \ =\ \ 202 \\
 & \Rightarrow 121\ \ +\ \ \left( 121\ \ -44\ \ +\ \ 4 \right)\ \ =\ \ 202 \\
 & \Rightarrow 121\ \ +\ \ 81\ \ =\ \ 202 \\
 & \Rightarrow 202\ \ =\ \ 202 \\
\end{align}\]
L.H.S = R.H.S
Hence, verified
This means that -11 is one of the correct values of ‘x’.
Now, we shall take ‘x’ as 9.
Case II: \[x\ \ =\ \ 9\]
\[\begin{align}
  & \Rightarrow {{\left( x \right)}^{2}}\ \ +\ \ {{\left( x\ \ +\ \ 2 \right)}^{2}}\ \ =\ \ 202 \\
 & \Rightarrow {{\left( 9 \right)}^{2}}\ \ +\ \ {{\left( 9\ \ +\ \ 2 \right)}^{2}}\ \ =\ \ 202 \\
 & \Rightarrow 81\ \ +\ \ \left( {{9}^{2}}\ \ +\ \ 2\ \times \ 9\ \times \ 2\ \ +\ \ {{2}^{2}} \right)\ \ =\ \ 202 \\
 & \Rightarrow 81\ \ +\ \ \left( 81\ \ +\ \ 36\ \ +\ \ 4 \right)\ \ =\ \ 202 \\
 & \Rightarrow 81\ \ +\ \ 121\ \ =\ \ 202 \\
 & \Rightarrow 202\ \ =\ \ 202 \\
\end{align}\]
L.H.S. = R.H.S.
Hence, verified

Note:-Let us now learn about the other factorization identities.
Here are some of them:-
1.\[{{\left( a\text{ }+\text{ }b \right)}^{2}}\ \ =\ \ {{a}^{2}}~\ +\ \text{ }2ab\text{ }\ +~{{b}^{2}}~\]
2.\[{{\left( a\text{ }-\text{ }b \right)}^{2}}\ \ =\ \ {{a}^{2}}~\ -\ \text{ }\ 2ab\text{ }+\ ~{{b}^{2}}\]
3.\[\left( a\text{ }+\ \text{ }b \right)\text{ }\left( a\text{ }\text{ }\ b \right)\ \ =\ \ {{a}^{2}}\ - \ \ ~\ \ {{b}^{2}}~\ \]
4.\[\left( x\text{ }+\text{ }a \right)\text{ }\left( x\text{ }+\text{ }b \right)\ \ =\ \ {{x}^{2}}~\ +\text{ }\ \left( a\text{ }+\text{ }b \right)\text{ }x\text{ }+\text{ }\ ab\text{ }\]
So these are some of the factorization identities that are also used for the solution of such questions in which factorization is required.