Question

The value of $4{{a}^{2}}+{{b}^{2}}$ can be written as which of the following given value:
\[\begin{align}
  & A.{{\left( 2a \right)}^{2}}+{{b}^{2}} \\
 & B.{{\left( 2a+b \right)}^{2}}-4ab \\
\end{align}\]

Answer 1

Hint: To solve this question, first step is to use ${{\left( ab \right)}^{2}}={{a}^{2}}{{b}^{2}}$ in ${{\left( 2a \right)}^{2}}$ then it would be ${{\left( 2a \right)}^{2}}=4{{a}^{2}}$ and for the second option, we will use the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ in ${{\left( 2a+b \right)}^{2}}$ so as to expand it and see if it is equal to $4{{a}^{2}}+{{b}^{2}}$ or not. We will assume $4{{a}^{2}}+{{b}^{2}}$ as x and then observe if option (a) and (b) are equal to x or not.

Complete step-by-step answer:
We have our expression given as \[4{{a}^{2}}+{{b}^{2}}\]
Let the value be \[x=\left( 4{{a}^{2}} \right)+{{b}^{2}}\]
We know that the value of ${{2}^{2}}=4$ then $4{{a}^{2}}$ can be written as ${{2}^{2}}{{a}^{2}}$
\[\Rightarrow 4{{a}^{2}}={{\left( 2a \right)}^{2}}\] and ${{b}^{2}}$ is ${{b}^{2}}$
\[\begin{align}
  & 4{{a}^{2}}+{{b}^{2}}={{\left( 2a \right)}^{2}}+{{b}^{2}} \\
 & x={{\left( 2a \right)}^{2}}+{{b}^{2}} \\
\end{align}\]
Which is option A so option A is correct answer.
Now, consider the second option B as
\[{{\left( 2a+b \right)}^{2}}-4ab\]
Now, we have an identity as ${{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$
Using this identity in ${{\left( 2a+b \right)}^{2}}$ using x = 2a and y = b in above we get:
\[\begin{align}
  & {{\left( 2a+b \right)}^{2}}={{\left( 2a \right)}^{2}}+2\times 2a\times b+{{b}^{2}} \\
 & \Rightarrow 4{{a}^{2}}+4ab+{{b}^{2}} \\
\end{align}\]
Then the value of \[{{\left( 2a+b \right)}^{2}}-4ab\], then substituting the value of ${{\left( 2a+b \right)}^{2}}$ as $4{{a}^{2}}+4ab+{{b}^{2}}$ in above we get:
\[\begin{align}
  & {{\left( 2a+b \right)}^{2}}-4ab=4{{a}^{2}}+4ab+{{b}^{2}}-4ab \\
 & \Rightarrow 4{{a}^{2}}+{{b}^{2}} \\
 & \Rightarrow x \\
\end{align}\]
Hence, the option B = x
Therefore, the value of \[4{{a}^{2}}+{{b}^{2}}={{\left( 2a+b \right)}^{2}}-4ab\] is also correct.
Therefore, both option A and B are correct.
So,
\[{{\left( 2a \right)}^{2}}+{{b}^{2}}=4{{a}^{2}}+{{b}^{2}}\]\[4{{a}^{2}}+{{b}^{2}}={{\left( 2a \right)}^{2}}+{{b}^{2}}\text{ and }4{{a}^{2}}+{{b}^{2}}={{\left( 2a+b \right)}^{2}}-4ab\]

Note: Another way to solve this question is by considering both options separately and making them equal to $4{{a}^{2}}+{{b}^{2}}$
Consider option A.
\[{{\left( 2a \right)}^{2}}+{{b}^{2}}=4{{a}^{2}}+{{b}^{2}}\] so this is correct.
Consider option B.
\[{{\left( 2a+b \right)}^{2}}-4ab\] using identity ${{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$
\[\begin{align}
  & \Rightarrow 4{{a}^{2}}+4ab+{{b}^{2}}-4ab \\
 & \Rightarrow 4{{a}^{2}}+{{b}^{2}} \\
\end{align}\]
Hence both are correct.