Question

What is the square root of ${{x}^{2}}+4$?

Answer 1

Hint: We solve this problem by representing the given function as the square of some other function. If we cannot represent it in that form in any case then we can conclude that there is no square root for the given function

Complete step-by-step answer:
We are asked to find the square root of ${{x}^{2}}+4$
Let us assume that the required value as,
$\Rightarrow {{S}^{2}}={{x}^{2}}+4$
We know that the number 4 can be represented as a square of 2.
By representing the number 4 as square of 2 then we get,
$\Rightarrow {{S}^{2}}={{x}^{2}}+{{2}^{2}}$
Here, we can see that we cannot represent the above function as a square of some other function.
We know that the standard exponent formula that is,
${{a}^{n}}\times {{b}^{n}}={{\left( ab \right)}^{n}}$
Here, we can see that the formula is applied when two terms in LHS are in multiplication.
But, we can see that in the given function the terms are in addition in which the formula cannot be applied.
Also, we can see that if we either multiply or divide the given function it cannot be converted as the square of some other function since the terms are in addition.
Therefore, we can conclude that there is no square root of given function or else we can directly represent the square root of ${{x}^{2}}+4$ as $\sqrt{{{x}^{2}}+4}$ that is,
$\therefore $ Square root of ${{x}^{2}}+4$ is $\sqrt{{{x}^{2}}+4}$

Note: We can explain this problem in other method using the negation statement.
We are asked to find the square root of ${{x}^{2}}+4$
Now, let us assume that the given function ${{x}^{2}}+4$ has a square root and is equal to $x+k$ that is,
$\Rightarrow \sqrt{{{x}^{2}}+4}=x+k$
Now, by squaring on both sides then we get,
$\begin{align}
  & \Rightarrow {{x}^{2}}+4={{\left( x+k \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}+4={{x}^{2}}+2kx+{{k}^{2}} \\
 & \Rightarrow 2kx+\left( {{k}^{2}}-4 \right)=0 \\
\end{align}$
Now, let us compare the coefficients of $'x'$ in LHS with RHS then we get,
$\begin{align}
  & \Rightarrow 2k=0 \\
 & \Rightarrow k=0 \\
\end{align}$
Now, by comparing the constant from LHS to RHS then we get,
$\begin{align}
  & \Rightarrow {{k}^{2}}-4=0 \\
 & \Rightarrow k=\pm 2 \\
\end{align}$
Here, we can see that there is no common vale for $'k'$ in the both comparisons which suggests that there is no value of $'k'$
But, we assumed that there is square root for the given function ${{x}^{2}}+4$ has a square root and is equal to $x+k$
So, we can say that our assumption is wrong which suggests that there is no square root of given function or else we can directly represent the square root of ${{x}^{2}}+4$ as $\sqrt{{{x}^{2}}+4}$ that is,
$\therefore $ Square root of ${{x}^{2}}+4$ is $\sqrt{{{x}^{2}}+4}$