Question

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

Answer 1

Hint: We solve this problem by representing each number and term in the given function as a square of some other function. Then we apply the square root to find the required answer. We need to keep in mind that $\sqrt{{{\left( a \right)}^{2}}}=\pm a$ to find the required answer correctly.

Complete step-by-step solution:
We are asked to find the square root of $4{{x}^{2}}$
Let us assume that the given function as,
$\Rightarrow p=4{{x}^{2}}$
Here, we can see that there is only one term and in that one term there are two numbers 4 and ${{x}^{2}}$
We know that ${{x}^{2}}$ can be represented as ${{\left( x \right)}^{2}}$
We also know that the number 4 can be represented as square of 2 that is ${{2}^{2}}$
Now, by using these representations in the given function then we get,
$\Rightarrow p={{2}^{2}}\times {{\left( x \right)}^{2}}$
We know that the standard identity of exponents that is ${{a}^{n}}\times {{b}^{n}}={{\left( ab \right)}^{n}}$
By using this standard formula of exponents in above equation then we get,
$\Rightarrow p={{\left( 2x \right)}^{2}}$
Now, let us apply the square root in both sides then we get,
$\Rightarrow \sqrt{p}=\sqrt{{{\left( 2x \right)}^{2}}}$
Now, we also know that the standard result of square roots that is $\sqrt{{{\left( a \right)}^{2}}}=\pm a$
By using this result in above square root equation then we get,
$\Rightarrow \sqrt{p}=\pm 2x$
Therefore, we can conclude that the required square root of $4{{x}^{2}}$ is given as $2x$ that is,
$\therefore \sqrt{4{{x}^{2}}}=2x$


Note: We need to note that if we take any number outside the square root then we need to give $'\pm '$ sign before it which is represented as $\sqrt{{{\left( a \right)}^{2}}}=\pm a$
Let us check our result. Here, we got the answer as $2x,-2x$
Let us take the square of both numbers then we get,
$\begin{align}
  & \Rightarrow {{\left( 2x \right)}^{2}}=4{{x}^{2}} \\
 & \Rightarrow {{\left( -2x \right)}^{2}}=4{{x}^{2}} \\
\end{align}$
This tells us that both the numbers $2x,-2x$ can be considered as square roots of $4{{x}^{2}}$