How do you write $3\sqrt {{x^2}} $as an exponential form?

Hint: Here we will apply the concepts of squares of positive or the negative term, also the square root and along with apply the identity that square and square-root cancel each other. Follow step by step simplification.

Complete step-by-step solution:
Take the given expression: $3\sqrt {{x^2}} $. Now, we know that the square of the positive number or the negative number always gives us the resultant value in positive as product of negative with the negative is positive. In other simple words, the product of minus with minus gives plus or the positive value.
Therefore, we can write the given expression as: $3\sqrt {{{( \pm x)}^2}} $
Square and square root cancel each other in the above equation.
$ = \pm 3x$
Hence, the required solution is –
$3\sqrt {{x^2}} = \pm 3x$


Note: The squares and the square roots are opposite to each other and so cancel each other. Perfect square number is the square of an integer, simply it is the product of the same integer with itself. For example - $25{\text{ = 5 }} \times {\text{ 5, 25 = }}{{\text{5}}^2}$, generally it is denoted by n to the power two i.e. ${n^2}$. The perfect square is the number which can be expressed as the product of the two equal integers. For example: $9$, it can be expressed as the product of equal integers. $9 = 3 \times 3$ square-root is denoted by $\sqrt {{n^2}} = \sqrt {n \times n} $ For Example: $\sqrt {{2^2}} = \sqrt 4 = 2$.

Similarly cube is the product of same number three times such as ${n^3} = n \times n \times n$ for Example cube of $2$ is ${2^3} = 2 \times 2 \times 2$ simplified form of cubed number is ${2^3} = 2 \times 2 \times 2 = 8$. and cube-root is denoted by $\sqrt[3]{{{n^3}}} = \sqrt {n \times n \times n} = n$ For Example: $\sqrt[3]{8} = \sqrt[3]{{{2^3}}} = 2$ Do not get confused between the square and square-root and also, similarly cubes and cube-root, know the concepts properly and apply accordingly.