Question

How would you simplify $\sqrt {\dfrac{{11}}{3}}$?

Answer 1

We will first of all rationalize the denominator by multiplying and dividing the whole fraction by the denominator of the given fraction and then do the rest as follows only to get the answer.

Complete step-by-step answer:
We are given that we need to simplify $\sqrt {\dfrac{{11}}{3}}$.
We know that, we have a formula which states that: $\sqrt {\dfrac{a}{b}} = \dfrac{{\sqrt a }}{{\sqrt b }}$.
Take a = 11 and b = 3 in the above formula to get the following expression:-
$\Rightarrow \sqrt {\dfrac{{11}}{3}} = \dfrac{{\sqrt {11} }}{{\sqrt 3 }}$
Now, we see that the given fraction has the numerator = $\sqrt {11}$ and the denominator = $\sqrt 3$.
$\Rightarrow \sqrt {\dfrac{{11}}{3}} = \dfrac{{\sqrt {11} }}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }}$
On multiply the mentioned calculations on the right hand side to obtain the following expression:-
$\Rightarrow \sqrt {\dfrac{{11}}{3}} = \dfrac{{\sqrt {11} \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }}$
Clubbing the denominator on the right hand side, we will then obtain the following expression:-
$\Rightarrow \sqrt {\dfrac{{11}}{3}} = \dfrac{{\sqrt {11} \times \sqrt 3 }}{{{{\left( {\sqrt 3 } \right)}^2}}}$
Now, we will just cut off the square root by the square in the denominator of the right hand side, we will then obtain the following expression:-
$\Rightarrow \sqrt {\dfrac{{11}}{3}} = \dfrac{{\sqrt {11} \times \sqrt 3 }}{3}$ ……………(1)
Since we know that $\sqrt {11} \times \sqrt 3 = \sqrt {33}$
Putting this above expression equation number (1) to obtain the following:-
$\Rightarrow \sqrt {\dfrac{{11}}{3}} = \dfrac{{\sqrt {33} }}{3}$

Hence, the simplification of $\sqrt {\dfrac{{11}}{3}}$ is $\dfrac{{\sqrt {33} }}{3}$.

Note:
The students must note that after equation (1), we actually used a formula to multiply two different numbers with square roots which is given by: $\sqrt a \times \sqrt b = \sqrt {ab}$. This is true for any a and b belonging to real numbers such that either $a \geqslant 0$ and $b \geqslant 0$ or a < 0 and b < 0. Here, we have the situation that either both a and b are greater than or equal to 0 or both a and b less than 0 because product of two positives is positive and product of negative is also positive. This is because we cannot take negatives inside a square root because here we are dealing with positives only and for that we require the product of a and b to be positive, no matter what.