Question

How do you simplify \[\sqrt{\dfrac{16}{3}}\] ?

Answer 1

Hint: For the given solution we are given that to solve the problem \[\sqrt{\dfrac{16}{3}}\] . We can see that a given rational number is not a perfect square, so we have to apply the square root for both numerator and denominator. Now rationalise the fraction to get the solution.

Complete step by step solution:
Let us consider and assume the equation as equation (1) and S.
\[S=\sqrt{\dfrac{16}{3}}......................\left( 1 \right)\]
First of all we have to rewrite \[\sqrt{\dfrac{16}{3}}\]as \[\dfrac{\sqrt{16}}{\sqrt{3}}\]
\[\Rightarrow S=\dfrac{\sqrt{16}}{\sqrt{3}}\]
Let us consider the above equation as equation (2), we get
\[\Rightarrow S=\dfrac{\sqrt{16}}{\sqrt{3}}......................\left( 2 \right)\]
Now we have to solve the above equation for the further step
In the equation (2)’s numerator we can see the number 16 which is a perfect square. So, now we rewrite the equation by rooting the number 16.
\[\Rightarrow S=\dfrac{4}{\sqrt{3}}\]
Let us consider the above equation as equation (3), we get
\[\Rightarrow S=\dfrac{4}{\sqrt{3}}........\left( 3 \right)\]
Now we have to multiply \[\dfrac{4}{\sqrt{3}}\]by \[\dfrac{\sqrt{3}}{\sqrt{3}}\] for rationalising the equation (3), we get
\[\Rightarrow S=\dfrac{4\sqrt{3}}{\sqrt{3}\sqrt{3}}\]
And now we have to use powers
\[\Rightarrow \dfrac{4\sqrt{3}}{{{\sqrt{3}}^{1}}{{\sqrt{3}}^{1}}}\]
And now we have to add both the powers
\[\Rightarrow \dfrac{4\sqrt{3}}{{{\sqrt{3}}^{2}}}\]
Now we to cancel the root on denominator
\[\Rightarrow \dfrac{4\sqrt{3}}{3}\]
Now, let us consider the equation as equation (4), we get
\[\Rightarrow S=\dfrac{4\sqrt{3}}{3}.........\left( 4 \right)\]
Therefore, the above equation is the exact solution for the given problem.

Note: This sum is based on the rooted one which is a difficult part of the whole mathematics. So all the students should have to be so careful while doing these rooted based sums. We all did the best and the simpler method to solve the given question which is much easier for the students to solve.