Question

How do you simplify $\sqrt{\dfrac{8}{9}}$.

Answer 1

Hint: We will first recall the concept of multiplication and division of the radical and also the square root of the fraction to solve the above question. When we square a fraction then the square root is distributed over the numerator and denominator. For example: Let us say that $\dfrac{a}{b}$ is a fraction whose square root is given as $\sqrt{\dfrac{a}{b}}$ and we can also write it as $\dfrac{\sqrt{a}}{\sqrt{b}}$by distributing the square root over numerator and denominator and we know that 4 and 9 perfect number so $\sqrt{4}=2\text{and }\sqrt{9}=3$

Complete step by step answer:
We will use the concept of square root of the fraction and multiplication and division by radical to solve the above question.
Since, we know that when we square root a fraction then the square root is distributed over numerator and denominator.
So, after distributing the square root over numerator and denominator of $\sqrt{\dfrac{8}{9}}$, we will get:
$\Rightarrow \sqrt{\dfrac{8}{9}}=\dfrac{\sqrt{8}}{\sqrt{9}}$
Now, we know that 9 is the perfect square number and is the square of 3, so we can write $\sqrt{9}=3$ .
$\Rightarrow \dfrac{\sqrt{8}}{\sqrt{9}}=\dfrac{\sqrt{8}}{3}$
Now, we will write 8 as $4\times 2$ and split the square root over multiplication. Then, we will get:
$\Rightarrow \dfrac{\sqrt{8}}{3}=\dfrac{\sqrt{4\times 2}}{3}$
$\Rightarrow \dfrac{\sqrt{4}\times \sqrt{2}}{3}$
Now, we know that 4 is the perfect square number and is the square of 2, so we can write $\sqrt{4}=2$ .
$\Rightarrow \dfrac{2\sqrt{2}}{3}$
Hence, the simplified form of $\sqrt{\dfrac{8}{9}}$ is equal to $\dfrac{2\sqrt{2}}{3}$.
This is our required solution.

Note: Students are required to note that we can split square root only over multiplication and division and we can not split it over addition and subtraction. For example: Let us say that a and b are two number and we perform the operation of addition, subtraction, multiplication and division and then we square root them, then $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$, $\sqrt{a-b}=\sqrt{a}-\sqrt{b}$ is not true and $\sqrt{a\times b}=\sqrt{a}\times \sqrt{b}$, $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ is true.