Question

How do you simplify $\sqrt{\left( \dfrac{-16}{25} \right)}$ ?

Answer 1

Hint: We are given $\sqrt{\left( \dfrac{-16}{25} \right)}$, we are asked to simplify it, to do so we will learn how to take term out of the square root to simplify them, then we will learn that we can distribute square root on the term that is we will use $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ then we factor 16 and 25 one by one then we will also learn what is the square root of $\sqrt{-11}$ , using all these information we form our solution

Complete step by step answer:
We are given term as $\sqrt{\left( \dfrac{-16}{25} \right)}$, we are having term as fraction inside the root, to simplify it we must know that to take terms out of square root, the terms must be in pair of 2, if there are same number 2 times then only we can take that term out.
Now as we have fraction in the square root, so we will start simplifying using a distribution of fraction.
We will use $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ .
Now as we have $\sqrt{\dfrac{-16}{25}}$ so using above identity we get –
$\Rightarrow \sqrt{\dfrac{-16}{25}}=\dfrac{\sqrt{-16}}{\sqrt{25}}$ .
Now we factor 16 and 25.
We know $16=2\times 2\times 2\times 2$ and $25=5\times 5$ .
So, we get –
$=\dfrac{\sqrt{-\left( 2\times 2\times 2\times 2 \right)}}{\sqrt{5\times 5}}$
We can see we have 2 pair of 2 and 1 pair of 5, so we take them out, we get –
$=\dfrac{2\times 2\sqrt{-1}}{5}$
As $2\times 2=4$ so, we get –
$\Rightarrow \sqrt{\dfrac{-16}{25}}=\dfrac{4}{5}\sqrt{-1}$
Now we know the complex, square root of -1 is denoted by I, called iota.
So, we get that $\sqrt{-1}=i$ .
Hence,
$\sqrt{\left( \dfrac{-16}{25} \right)}=\dfrac{4}{5}i$
Sp, we get our solution as –
$\sqrt{\left( \dfrac{-16}{25} \right)}=\dfrac{4i}{5}$

Note:
In the real line we cannot simplify square roots consisting of negative terms, the only possible way to simplify is using complex numbers.
The complex numbers are the numbers of the form $a+ib$ , where I is iota, given as $i=\sqrt{-1}$ and ‘a’ and ‘b’ are real numbers.
Also remember that, if we have a term whose pair is not available then that term will stay inside the square root.
For example: square root of 18
As $18=2\times 3\times 3$
So, there is a pair of 3, but 2 is not available in pairs. So it cannot be taken out.
So, we get –
$\sqrt{18}=\sqrt{2\times 3\times 3}$
$=3\sqrt{2}$
2 will stay inside the roof after simplifying. Hence we will get $\sqrt{18}$ as $3\sqrt{2}$ .